Hermitian and expectation values.... imaginary?

[learn.steadfast]

I've been studying quantum mechanics, and working problems to get a feel for expectation values and what causes them to be real.

I was working the problem of finite 1D wells, when I came across a situation I did not understand.

A stationary state solution is made up of a forward and reverse traveling wave of the same amplitude in regions where the potential is constant.

When I try to compute the expectation values in these areas, I get a different answer depending on whether or not I sum up the waves before or after I apply the momentum operator. I am not sure what I am doing wrong ... or why the method works one way, but not the other, as the books I've read aren't clear on this point:

Consider a forward traveling wave with defined momentum:
$\psi_f(x) = e^{ i k x}$

I can compute the contribution of momentum of this wave to a 1D well, using the following generic formula:
$< p > = \int_{x_1}^{x_2} \psi_f^{*} (- i \hbar ){ \partial \over { \partial x }} \psi_f$

For the forward wave, I compute a positive momentum:
$< p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x }$
$< p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x }$
$< p > = \int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar )( i k )A e^{ i k x }$
$< p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (- i \hbar )( i k )$
$< p > = A^2 \hbar k ( x_2 - x_1 )$

If I were to use a reverse traveling wave, instead $\psi_r(x) = e^{- i k x }$, the solution is the same but with reversed sign:

$\int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar ){ \partial \over { \partial x }} e^{ -i k x }$
$< p > = \int_{x_1 }^{x_2 } e^{ -i k x } (- i \hbar )( i k ) A e^{ i k x }$
$< p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (i \hbar )( i k )$
$< p > = - A^2 \hbar k ( x_2 - x_1 )$

If A is the normalisation constant for an infinite well, over all space, then $A={1 \over \sqrt{ x_2 - x_1 }}$ and $<p>=- \hbar k$
That's the obvious, and correct result.

Since these solutions are all linear and hermitian, I'm also supposed to be able to add together solutions to make another solution... but, that's where I get into trouble!

The sum of the solved expectation of momentums is clearly 0. The forward and reverse waves cancel each other's momentum out. A stationary state neither moves (on average) to the right or to the left, but simply oscillates in place. So, the sum of expectation values gives the intuitive result.

However, when I attempt to solve for momentum using the sum of the forward and reverse wave-functions $\psi = \psi_f + \psi_r$ ... I don't get a real number.

I thought that momentum being a "linear" Hermitian operator, meant it would produce the same result whether working on individual waves, and summing the results ... or a superposition of waves with a single result. Did I misunderstand something, or did I make a mistake in the following computation; because I get a purely imaginary result:

$< p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar \partial \over { \partial x } } A (e^{ i k x } + e^{ -i k x } )$

$< p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar } A ( i k e^{ i k x } - i k e^{ -i k x } )$

$< p > = A^2 \hbar k \int_{x_1}^{x_2} ( e^{ -i k x } + e^{ i k x } ) ( e^{ i k x } - e^{ -i k x } )$

$< p > = A^2 \hbar k \int_{x_1}^{x_2} (e^{ 2 i k x } - e^{ -2 i k x } )$

$< p > = A^2 \hbar k \int_{x_1}^{x_2} (i 2 sin( 2 k x) )$

Since the right hand side of the equation is is purely imaginary, then the integral must be purely imaginary ... and so must the result.

The issue seems to be that sum of products, is not the same as product of sums:
( a + 1/a ) (-1/a + a) = a*a - 1/(a*a)
(a*-1/a) + (1/a*a) = 0

But momentum operator notation sets the problem up differently depending on whether it's passed individual wave components or sums.

How am I supposed to correctly do operatior notation on a superposition of solutions to the wave equation?

 

[vanhees71]

To evaluate expectation values in the position representation, you have to use the complete wave function and integrate over the entire axis, i.e., if the particle is prepared in a pure state, which is represented by the wave function $\psi(x)$ in position representation, then the expectation value of the momentum observable is
$$\langle p \rangle = \int_{-\infty}^{\infty} \mathrm{d} x \psi^*(x) (-\mathrm{i} \hbar \partial_x) \psi(x).$$
Note that for the infinite square well there is no momentum observable, and thus it doesn't make any sense to evaluate momentum expectation values. For the finite square well there is a momentum observable, and it makes sense to ask for an expecation value of the momentum. Note, however, that energy eigenstates in the continuous part of the spectrum are generalized eigenvectors and thus do not represent states and consequently the momentum expectation value won't exist in general.

 

[learn.steadfast]

To evaluate expectation values in the position representation, you have to use the complete wave function and integrate over the entire axis,

Even though I did not integrate over the entire axis, I did integrate over the entire infiinite well region. The integral over the whole axis, is merely the sum of three integrals;

$\langle p \rangle = \int_{-\infty}^{x_1} \mathrm{d} x \psi^*(x) (-\mathrm{i} \hbar \partial_x) \psi(x) + \int_{x_1}^{x_2} \mathrm{d} x \psi^*(x) (-\mathrm{i} \hbar \partial_x) \psi(x) + \int_{x_2}^{\infty} \mathrm{d} x \psi^*(x) (-\mathrm{i} \hbar \partial_x) \psi(x)$

There are two "wall" integrals, and one free particle integral.
The $\psi$ derivative in either wall will be zero, rendering the integral of both walls identiacally zero. Therefore, the integral I performed is still the same as integrating over the whole axis. The only possible error would be if a discontinuity existed such that I was integrating a Dirac delta, but no such discontinuity exists in an infinite well's solution.

Note that for the infinite square well there is no momentum observable, and thus it doesn't make any sense to evaluate momentum expectation values. For the finite square well there is a momentum observable, and it makes sense to ask for an expecation value of the momentum.

Why would that be the case? Why is the infinite square well different from a finite square well when observing the particle or making measurements on it? ( note: the original 1D problem I was working on was the harmonic oscillator, I reduced the problem to 1D finite *asymmetrical* wells, and then the infinite well ... trying to figure out what was wrong. I wanted to reduce the problem to the simplest one where it existed, to make understanding the problem easier. )

I think Zero is what the expectation value of momentum ought to be, even if it's meaningless...?

