Why is E not equal to p^2/2m for a particle in a box?

[Happiness]

Consider a particle in the ground state of a infinite well with sides at $x=0$ and $x=a$. The ground-state energy $E_1=\frac{\hbar^2\pi^2}{2ma^2}$. The energy of the particle is entirely kinetic. Hence, $E_1=\frac{p^2}{2m}$. Solving for $p$, we get $p=\pm\frac{\hbar\pi}{a}$. So the momentum $p$ of the particle can only take on one of these two values. In other words, we expect the wave function $\phi(p)$ in momentum space to peak at these two values: $\phi(p)=\frac{1}{2}\delta(p+\frac{\hbar\pi}{a})+\frac{1}{2}\delta(p-\frac{\hbar\pi}{a})$, where $\delta$ is the Dirac delta function. But $\phi(p)$ is calculated to be $\phi(p)=e^{-ipa/2\hbar}\frac{2\pi/a}{\sqrt{\pi\hbar a}}\frac{cos(pa/2\hbar)}{(\pi/a)^2-(p/\hbar)^2)^2}$ (see 3rd equation in the picture¹).

We are led to conclude that $E\neq\frac{p^2}{2m}$, or equivalently, the energy of the particle is not entirely kinetic. But why?

¹There is a typo in the text: an $i$ is missing in the exponent in the expression for $\phi(p)$ and there shouldn't be a negative sign. Reference: http://en.wikipedia.org/wiki/Uncertainty_principle#Particle_in_a_box

Source: Quantum Physics, 3rd Edition by Gasiorowicz, p. 57.

 

[dextercioby]

As the walls are removed, the „outside” potential drops from +infinity to 0, thus is no longer constant. How could energy be conserved? Side note: I congratulate you for opening this topic.

 

[DrDu]

p is not a self-adjoint operator in case of the particle in a box problem. Hence your argument breaks down. However, p^2 remains well defined. This is treated in most texts on functional analysis. My preferred reference is Akhieser, Glazman,
Theory of Linear Operators in Hilbert Space (Dover Books on Mathematics)

 

[dextercioby]

In the other thread I mentioned the Bonneau article, here I quote the one by Gieres. It all has to do with the so-called "physical boundary conditions" in which the Hamiltonian is a self-adjoint operator, but the momentum not.

 

[Happiness]

View attachment 205079

Any physical reasons why momentum $p$ does not represent an observable? If we measure the momentum of the particle, what would be observed? Or what makes it physically impossible to measure the momentum?

 

[Happiness]

p is not a self-adjoint operator in case of the particle in a box problem. Hence your argument breaks down. However, p^2 remains well defined. This is treated in most texts on functional analysis. My preferred reference is Akhieser, Glazman,
Theory of Linear Operators in Hilbert Space (Dover Books on Mathematics)

Since $p^2$ is well defined, it is always $2mE_1$ when measured. But yet $p$ is not always $\pm\sqrt{p^2}$ according to the expression for $\phi(p)$. How do you make sense of this physically?

 

[DrDu]

Since $p^2$ is well defined, it is always $2mE_1$ when measured. But yet $p$ is not always $\pm\sqrt{p^2}$ according to the expression for $\phi(p)$. How do you make sense of this physically?

Maybe it is not lucky to say that $p^2$ is well defined. What is well defined is the operator $- d^2/dx^2$ on the domain 0<x<a. This is not the square of p, which has to be defined on $-\infty <x <+\infty$. Maybe it is more physical to consider instead the following problem:

Replace the infinite potential with delta functions of infinite strength at the boundary. The corresponding hamiltonian has a continuous spectrum ranging from 0 to infinity, just like for a free particle. For most energies, the wavefunction has support only outside the box, but for special energies $E_b$, which coincide with the energies of the infinite box problem, the wavefunction can also reside inside the box. The general wavefunction for these energies is $\psi(E_b)= \psi_i(E_b)+a \psi_e(E_b)$ where a is an arbitrary coefficient and i and e stand for inside and outside, respectively. For one special value of a, $\psi(E_b)= \cos(\sqrt{2mE_b}x)$ and exactly this function also an eigenstate of the free particle hamiltonian $\propto p^2$. For this function, your argument regarding the delta functions wrt k hold true.
We can use this free hamiltonian conveniently to calculate the possible values of $E_b$, and also the internal eigenfunctions $\psi_i(E_b)$, but it won't yield the correct solution outside the box which should vanishe before we remove the boundary.