My intuitive thought is this: SInce the particle of interest is (by definition) completely trapped, and the trapping potentials are time-invariant, the particle must move to the right and left with exactly the same momentum (on average) in the rest of the well; The particle must traverse every well bottom point (whether from the right or the left) with equal probability because the particle must go to the right in order to come back left, or vice versa.

Therefore, the total momentum expectation value ought to cancel and be zero in any 1D finite or infinite well where the particle is trapped and the potentials are static.

What I think is happening in the integral over the infinite well, is that imaginary results of the momentum integrals *appear* to measure how much of the wave is "standing" still, and oscillating in place.

eg: The real potion of the momentum measurement in an infinite well, is "zero"; but the imaginary portion is non-zero.

Note, however, that energy eigenstates in the continuous part of the spectrum are generalized eigenvectors and thus do not represent states and consequently the momentum expectation value won't exist in general.

In order for a particle to be expected to "move", it has to be a combination of two or more eigenstates. For example, I could chose a forward wave with n=2 and a reverse wave with n=1 (standard solutions in a finite well), add the two waves together equally ... and the result will be a wave packet with net momentum identical to the n=1 state, in the positive direction. If we graph the time varying soltuions, (complete), we will "see" a modulation of the waves travel to the right with a velocity proportional to the momentum. Thus we will have created a minimialistic, discrete, wave packet.

I get that part of what you're saying.

I've been reviewing QM, online. But, there are many realistic situations which mimmic 1D potential wells, or the harmonic osciallator. For example, the 1S orbital of a hydrogen atom is a stationary state. We can not determine the "motion" of a 1S orbital, because there is none (on average). The spectrum of orbitals is discrete, just as the specturm of a 1D well is discrete.

1D discrete problems are the only kinds of problems which interest me at the moment, so I'm not really worried about "continuous parts" of the spectrum, yet.

Keeping that point in mind; there are many physics courses and online homework assignments where the teacher asks students to calculate the momentum for a position representation of solutions. So, the general physics community seems to think that such calculations are at least possible; eg: teachers and books are not carefully delineating when such solutions are "meaningless".

I'll wait for your response before constructing a finite-well, and re-examining the problem.

But, consider ... the momentum operator, as you show it, is going to produce an imaginary result in any region of the axis where $\psi$ is purely real. eg: where $\psi^* = \psi$. That means imaginary results are computed in *parts* of the axis where the particle is tunneling into a finite "wall". Therefore, In order for the total expectation value of the whole axis to be "real", the imaginary value generated in the tunneling parts of a finite well must cancel with another region's opposite imaginary value. eg: the imaginary tunneling momentum generated on the "right" side of the well might cancel with the tunneling value on the "left" side of the well.

BUT: If the Hermitian properties of the momentum operator guarantee that the result is purely "real"; a moments thought will show that the summation of opposed right and left wall's imaginary values is not sufficient to guarantee a real value. eg: ; because wells can be made a-symmetrical in potential. Therefore, if Hermiticity truly guarantees a "real" result, then the reason must be that the imaginary value generated in the tunneling region of a wall *MUST* cancel with the imaginary value generated in the oscillatory region immediately adjacent to the wall.

I do not know why that would be the case, but have I deduced the requirement correctly?

 

[PeroK]

@learn.steadfast There is an imaginary number which is also real. Namely 0.

If you have a real wave-function, then in fact the expected value of momentum must be zero.

 

[learn.steadfast]

@learn.steadfast There is an imaginary number which is also real. Namely 0.

If you have a real wave-function, then in fact the expected value of momentum must be zero.

Yes. That's exactly what I think. A purely real wave function is a standing wave, and a pure standing wave is not moving and has zero net momentum.

In the problem I set up, the infinite wall size is the limiting case of a very deep but finite well with a purely real wave function.
However, I don't see how to prove/guarantee I get zero when I integrate the momentum using Hermitian operator notation as taught in my QM textbook.

When $\psi = \psi^*$, ( A real ) and psi is both continuous and finite ... a partial derivative operator on the wavefunction always produces a real value. Therefore, the operator notation sets up an integral with the following form:
$\int \real(x) \cdot (- i \hbar k ) \cdot \real(x)$

Therefore, the result is purely imaginary at every point, x. The only way it can integrate to zero, is if there is as much negative area as positive area under the integral.

With very large well wall potentials, the magnitude of the imaginary values computed in the tunnneling regions become vanishingly small; and I think it can be ignored. However, the imaginary values computed in the standing wave region of the box are finite and can't be ignored.

$<p>=A^2ℏk \int_{x1}^{x2} i 2sin(2kx)$

The issue that I see, is: I can legitimately place the walls, x1 and x2, at any boundary where the $\psi(x)$ standing wave is zero. The continuity / boundary conditions will be met, and therefore Schrodinger's can be solved. But, that putting walls at zero crossing condition allows us to set walls in places that trap areas of two positive sinusoidal peaks and only one negative sinusoidal peak (or vice versa.). In such cases, the integral for area under the sine curve has a nonzero imaginary results.
Did I make a mistake in my original post's calculus, when I derived the above area equation?

 

[PeterDonis]

Since the right hand side of the equation is is purely imaginary, then the integral must be purely imaginary

But, as @PeroK has pointed out, zero counts as "purely imaginary", and the result of your integral is zero. Why? Because the bounds of the integral must be points where $\psi = 0$, i.e., the range of integration must be an integral number of half cycles; but with the double angle in the integrand, that means you are integrating the sine function over an integral number of full cycles, which is identically zero.

 

[PeterDonis]

putting walls at zero crossing condition allows us to set walls in places that trap areas of two positive sinusoidal peaks and only one negative sinusoidal peak

That's true for the function $\sin kx$. But your integral is over the function $\sin 2 kx$. Doubling the frequency adds a peak of opposite sign for each peak, so you automatically have an equal number of positive and negative peaks.

 

[Ibix]

But, as @Ibix has pointed out

@PeroK, in fact.

 

[PeterDonis]

@PeroK , in fact.

Oops, sorry. Fixed.

 

[learn.steadfast]

That's true for the function $\sin kx$. But your integral is over the function $\sin 2 kx$. Doubling the frequency adds a peak of opposite sign for each peak, so you automatically have an equal number of positive and negative peaks.