 

[stevendaryl]

p is not a self-adjoint operator in case of the particle in a box problem. Hence your argument breaks down. However, p^2 remains well defined. This is treated in most texts on functional analysis. My preferred reference is Akhieser, Glazman,
Theory of Linear Operators in Hilbert Space (Dover Books on Mathematics)

Why isn't $p$ self-adjoint? Isn't the definition of self-adjoint the following:

$\langle A | p B \rangle = \langle p A | B\rangle$

Using $p = -i \partial_x$, then:

$\langle A |p B\rangle = \int_0^L A(x)^* (-i \partial_x B(x)) dx = -i \int_0^L A(x)^* ( \partial_x B(x)) dx$
$\langle p A | B\rangle = \int_0^L (-i \partial_x A(x))^* B(x) dx = +i \int_0^L (\partial_x A(x))^* B(x) dx$

So $\langle A | p B \rangle - \langle p A|B\rangle = -i \int_0^L \partial_x (A(x)^* B(x)) dx = -i (A(L)^* B(L) - A(0)^* B(0)) = 0$

So why isn't $p$ self-adjoint? Is it because $p A(x)$ is undefined at $x=0$ and $x=L$?

 

[stevendaryl]

Why isn't $p$ self-adjoint? Isn't the definition of self-adjoint the following:

$\langle A | p B \rangle = \langle p A | B\rangle$

Using $p = -i \partial_x$, then:

$\langle A |p B\rangle = \int_0^L A(x)^* (-i \partial_x B(x)) dx = -i \int_0^L A(x)^* ( \partial_x B(x)) dx$
$\langle p A | B\rangle = \int_0^L (-i \partial_x A(x))^* B(x) dx = +i \int_0^L (\partial_x A(x))^* B(x) dx$

So $\langle A | p B \rangle - \langle p A|B\rangle = -i \int_0^L \partial_x (A(x)^* B(x)) dx = -i (A(L)^* B(L) - A(0)^* B(0)) = 0$

So why isn't $p$ self-adjoint? Is it because $p A(x)$ is undefined at $x=0$ and $x=L$?

I see that this is addressed here: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=7&ved=0ahUKEwiH073Kjq7UAhWCoD4KHSCICyEQFghLMAY&url=http://cds.cern.ch/record/493028/files/0103153.pdf&usg=AFQjCNENt2Or59M7a7d_U9kd9DjPJIWaiw&cad=rja

But I'm not sure I understand the argument.

 

[vanhees71]

Why isn't $p$ self-adjoint? Isn't the definition of self-adjoint the following:

$\langle A | p B \rangle = \langle p A | B\rangle$

Using $p = -i \partial_x$, then:

$\langle A |p B\rangle = \int_0^L A(x)^* (-i \partial_x B(x)) dx = -i \int_0^L A(x)^* ( \partial_x B(x)) dx$
$\langle p A | B\rangle = \int_0^L (-i \partial_x A(x))^* B(x) dx = +i \int_0^L (\partial_x A(x))^* B(x) dx$

So $\langle A | p B \rangle - \langle p A|B\rangle = -i \int_0^L \partial_x (A(x)^* B(x)) dx = -i (A(L)^* B(L) - A(0)^* B(0)) = 0$

So why isn't $p$ self-adjoint? Is it because $p A(x)$ is undefined at $x=0$ and $x=L$?