OH !

I see. Thank you, I didn't notice that. That solves half the problem, right there!
Even the infinite well produces zero momentum in that case. :)

So, that only leaves checking the case of a-symmetrical wells where the imaginary value of the integral inside each wall is different ...

In order for that case to integrate up to exactly zero, each tunneling portion must cancel with something in the oscillatory wave.

Note: when a wave tunnels into a wall, it hits the wall at some phase just shy of the zero crossing point of a pure sine-wave with no wall present; and that missing part of the sine-wave unbalances the paired cancellation of areas you spoke of. Therefore, the imaginary value due to integrating exponential decay regions must exactly replace the area missing from the sine wave because the wall exists there.

To test this idea, I'll set up a crude problem with a single wall:
Given that a free particle wave function would normally cross zero at x=z, where z is an arbitrary point on the axis; but instead, we have a finite height wall at x=0... therefore the particle tunnels into the wall at x=0, from a phase of -k z; Therefore, we can compute a (normalization independent) solution and see if the imaginary area lost from the sinewave is replaced by the area created under the exponential decay inside the wall.

This is what I imagine the wave function must look like:

$\psi (x) = sin( k (x-z) )$ :: for x<0 the particle is free and oscillatory

$\psi (x) = C e^{ -D x }$ :: for x >= 0 the particle is tunneling into the wall

Constants, C and D, are found by matching boundary conditions. They define the exponential decay (incidentally, they also tell us how high the wall's potential is..., but we don't really care.)

$C e^{ 0 } = sin( k(-z) )$ :: Because the height of a wave function must be continuous.
and
$C(-D) e^{ -D( 0) } = k \cdot cos ( k(-z) )$ :: Because the slope of a wave function must be continuous.

Therefore, $C=sin(-k z)$ and $D=-k \cdot cot( -k z )$

The area missing from the sinewave is $\int_{-z}^0 sin( kx )(-i \hbar) k \cdot cos( kx )$
The area created underneath the exponential decay is$\int_{0}^{ \infty } sin(-k z) e^{ - k \cdot cot( -k z ) x } (- i \hbar ) { sin( -k z ) \over tan( - k z )} k e^{ -{k \over tan ( -k z ) } x }$

I'm not quite sure how to show these two integral values are exactly the same, but I think they pretty much have to be in order for the imaginary value to cancel in all possible (asymmetrical) finite depth wells. The normalization constant that I ignored is irrelevant, because it would end up multiplying both integrals by the same amount. We don't care about actual values, but only whether or not the integrals will produce the same area ... so the normilzation constant shouldn't matter.

 

[vanhees71]

Again: For the infinite square well it doesn't make sense to evaluate momentum expectation values, because momentum is not defined as an observable for this system. However, there's no infinite square well in nature. It's a model approximation for very tightly bound particles in some deep (but always) finite potential.

For the mathematical reasons that for the infinite square well the momentum observable cannot be defined, see earlier postings in this forum:

https://www.physicsforums.com/threa...ing-momentum-space.958402/page-2#post-6083298

 

[stevendaryl]

I'm not quite sure how to show these two integral values are exactly the same, but I think they pretty much have to be in order for the imaginary value to cancel in all possible (asymmetrical) finite depth wells. The normalization constant that I ignored is irrelevant, because it would end up multiplying both integrals by the same amount. We don't care about actual values, but only whether or not the integrals will produce the same area ... so the normilzation constant shouldn't matter.

Well, the proof that the integral is zero doesn't require knowing how the positive and negative contributions balance:

$$\langle p \rangle = \int_{-\infty}^{+\infty} \psi (-i\hbar \frac{\partial}{\partial x}) \psi dx = \frac{-i \hbar}{2} \int_{-\infty}^{+\infty} \frac{\partial}{\partial x} (\psi)^2 dx = \frac{-i \hbar}{2} lim_{L \rightarrow \infty} (\psi^2(L) - \psi^2(-L)))$$

Since $\psi \rightarrow 0$ at infinity, this evaluates to 0. In the case of an infinite square well, you don't take the integral to infinity, but to the edge of the well, but it's also true that $\psi$ goes to zero there.

So a real wave function always has an average momentum of zero. The reason is that the wave function has to go to zero at the boundary. (In 3 dimensions, instead of the last step being $\psi^2$ evaluated at $\pm L$, there will be a surface integral over a surface enclosing all of space, and the surface integral goes to zero.)

As far as interpreting this as there being equal and opposite regions where $\psi (-i\hbar \frac{\partial}{\partial x}) \psi$ is positive imaginary and negative imaginary, I guess that has to be true, but I don't immediately see the significance.

 

[learn.steadfast]

Again: For the infinite square well it doesn't make sense to evaluate momentum expectation values, because momentum is not defined as an observable for this system.
https://www.physicsforums.com/threa...ing-momentum-space.958402/page-2#post-6083298

What I see is that momentum (mathematically) is an operator, and can be computed on a Schrodinger equation (psi) when the boundary conditions are met such as to make a valid solution. In gerenal, that's when the result is square integratable and normalisable solution to Schrodinger's equation.

When I read the thread, you mention the Hermiticity and a requirement of "self adjointness"; I read a text ( Introduction to QM, Second edition, David J. Griffiths) and that was not listed as a requirement for the result of the momentum operator to be valid.

He says, "It's sufficient for our purpose to posulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position."

Hence, it follows that if the expectation value of position is well defined, then so is momentum. eg: in Intro to QM, mass is not relativistic and is simply a constant.
If we have a valid expectation value of position, vs time, then being able to take the derivative of that position is the same as being able to compute a scalar equivalent to physical momentum.

Griffiths then says,"Actually it is customary to work with momentum (p=mv),, rather than velocity."

At which point Griffiths shows how to arrive at a QM operator for the expectation value of momentum rather than velocity; and the text clearly shows that it's merely an earlier formula for the position expectation value as it evolves in time.
$<p> = -i m { d<x> \over d<t>}$

After substitution of variables, the "mass" is removed in favor of "-ih" ; but they are mathematically equivalent.

So, if the position expectation value exists and is a well behaved function of time ... then I understand the momentum expectation value does too.