What you've shown is Hermiticity but not self-adjointness. For an operator to be self-adjoint its co-domain must be in the Hilbert space and you should have corresponding eigenvectors. The eigenvectors of $\hat{p}=-\mathrm{i} \mathrm{d}_x$ are $u_p(x)=N \exp(-\mathrm{i} p x)$, but they are not in the Hilbert space discussed since the boundary conditions are not fulfilled for any $p \in \mathbb{R}$.

Note that for periodic boundary conditions the operator is self-adjoint on the corresponding Hilbert space leading to the exponential form of the Fourier series.

 

[stevendaryl]

Does being a self-adjoint operator imply that $\hat{p} |\psi\rangle$ must be an element of the Hilbert space whenever $|\psi\rangle$ is? In that case, neither $\hat{p}$ nor $\hat{p}^2$ is self-adjoint on the space of square-integrable functions $\psi(x)$ with $\psi(0) = \psi(L) = 0$. That can be seen by considering the functions

$\psi(x) = x$ from $x=0$ to $x=L/2$
$\psi(x) = L - x$ from $x=L/2$ to $x=L$

When $\hat{p}$ acts on this function, it produces a function that is not zero at $x=0$ or $x=L$.

By using a quadratic, we can get a function where $\hat{p}^2 \psi$ similarly fails to vanish at the boundary.

 

[DrDu]

No, p isn't bounded, i.e. it is not defined on all of hilbert space.
For an unbounded operator to be self adjoint, it has to be:
1. Hermitian
2. Defined on a dense subset of H
3. This domain has to coincide with the domain of the adjoint operator.

 

[kith]

Any physical reasons why momentum $p$ does not represent an observable? If we measure the momentum of the particle, what would be observed? Or what makes it physically impossible to measure the momentum?

If you imagine a standard momentum measurement with high accuracy, the uncertainty in the position becomes larger than the width of the box due to the HUP. So such a measurement would lead to the possibility that the particle escapes the box.

 

[dextercioby]

If you imagine a standard momentum measurement with high accuracy, the uncertainty in the position becomes larger than the width of the box due to the HUP. So such a measurement would lead to the possibility that the particle escapes the box.

There is a problem with the HUP particularly for this problem which has its mathematical limitations, hence it's illegal to invoke it to disprove that the momentum is an observable. In blank, the axiom of QM says: (in the absence of superselection rules) any observable is described by a linear self-adjoint operator acting in the complex, separable Hilbert space of (pure) states and viceversa. By this axiom, if p is selfadjoint, then it must be (an) observable.

For a particle in a box, as soon as the proper boundary conditions are placed, the momentum operator described by the derivative wrt to coordinate has an infinity of self-adjoint extensions parametrized by a U(1) phase factor theta. This is explained in the article by Bonneau (uploaded by me in the other current thread and linked to here by stevendaryl). The true self-adjoint operator p_theta admits true eigenfunctions in the Hilbert space, hence it's possible to pick any of these wavefunctions to have $\Delta p =0$. This is explained on page 9, point 3 on the top of the page. So, there's a limitation of the HUP usage and the explanation is simple: the coordinate operator is bounded (because the particle is constrained in the box), the momentum otoh is not, both are self-adjoint, but the domain of their commutator cannot be defined (x psi_theta is not in the domain of p). This is explained by Brian Hall in his book: "Quantum Theory for Mathematicians", section 12.2 (where the particle in the box is called "counterexample" to the HUP).

 

[DrDu]

I also like to recommend the following article for the problem at hand:
Zhu, Chengjun, and John R. Klauder. "Classical symptoms of quantum illnesses." American journal of physics 61.7 (1993): 605-611.