I only vaguely grasp the point you are trying to make in the other thread.
There are some mathematical operations that are not proven to be defined in a given framework, unless certain conditions are met. I recall pointing out in a level 100 math class, that the derivation we had for the function exp(y) was undefined at the point "y=0", based on the axioms and lemmas given in the book up to that chapter;
In a final exam problem the teacher asked us to evaluate the infinite series of exp(x) at zero. It caused quite a stir because even though the proof did not guarantee a solution, everyone knew by inspection that the result converged at zero. Hence the whole class did the problem and ignored the strict requirement that the teacher had laid on me as a "lesson", to mark any problem insoluble that did not strictly fall within the lemmas proven in the textbook. The teacher had mocked me on the first day of class over a quibble. The result was that he tried to mocked me in front of class again because he had forgotten the "lesson" and tried to ridicule me and give me a "D" for missing one problem out of 3 that the whole class supposedly did correctly. I ended up making a fool of him by his own words on the chalkboard as I followed his strict instructions and showed that the derivation failed at zero -- even though the answer was right when evaluated.

There is a very big difference between not being able to prove something fits within a mathematical definition (eg: a process is not provably correct); and a result being wrong or truly non-extant. I don't (for example) say that everyone's solution in the class was wrong, it's just that it couldn't be arrived at using the tools presented to us in our textbook in a "provable" way. The class had committed "folly" of sorts in what they did, not in the answer to exp(0)=1.

So, I'm not really sure what you mean when you say the "momentum" doesn't exist / isn't defined.

When I look at your comment that the eigenvectors cos(kx), and sin(kx), should have the same co-domain; I am a bit lost in the details of exactly what is being argued.
The derivative of cos(kx) is just sin(kx)*k (a constant.); that seems to be a scalar multiple of the other eigenvector. The domain, in normal mathematics is whatever 'x' is, and that domain is the same in both cos() and it's derivative. The range is also a pure real in both cases...

So, I'm missing something in the argument about whatever is a "co-domain"?

I know that Observables require Hermitian operators (and why), also square inegratable, but nothing else was explicitly stated in the book as a requirement.

I'm lost.
I don't see any reason why momentum "can" not or "does" not exist in a rigid (infinite wall) box.

I infer, that there is something similar to point discontinuity or divide by zero type of error; where mathematically, a particular form of equation usually renders a wrong result. But, in some cases (due to L'hospital's rule) or the taking of limits, such errors can in fact cancel themselves out in many practical cases ... and so that an equation may give a correct result, although strictly speaking there is a potential error in the way it was arrived at. ( I'm trying to give an analogy, as I'm not sure what you're trying to describe. )

I notice that Sin and Cos abruptly terminate at the infinite walls, so there is the possibility of a point discontinuity .. or something?

 

[learn.steadfast]

As far as interpreting this as there being equal and opposite regions where $\psi (-i\hbar \frac{\partial}{\partial x}) \psi$ is positive imaginary and negative imaginary, I guess that has to be true, but I don't immediately see the significance.

Much of what I'm doing it to gain familiarity with QM, again. Significance is often discovered in the future.

Sometimes knowing that something is true over a region vs. at every point allows a person to solve equations by a short cut method where less information is available than a person might like. Real life situations often doesn't give us all the information we want about a real problem we may be required to solve.

I had thought, originally, that the imaginary values of the equation during the integration might cancel only due to symmetry of wells as given in typical textbooks; (which is wrong, it cancels within one wavelength in all cases and weird shaped wells.)

There are many cases where we only know what half of an well looks like in semiconductor physics, etc. (I'm an EE). So, how I write numerical software to attack a problem has much more flexibility if the details balance out much closer together.

There are also issues of interprtation of the wave function (Copenhagen, and others) that I'm not sure of; and the quantum well seems like a decent 1D problem to explore some of my reservations in.

There are various reasons that a particle might not be detected; and I think Heisenburg was quoted by Born as having the idea that maybe electrons were "no where" in between reactions. (AKA collapse of the wave packet.)

There are different possibilities as to how the "probability" of a wave might correspond with reality, etc. If I know that the real part of the integral is the momentum, and the imaginary cancellation is what assures it's a physical momentum; I have one more tool when it comes to estimating momentum and testing ideas, as there might be ways to numerically integrate smaller solution regions and still be able to estimate the total momentum from the mathematical sample.

I might, for example, try assuming that if I integrate from a finite region of psi such that the result is real; that I have a sample of some portion of the electron's mass and it's motion. eg: I integrate the probability for existence to know "how" much of the electron's momentum I've likely measured; then I might estimate the momentum of the particle within some standard deviation. eg: If I have integrated half the particle's probablility, I might try estimating that my value of momentum is off by less than 50% (on avarage for an ensemble of experiments.)

I think it was max Born, who's book I scanned through last week; who discusses the stationary state of hydrogen as "if" the electron was diffusing back and forth (in and out) of the nucleus with some RMS speed determined by the stationary state wave function. And another author who spoke about Heisnburg? thinking that "what if an electron was "no-where" between interactions?"

This got me thinking about the interpretation of the wave function of a particle in a box, again. There are nodes, where the particle "never" is found. So, even though we don't know where a particle is ... we *seem* to know where it is not. I recall my QM teacher, 20+ years ago, asking us "how does the particle get from one probable place to the next without going through the node?" The teacher was surnamed "Young", the very son of the famous Dr. Young. I have a few theories ... incomplete; but tractable.

As an EE, his comment bothers me for it matches an experience related to electromagnetic waves; where nodes form; eg: as a microwave travels forward and backward in a tube or on the skin of a wire making a standing wave.

I can take a thin antenna wire, and place it at the nodes perpendicular to the waveguide or microwave mains wire, and no electricity will be conducted into the antenna wire ... but the wave will continue right through it as if the antenna wasn't there.

In essence, the energy goes through a point ... but can not be detected. It seems to me that the wave statistical interpretation saying that the particle is "never" at a point, may be very wrong. Even the idea that a particle "must" be somewhere ... is also subtly wrong; I think the situation may be more like there are some places a particle *can* be where it is unable to interact with other matter that is perpendicular to it for a short time. ( for a microwave will interact with a wire coaxial with mains wire, but not a wire perpendicular to it at a node.)
Energy WILL be conducted away (perpendicularly), at a peak along the standing wave.