 

[DrDu]

For a particle in a box, as soon as the proper boundary conditions are placed, the momentum operator described by the derivative wrt to coordinate has an infinity of self-adjoint extensions parametrized by a U(1) phase factor theta. This is explained in the article by Bonneau (uploaded by me in the other current thread and linked to here by stevendaryl). The true self-adjoint operator p_theta admits true eigenfunctions in the Hilbert space, hence it's possible to pick any of these wavefunctions to have $\Delta p =0$. This is explained on page 9, point 3 on the top of the page. So, there's a limitation of the HUP usage and the explanation is simple: the coordinate operator is bounded (because the particle is constrained in the box), the momentum otoh is not, both are self-adjoint, but the domain of their commutator cannot be defined (x psi_theta is not in the domain of p). This is explained by Brian Hall in his book: "Quantum Theory for Mathematicians", section 12.2 (where the particle in the box is called "counterexample" to the HUP).

The point is that the momentum operator by definition generates a translation. For a hilbert space of functions defined on a limited range, this is only possible if the space is assumed to be periodic. However for a box with vanishing wavefunctions at the boundary this is not the case and any translation will lead to a violation of these boundary conditions, hence momentum operator cannot exist.

 

[PeroK]

Is one part of the answer that an infinite potential is physically impossible? The infinite well must be replaced by a finite well with very large potential and then things settle down?

 

[Happiness]

If you imagine a standard momentum measurement with high accuracy, the uncertainty in the position becomes larger than the width of the box due to the HUP. So such a measurement would lead to the possibility that the particle escapes the box.

Would it be physically impossible too to measure the temperature of a gas in a box? Suppose we have an ensemble of monoatomic ideal gas particles in a box. The average internal energy of a particle $=\frac{3}{2}kT=\frac{<p^2>}{2m}=\,<E>\,=E=\frac{\hbar^2\pi^2}{2ma^2}$, from which we get $T=\frac{\hbar^2\pi^2}{3kma^2}$.

If so, does it mean that $E=\,<E>\,=\frac{<p^2>}{2m}\neq\frac{p^2}{2m}$? Do the particles have different $p$ but the same $E$? How is this possible?Does $E\neq\frac{p^2}{2m}$ changes the way we interpret the kinetic theory of gases? Do the particles have some definite (although non-measurable) momentum when moving in a box and when colliding with other particles?

 

[DrDu]

Is one part of the answer that an infinite potential is physically impossible? The infinite well must be replaced by a finite well with very large potential and then things settle down?

No, momentum doesn't care about a potential term in the hamiltonian, whether finite or not. Even with an infinite potential, momentum remains well defined.
The peculiarities are only due to restricting the whole Hilbert space to functions being defined on the range 0<x<a. In this new hilbert space, momentum is no longer defined, but the hamiltonian looks much nicer. So in some sense you can't have the apple and eat it.

 

[dextercioby]

The point is that the momentum operator by definition generates a translation. For a hilbert space of functions defined on a limited range, this is only possible if the space is assumed to be periodic. However for a box with vanishing wavefunctions at the boundary this is not the case and any translation will lead to a violation of these boundary conditions, hence momentum operator cannot exist.

In other words, the particle in a box of infinite walls breaks the spatial isotropy (an assumption to have momentum as generator of the group of spatial translations) which can only be achieved, iff the particle is free to move along the whole real axis. For half line, it's even worse, the momentum has deficiency indices (1,0) hence it can't be extended to a s-adj operator.

 

[kith]

Would it be physically impossible too to measure the temperature of a gas in a box?

I don't understand what you write after that (maybe it would be good to distinguish clearly between operators, expectation values of operators and simple scalar quantities). But even if I did, I'm not sure if I have good answers to your questions.

Strictly speaking, the infinite square well is unphysical. As @DrDu pointed out, using it as an approximation has the effect that some things get easier and some things get more complicated. The merit of such approximations is that they help us to understand real physical situations. For me, dwelling on the things which get more complicated is also interesting but if I get strange conclusions for e.g. what happens when I try to measure a certain observable, I don't want to lose sight of the fact that measurements happen in the real world.

 

[Paul Colby]

Just a non-mathematical comment. The particle applies a pressure to the wall. Moving the wall to infinity would do work on the particle.