A voltage on a wire (then) is definitely an example of the potential of detecting a electron leaving the wire at 90 degrees to the standard direction of travel. But the electrons moving forward and backward in the wire, can NOT be detected doing their motion at the node points (a magnetic field exists only, and a loop of wire around the point can detect that ... but not a wire *AT* the point.)

So, I'm exploring.
In the harmonic well, there is a tractable situation to explore the relationship between wavelength, detectability, and classical particle motion and ideas related to classical physics;

For example, (classicly), probability is inversely proportional to speed at a point; for the probability of "catching" a particle is proportional to how long it is available at a point times how "often" a particle goes through that same point (repetition x times). In any closed path, where a particle must go through all points equally; repetition is irrelevant and probability is strictly based on velocity; Therefore, classically a particle is more probably caught where it moves slow vs. where it moves fast.

This is a feature preserved in the harmonic well. The amplitude is large near the turning points, and smallest at the center where kinetic energy is highest.

In theory, if one were to multiply the normalized velocity of a particle by the wave function's amplitude, you ought to get a probability density that is unity everywhere on average (because of Ehrnfests theorem).

But, the limited attempts I've made so far suggest that the average velocity indicated by the wave and integration ... is a stair step of discrete velocities; one for each zero crossing of psi.

I can only think of a couple of reasons why that would be the case.
One, is that the particle exists only between one pair of nodes when in a stationary state during any single experiment; but we don't know which peak is the trapped particle.

Dr. Young's question is answered this way ... the particle was trapped between one and only pair of nodes during the set up of each experiment; It no longer travels between nodes until the experiment is over, unless there is a disturbance to it's energy.
Since we do not know WHICH nodes trapped the particle, it is statistically possible to be between any pair inversely proportional to the momentum being trapped.

The other idea, is to hard to explain right now.

But, I'm seriously beginning to wonder if the "classical" probability model used to compare the harmonic well is in fact over-simplified. eg: all QM particles have spin; and a classical particle which is in a gravity well with walls made in various ways could easily spin. The imaginary part of the QM math, can in some ways, be made to correlate to oscillatory or rolling motion.

There are issues about Ehrnfast's theorem, and "what if's."
A ball in a frictionless half circle track could be made to only have friction on certain parts, and the track could have friction added at the classical stopping point;
The result would be that at all non-friction points, the rotating motion would be preserved but the translational energy would be converted between kinetic and potential as the ball traveled up and down the slopes. But at the end-points, the ball would have a finite amount of rotational energy that could be converted to *translational* motion in order to "jump" beyond the "classical" stopping points. If the friction is only on "part" of the ball, the the jump happening, height, etc. would depend on the phase of the ball when it hit the friction ramp. I hope that's not to convolued a thought ...

You can just laugh and call these ideas the Andrew Robinson of Scappose modifications to the Copenhagen interpretation.

But, there are a lot of questions I have about the "tunneling" ends of the QM waveform, and whether or not there is a classical analog in spinning objects that is ignored because people are "dumbing down" classical physics.

 

[vanhees71]

What I see is that momentum (mathematically) is an operator, and can be computed on a Schrodinger equation (psi) when the boundary conditions are met such as to make a valid solution. In gerenal, that's when the result is square integratable and normalisable solution to Schrodinger's equation.

When I read the thread, you mention the Hermiticity and a requirement of "self adjointness"; I read a text ( Introduction to QM, Second edition, David J. Griffiths) and that was not listed as a requirement for the result of the momentum operator to be valid.

He says, "It's sufficient for our purpose to posulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position."

Hence, it follows that if the expectation value of position is well defined, then so is momentum. eg: in Intro to QM, mass is not relativistic and is simply a constant.
If we have a valid expectation value of position, vs time, then being able to take the derivative of that position is the same as being able to compute a scalar equivalent to physical momentum.

Griffiths then says,"Actually it is customary to work with momentum (p=mv),, rather than velocity."

At which point Griffiths shows how to arrive at a QM operator for the expectation value of momentum rather than velocity; and the text clearly shows that it's merely an earlier formula for the position expectation value as it evolves in time.
$<p> = -i m { d<x> \over d<t>}$

After substitution of variables, the "mass" is removed in favor of "-ih" ; but they are mathematically equivalent.

So, if the position expectation value exists and is a well behaved function of time ... then I understand the momentum expectation value does too.

I only vaguely grasp the point you are trying to make in the other thread.
There are some mathematical operations that are not proven to be defined in a given framework, unless certain conditions are met. I recall pointing out in a level 100 math class, that the derivation we had for the function exp(y) was undefined at the point "y=0", based on the axioms and lemmas given in the book up to that chapter;
In a final exam problem the teacher asked us to evaluate the infinite series of exp(x) at zero. It caused quite a stir because even though the proof did not guarantee a solution, everyone knew by inspection that the result converged at zero. Hence the whole class did the problem and ignored the strict requirement that the teacher had laid on me as a "lesson", to mark any problem insoluble that did not strictly fall within the lemmas proven in the textbook. The teacher had mocked me on the first day of class over a quibble. The result was that he tried to mocked me in front of class again because he had forgotten the "lesson" and tried to ridicule me and give me a "D" for missing one problem out of 3 that the whole class supposedly did correctly. I ended up making a fool of him by his own words on the chalkboard as I followed his strict instructions and showed that the derivation failed at zero -- even though the answer was right when evaluated.

There is a very big difference between not being able to prove something fits within a mathematical definition (eg: a process is not provably correct); and a result being wrong or truly non-extant. I don't (for example) say that everyone's solution in the class was wrong, it's just that it couldn't be arrived at using the tools presented to us in our textbook in a "provable" way. The class had committed "folly" of sorts in what they did, not in the answer to exp(0)=1.

So, I'm not really sure what you mean when you say the "momentum" doesn't exist / isn't defined.

When I look at your comment that the eigenvectors cos(kx), and sin(kx), should have the same co-domain; I am a bit lost in the details of exactly what is being argued.
The derivative of cos(kx) is just sin(kx)*k (a constant.); that seems to be a scalar multiple of the other eigenvector. The domain, in normal mathematics is whatever 'x' is, and that domain is the same in both cos() and it's derivative. The range is also a pure real in both cases...

So, I'm missing something in the argument about whatever is a "co-domain"?

I know that Observables require Hermitian operators (and why), also square inegratable, but nothing else was explicitly stated in the book as a requirement.

I'm lost.
I don't see any reason why momentum "can" not or "does" not exist in a rigid (infinite wall) box.

I infer, that there is something similar to point discontinuity or divide by zero type of error; where mathematically, a particular form of equation usually renders a wrong result. But, in some cases (due to L'hospital's rule) or the taking of limits, such errors can in fact cancel themselves out in many practical cases ... and so that an equation may give a correct result, although strictly speaking there is a potential error in the way it was arrived at. ( I'm trying to give an analogy, as I'm not sure what you're trying to describe. )

I notice that Sin and Cos abruptly terminate at the infinite walls, so there is the possibility of a point discontinuity .. or something?

Sigh, Griffiths's QM book again... It's quite sloppy at some points and it obviously leads to some misunderstanding due to this sloppiness.

Again: Within standard quantum theory there'd be no momentum observable for a particle in an inifinitely high potential well. Of course, no such thing exists, and that's why this is a oversimplified example in some quantum-physics textbooks to exemplify the ideas about eigenstates (in this case energy eigenstates but no momentum eigenstates, because the Hamiltonian, i.e., energy is well-defined and represented by a self-adjoint operator, while this is not possible for momentum).

For physical cases, of course a massive particle has position and momentum as well-defined observables, which are represented by self-adjoint operators. One way is the position representation, where the Hilbert space is the space of square-integrable functions $\psi:\mathbb{R}^3 \rightarrow \mathbb{C}$, $\mathrm{L}^2$ and the operators for position and momentum are
$$\hat{\vec{x}} \psi(\vec{x})=\vec{x} \psi(\vec{x}), \quad \hat{\vec{p}} \psi(\vec{x})=-\mathrm{i} \hbar \vec{\nabla} \psi(\vec{x}).$$
Now one has to note that both operators are not defined everywhere on $\mathrm{L}^2$, because there are plenty of square integrable functions, where $\hat{x}$ leads out of this function space or $\hat{p}$ isn't even defined at all, because $\psi$ needs not to be differentiable at all. But that's fortunately not necessary either. You must only find a dense subspace of the Hilbert space, where these operators are well-defined (the domain of the operators) and don't lead out of this dense subspace when applied to a member of the domain, i.e., $\hat{x} \psi$ and $\hat{p} \psi$ (building the co-domain when applied to all functions in the domain). For position and momentum such a dense subspace is the Schwartz space of quickly falling functions that are arbitrarily often continuously differentiable.

For details on what a self-adjoint operator is and also pointing out some pitfalls when not remembering the caveats of the physicists' sloppy use of mathematical ideas, see

https://arxiv.org/abs/quant-ph/9907069

 

[learn.steadfast]

Sigh, Griffiths's QM book again... It's quite sloppy at some points and it obviously leads to some misunderstanding due to this sloppiness.

Yes ... I'm sure that's the case. I wasn't aware that it had specifically annoyed you before, now. It's not my intent to bring up a peeve...
I'm just glad that Griffiths leads to "some" understanding, so that I'm able to grasp part of the problem.

Now one has to note that both operators are not defined everywhere on $\mathrm{L}^2$, because there are plenty of square integrable functions, where $\hat{x}$ leads out of this function space or $\hat{p}$ isn't even defined at all, because $\psi$ needs not to be differentiable at all.

And you've lost me again ... the reasons why the things you say are true is not immediately apparent to me. I'm sure I will figure out what you mean, after spending a few hours looking up terms; and doing my best to contrive a simple example. But you're quoting a lot of formalism which I think are based upon examples that you've worked out yourself, in the past, and aren't sharing. Most of us recall a specific example in our minds that we've experienced learning when citing an abstract rule. So, it's often easier for someone with less experience to follow the abstract thoughts of someone with more experience when they cite examples.

For details on what a self-adjoint operator is and also pointing out some pitfalls when not remembering the caveats of the physicists' sloppy use of mathematical ideas, see

https://arxiv.org/abs/quant-ph/9907069

Thank you!
I've only read the first three pages, but so far ... the outline given of the nature of the problem, makes sense. The author has kind of "nailed" this "student's" thought:
"One has to wonder what remains in the mind of a student ... It should be to believe that today’s physicists are only at ease in the vagueness, the obscure and the contradictory."

I am not sure that "gibberish" is entirely fair to judge the QM text I'm reading; but in affirmation of arxiv article; I've always been amazed by how much the idea of "uncertainty" in position is emphasised as a strict rule; but then we are taught to solve problems with potential energy fields that are precisely located. eg: The walls of the well, even a finite one, are defined with infinite mathematical precision as to "where" they are... :D

I'll get back to you after digesting the arxiv article in full. Thanks for the link.

 

[learn.steadfast]

Again: Within standard quantum theory there'd be no momentum observable for a particle in an inifinitely high potential well. Of course, no such thing exists, and that's why this is a oversimplified example in some quantum-physics textbooks to exemplify the ideas about eigenstates (in this case energy eigenstates but no momentum eigenstates, because the Hamiltonian, i.e., energy is well-defined and represented by a self-adjoint operator, while this is not possible for momentum).

Momentum is a self adjoint opererator depending on the boundary conditions picked.
If the *operator* operates on L^2[0:1], f(0)=f(1)
Then, I can prove it's adjoint.

I'm still not quite sure what you're trying to argue ...
The potential solution of the Schrodinger equation, and the domain of an operator seem to be distinct things to me.

The particular solutions to the infinite well are a subset of the solutions to a periodic boundary condition. But, if we only require the domain of the *operator* to chosen by saying the function value at the walls are the *same*, then the operator ought to be self adjoint for *all* solutions which meet that boundary condition. The result is, when I plug a schrodinger equation solution into the operator, if the value of f() *happens* to be 0 for a particular solution, the operator still works on it. I'm not saying the operator *must* only work when the well vanishes to zero at the walls ... the operator's domain can be bigger than that, and still give a correct result when psi happens to vanish at the walls.