 

[Happiness]

maybe it would be good to distinguish clearly between operators, expectation values of operators and simple scalar quantities

Let (1) an operator, such as the hamiltonian operator, be represented by a hat on top: $\hat{H}$,
(2) the expectation value of the hamiltonian operator be represented by angle brackets: $<\hat{H}>$, and
(3) a simple scalar quantity, such as temperature, be represented just by a letter: T.

Would it be physically impossible too to measure the temperature of a gas in a box? Suppose we have a collection of identical monoatomic ideal gas particles in a (one-dimensional) box, with each particle in the ground state of energy $E$. The average internal energy $\bar{U}$ of a particle $=\frac{3}{2}kT=$ the average kinetic energy of a particle $\frac{<\hat{p^2}>}{2m}=\,<\hat{H}>\,=E=\frac{\hbar^2\pi^2}{2ma^2}$, from which we get $T=\frac{\hbar^2\pi^2}{3kma^2}$.

If so, does it mean that $E=\,<\hat{H}>\,=\frac{<\hat{p^2}>}{2m}\neq\frac{p^2}{2m}$ (where $E$ and $p$ are the energy and the magnitude of the momentum of a particle respectively)? Do the particles have different $p$ but the same $E$? How is this possible?

How does $E\neq\frac{p^2}{2m}$ change the way we interpret the kinetic theory of gases? Do the particles, when moving in a box and when colliding with other particles, have some definite momentum or some definite range/distribution (e.g. when in a superposition of momentum states) of values of momentum (although non-measurable)?

 

[kith]

Suppose we have a collection of identical monoatomic ideal gas particles in a (one-dimensional) box, with each particle in the ground state of energy $E$. The average internal energy $\bar{U}$ of a particle $=\frac{3}{2}kT=$ the average kinetic energy of a particle $\frac{<\hat{p^2}>}{2m}=\,<\hat{H}>\,=E=\frac{\hbar^2\pi^2}{2ma^2}$, from which we get $T=\frac{\hbar^2\pi^2}{3kma^2}$.

That's not how statistical mechanics works. The temperature is only defined if the system is in thermodynamic equilibrium, i.e. if your state is given by a density matrix of the form
$$\rho = \sum_i \exp \left( -\frac{E_i}{kT} \right) |E_i\rangle \langle E_i|.$$

 

[Happiness]

Quantum mechanics says when a particle is in a box, $E\neq\frac{p^2}{2m}$. But the kinetic theory of gases says internal energy $U =$ kinetic energy $\frac{p^2}{2m}$, even though the gas particles are in a box, that is, the probability that the gas particles are found outside the box is zero. Isn't there a contradiction?

 

[kith]

It is still not clear to me, what you mean by writing $p$.

But it doesn't matter much: there can't be a contradiction, because the kinetic theory of gases is derived from statistical mechanics where you need to use the same operators as in the eigenvalue problem for the energies of the box. So if there's a problem with the momentum operator, it shows up in both cases. And if there's a cure (like in the article to which @dextercioby linked in the other thread), it works in both cases.

You have the impression that there's a problem because you incorrectly use two different notions of momentum.

 

[Happiness]

It is still not clear to me, what you mean by writing $p$.

Since operator $\hat{p}$ does not admit any eigenvalues as stated in post #4:

View attachment 205079

, operator $\hat{p}$ is not an observable, and the momentum of a particle in a box would be undefined. Thus, it would be wrong to visualise a particle moving in a box with a certain momentum. And to visualise the particles (of an ideal gas in a box) colliding with one another and the walls of the box would be wrong too. Nothing moves. Each particle is just found to be at a certain location when measured, with a certain probability prescribed by its wave function $\psi(x)$.

Is this interpretation correct? And if nothing moves, there would be no pressure exerted on the walls of the box. Right?

Or should we visualise the particles moving with a certain momentum, each with a certain probability of moving at that certain momentum prescribed by its wave function $\phi(p)$? In other words, we cannot physically measure at what momentum they are moving, but they do move. Suppose we accept this visualisation. Let's denote the magnitude of the momentum of a particle by $|\mathbf{p}|$. The kinetic energy of a particle would be $\frac{|\mathbf{p}|^2}{2m}$. Since the value of $|\mathbf{p}|$ follows the distribution prescribed by the wave function $\phi(p)$, it may not always be equal to $\sqrt{2mE}$, where $E$ is the ground-state energy of the box. In other words, the kinetic energy $\frac{|\mathbf{p}|^2}{2m}$ of a particle may not always be equal to $E$. But since we are visualising ideal-gas particles, the potential energy of the particles is $0$. And so the total energy of a particle equals its kinetic energy, which may not be equal to $E$. On the other hand, since the particle is in ground state, its total energy is $E$. And we have a contradiction.

 

[Jilang]

Happiness, momentum is a vector and has a direction as well as a magnitude. One may be sure of the magnitude but unsure as to the direction.

 

[Happiness]

Happiness, momentum is a vector and has a direction as well as a magnitude. One may be sure of the magnitude but unsure as to the direction.

That is true. But that still doesn't resolve the contradiction for me in the second visualisation. From the calculated expression for $\phi(p)$, we can't be sure of the magnitude of the momentum too, but yet we can be sure of the energy.

 

[Jilang]

The calculated expression looks like the double delta function though as the potential increases. It's only when it becomes infinite and you are trying to divide zero by zero it looks a bit odd.

 

[Happiness]

The calculated expression looks like the double delta function though as the potential increases. It's only when it becomes infinite and you are trying to divide zero by zero it looks a bit odd.

Could you elaborate clearly? I plotted and I don't get a double delta function: http://www.wolframalpha.com/input/?i=plot+y=e^(-i10000x)/100*cos(10000x)/(1-(10000x)^2), taking $\hbar=10^{-4}$ and $a\approx\pi\approx2$.

 

[Jilang]

I was talking about the final equation. Have you tried plotting that?

 

[Happiness]

I was talking about the final equation. Have you tried plotting that?

$|\phi(p)|^2$? Yes, it's not a delta function too.

 

[vanhees71]

Let (1) an operator, such as the hamiltonian operator, be represented by a hat on top: $\hat{H}$,
(2) the expectation value of the hamiltonian operator be represented by angle brackets: $<\hat{H}>$, and
(3) a simple scalar quantity, such as temperature, be represented just by a letter: T.

Would it be physically impossible too to measure the temperature of a gas in a box? Suppose we have a collection of identical monoatomic ideal gas particles in a (one-dimensional) box, with each particle in the ground state of energy $E$. The average internal energy $\bar{U}$ of a particle $=\frac{3}{2}kT=$ the average kinetic energy of a particle $\frac{<\hat{p^2}>}{2m}=\,<\hat{H}>\,=E=\frac{\hbar^2\pi^2}{2ma^2}$, from which we get $T=\frac{\hbar^2\pi^2}{3kma^2}$.

If so, does it mean that $E=\,<\hat{H}>\,=\frac{<\hat{p^2}>}{2m}\neq\frac{p^2}{2m}$ (where $E$ and $p$ are the energy and the magnitude of the momentum of a particle respectively)? Do the particles have different $p$ but the same $E$? How is this possible?

How does $E\neq\frac{p^2}{2m}$ change the way we interpret the kinetic theory of gases? Do the particles, when moving in a box and when colliding with other particles, have some definite momentum or some definite range/distribution (e.g. when in a superposition of momentum states) of values of momentum (although non-measurable)?

If a particle is in an eigenstate of the Hamiltonian it's not in thermal equilibrium with a heat bath (except at 0 temperature).

 

[Jilang]

$|\phi(p)|^2$? Yes, it's not a delta function too.

Can you link to It?