I mean, I don't see how the momentum operator *knows* the difference in a physical sense of whether the schrodinger equation soltuion comes from a infinite well or from a circular boundary, or whatever.

Schrodinger, from Robert Weinstock's review (Calculus of Variations), only required that the wave function be twice differentiable and vanish at infinity. Those were the basic requirements he put on the function. For reasons that are separate from his double analogy, I came to understand that the the solutions must also have continuous value and first derivative in physical situations which are meaningful.

The vanishing at infinity, is generally supposed when the function is square integrateable in my QM book ... although the link you give shows some interesting counter examples. The comb function, however, is really preserving of area by shrinking the widths and raising the height; so I see that as a kind of cheating by making a Dirac delta kind of thing as the "limit" goes to infinity of the comb's position. That's not a situation we were discussing in the infinite well; and I think I mentioned that there's no dirac delta issue... or at least, I thought there wasn't.

Why do you think it's impossible for momentum operator to be self adjoint ?
I'm not a math major, and I admit I rely more on intuition than rigorous proofs at times;
but I still wonder if there's some kind of error which might be canceled out by merely looking at the problem another way.

 

[PeroK]

The vanishing at infinity, is generally supposed when the function is square integrateable in my QM book ... although the link you give shows some interesting counter examples. The comb function, however, is really preserving of area by shrinking the widths and raising the height; so I see that as a kind of cheating by making a Dirac delta kind of thing as the "limit" goes to infinity of the comb's position. That's not a situation we were discussing in the infinite well; and I think I mentioned that there's no dirac delta issue... or at least, I thought there wasn't.

Griffiths himself at some point in the book says something like "any decent mathematician can furnish you with a counterexample", and focuses on functions that are in some sense physically possible, without formally specifyng precise criteria. In my view, for an introduction to QM this is fair enough. There are a number of areas where there are mathematical subtleties that an introduction to QM does not have the time to explore: Dirac Delta function, allowable subset of square integrable functions; the non-square-integrabiility of position and momentum eigenfunctions etc.

One of the good things about Griffiths is that he gets on with the subject of QM and doesn't get bogged down in mathematical subtleties. On the other hand, he is quite precise about the physics. If your goal is to learn QM, then I believe it is a good text.

Are you studying this on your own?

 

[vanhees71]

Just to clarify: I'm annoyed with Griffiths's QM book, because it's "more sloppy" than allowed. This is the more ununderstandable as Griffiths is an expert in the foundations of QT, and particularly when it comes to the foundations one needs a sufficiently rigorous mathematical basis not to be hindered by mathematical mistakes.

Of course in the QM 1 lecture we want to provide the students with an as clear physics picture what QM is about and not with the mathematical finesses and details of the "Rigged Hilbert Space", which is the mathematically rigorous version of the physicists' pragmatic and sometimes sloppy treatment of self-adjoint and unitary operators on Hilbert spaces. Nevertheless the math must be made clear enough so that the student is not confused by apparent paradoxes. I think the quoted paper by Gieres is a nice reading also for QM 1 students (in about the middle of the lecture, I guess) to get some idea about the mathematical pitfalls quoted there.

There are plenty of very good introductory QM 1 textbooks. My favorite is still the one we used in my own QM 1 lecture:

J. J. Sakurai, F. S. Tuan, Modern Quantum Mechanics, Addison Wesley (1994), Revised Edition

The newer edition of this book edited by Napolitano is also nice but contains unfortunately an outdated chapter about a socalled "Relativistic Quantum Mechanics", which I don't favor as an introduction to relativistic QT at all. Relativistic QT should be taught right away as relativistic QFT. We live in the 21st, not the first half of the 20th century, where Dirac's hole-theoretical formulation of QED was still modern, but it's more difficult than the modern QFT formulation of QED, but that's another story.

 

[learn.steadfast]

Are you studying this on your own?

Yes. I'm "going it alone!"

I took properties of materials 25+ years ago for engineering, but happened to have Dr. Young for physics 205. So, Dr. Young ... being the son of the famous double slit doctor experimenter of the same name ... worked QM problems into physics class, but not with operator notation or Dirac's approach. At the time, QM wasn't taught, specifically, to Electrical Engineers. Our knowledge of quantum effects was pretty much restricted to what is taught in the book "Lectures on the Electrical properties of Materials, Solymar & Walsh." LEPM is decent for solving simple equations and getting a feel for solid state physics of transistors, but it's miserable when working on details associated with magnetic effects. Dr. Young was mostly trying to teach us enough to understand the "dispersion" relation, and that doesn't require understanding magnetic effects. We only had the classical foundations leading to Maxwells' equations taught; and I recall mention being made that "magnetic fields can affect a particle's trajectory but not it's energy." Strictly speaking, that the proof given at the time only applies to the net energy of non-magnetic particles. Electrons which have magnetic moments, can be affected in both trajectory and energy. So, classical physics with "wave equations" was a little misleading (and a review of physics on Youtube, shows it still is taught, over-simplified.)

Griffith's was the textbook being used for engineering two years ago for new students, for apparently University of Portland finally decided that EE's need a formal course of some kind in QM. So, I picked Griffith's up when I dropped my son off to college. I wanted to see what the new EE's were learning.

In my back PF threads, you might notice ... I was trying to understand cyclotron experiments of effective mass measurements for semiconductors. This is one of the places where Solymar & Walsh was terribly deficient. Apparently, a very large proportion of the semiconductor business out there, is unaware that the magnetic strength of the cyclotron field actually affects the mass measurement. eg: I think "Greens" functions that claim to fit the effective mass of the electron over temeperature is grossly in error, and so is the "Minmos" simulator based on his curve fits because of the oversight. Green actually accuses the experiments of being in error! I didn't post my discovery, but I came across an article of a hetero-junction where the author both simulated and measured the effective mass of the electron as affected by a strong magnetic field vs. as if no field was present. The agreement was eye-opening.

 

[learn.steadfast]

Nevertheless the math must be made clear enough so that the student is not confused by apparent paradoxes. I think the quoted paper by Gieres is a nice reading also for QM 1 students (in about the middle of the lecture, I guess) to get some idea about the mathematical pitfalls quoted there.

It definitely is interesting. But, almost every student is prone to making mistakes ... and if they come up with a contradiction, they are more likely to think they made a mistake or overlooked something ... than to believe the contradiction they came up with is proof that QM is false in some way.

The problem I'm seeing is that the proofs are being appealed to as if they are more fundamental than the relationships (experimental) which led to the formulation. I mean, the way you're talking abstract algebra and domains, and co-domains, makes them seem as if those are somehow rigorously correct and it's impossible that they are "sloppy."

My brother is a mathematician, and I recall that after studying GF2 ... that he studied another Galois Field subject that included GF2 as one possible application. On the first test, out of 30 students ... My brother was the only one to notice that one of the proofs for the new subject made an assumption that actually failed for GF2 (It worked in other GF's). The teacher, after looking at the proof ... threw out the textbook and canceled class.

Every subject, and even the proofs, can have subtle failures which may or may not affect the results of what a person is trying to compute; An error and a failure depending on the particular way an imperfect proof is applied.

I am not used to computing domains by using functions, I am used to numbers being put in matricies ... determinants taken, transposes hand computed, etc. and domains found by inspection of what "can" be plugged in. But, I can still see that Hermitian vector language extends the old ideas in an analogous way, albeit confusing to me. I'm not sure it's really helpful.

I have an intuitive "wait a minute", let's check feeling, though, when I see a mathematical rigorist asserting that "momentum" doesn't exist when a mathematical operator fails to have a consistent domain, co domain, etc. computed; and the only thing they are pointing at is abstract proofs.

My reasoning is only based on numerical matricies and algebra:
But: With matricies ... a student can over specify a problem, leading to undefined results unless special care is taken: EXAMPLE:

If I have a problem with four undetermined variables, I will generally write a 4x4 square matrix equated with a column vector for boundary conditions; eg: I need exactly one row for each unknown to solve for. If I try to make a 5 row matrix, the extra equation will interfere with my ability to solve the matrix using simple mindless matrix diagonalization. eg: The "brute force" approach, won't work ... but that doesn't mean the problem is insoluble, it just means I need to eliminate one row of the matrix in order for "brute force" diagonalization to work in a case of 5 row matrix solves for 4 variables.

A computer can even be programmed to diagonalize the matrix and eliminate a redundant equation when detected (automatic). So this is a well known math problem, with a well known solution. A "rigourous" computer, would also check that the solution is the same with either of the redundant equations used; so as to prove neither is a contradictory boundary condition.

However, proofs in physics ... often depend on a human mind recognizing when a problem is over-specified, and hand eliminating a superfluous equation. The "automatic" correction for the case of over-specification is NOT included in the proof. Therefore, a mathematician can get a false sense of confidence in blindly applying proofs.

Under specification of boundary conditions is the more usual bane of mathematicians leading to errors, but they can over-compensate for that kind of failure by going too far in counter-proofs and over-specification! ( OR they can be sloppy writing proofs, and forget to note another boundary condition was already assumed.)

Schrodinger specified that solutions which satisfy his equation must vanish at infinity. Griffiths specifies that a function must be square integratable. Therefore, if I have a solution to schrodinger's equation (meeting his requirement) and one meeting Griffith's requirement for Hermaticity, I automatically have weeded out cases like the "comb" function you gave me in the link. It's an illegal "counter example", for it doesn't solve Schrodinger's equation as I was taught to do it! So, the ARXIV article is just as guilty of overlooking boundary conditions as the people it accuses of overlooking boundary conditions. It's just finger pointing...

De'broglie's Hypothesis, that matter has a ficticious "wave" length ... (later to be proven to be non-ficticious! ) was speicfied only for a test case of an object with constant momentum. De'broglie does not really attempt to calculate wavelengths of accelerating objects. But we should be VERY confident that momentum to wavelength conversion works when the energy of an object is constant, kineticially. That's what experiments have proven.

The characteristic mathematical test for an object with constant momentum, is that it has constant kinetic energy. (often called by letter "T").
So, we are mathematically guaranteed that in finite regions where delta or derivative of T is zero ... that momentum must be constant and computable by simple algebra (no integrals required). All bets are off in regions where T varies with distance: d T(x) / d x != 0

In square wells, there is a *region* of constant T, therefore the wavelength is well known; and the distance between zero crossings of phi can be accurately determined. Once we reach the walls, all bets are off... because the particle decelerates by tunneling into a wall. But, in the bottom of a "flat" well ... I don't need an integral to compute the wavelength and momentum of the particle. I can do it with algebra.

So, I find it very strange ... that integration, which attempts to extend de'broglie's hypothesis to "accelerating" particles, would fail in a region where the particle is not accelerating;

I think it more likely, that some proof ... which a mathematician or physisict used to extend De'Broglie's hypothesis to arbitrary potentials; is over-looking the effects of a boundary condition on the proof and (sometimes) following those proofs is creating a false sense of error, where there really can't be one.

eg: We know that when a particle goes through a double slit, in a double slit experiment ... that the particle must be found in one, and only one location. However, if we plug that boundary condition into the problem ... and insist that the particle be "probably" found in one, and only ONE, slit ... then the equation will fail to produce the interference pattern found by Scrhodinger's equation.
eg: The "probable" boundary condition, over-specifies the problem and destroys the result.

By analogy: Even though we know that in the case of the infinite well, that psi must be identically zero at the wall ... plugging that boundary condition in, apparently destroys the results of the momentum measurement or it's domain calculation. I could carefully work through the mathematical proofs, and locate what I already know by intuition ... one of the mathematical proofs for Hermiticity, Adjointness, vector spaces, etc. is making an invalid assumption in the case that the value is both zero and the same time, periodic; (and some Mathematician should probably *do* the check!), but I don't think I should believe a proof that denies what every experiment ever done has repeatedly shown; Where kinetic energy is constant and real, wavelength and therefore momentum are constant and real. The zero crossings of "psi" are a fixed distance apart (periodic, even) along an axis.