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

[kith]

@Happiness, I get the impression that you are not trying to understand what the people here are telling you.

1. If you want to talk about movement or its absence in QM, you need to make the notion precise by using a corresponding self-adjoint operator. In the problem of the infinite potential well, the usual momentum operator isn't self-adjoint. A paper has been cited where a self-adjoint extension of the momentum operator is constructed. Very cool! Now you can explore how movement works in this problem. So why do you keep talking about vaguely defined classical notions of momentum and movement?

2. The momentum space wavefunction $\phi(p)$ of the OP is only valid after the walls have been removed. I have asked you repeatedly what $p$ stands for and it stands for an eigenvalue of the ordinary momentum operator. So you cannot use it to reason about the situation with the walls in place.

I have told you in my last post that you are incorrectly using two different notions of momentum to reason about the same thing and that this is very likely the reason for your perceived contradiction. And in your next post? You use the same incompatible notions of momentum again.

 

[stevendaryl]

Maybe it should be a separate thread (or perhaps it already is), but I still am not clear about the reasons that $\hat{p}$ is not self-adjoint. I understand that the spectral theorem tells us that if $\hat{p}$ is self-adjoint, then its eigenfunctions form a complete basis. But since $\hat{p}$ has no eigenfunctions (it has functions $\psi(x)$ satisfying $\hat{p} \psi(x) = \lambda \psi(x)$, but such functions are not elements of the Hilbert space), it can't be self-adjoint.

But the definition of self-adjoint is (if I have this right):

1. $\hat{p} |\psi \rangle = \hat{p}^\dagger |\psi \rangle$ whenever both sides are defined (that is, it's symmetric)
2. $dom(\hat{p}) = dom(\hat{p}^\dagger)$ where $dom$ means the domain (which I take to mean that $dom(A)$ = those $|\psi\rangle$ such that $A |\psi\rangle$ is an element of the Hilbert space)

The first is easily proved using integration by parts. So if $\hat{p}$ is not self-adjoint, it must be that $dom(\hat{p}^\dagger) \supset dom(\hat{p})$. That in turn means that there is some element $|\psi\rangle$ that is in the domain of $\hat{p}^\dagger$ but not in the domain of $\hat{p}$. So what's an example of such a $|\psi\rangle$?

 

[Happiness]

Now you can explore how movement works in this problem. So why do you keep talking about vaguely defined classical notions of momentum and movement?

you are incorrectly using two different notions of momentum to reason about the same thing and that this is very likely the reason for your perceived contradiction. And in your next post? You use the same incompatible notions of momentum again.

It should be apparent that I'm asking these questions because I am a beginner, who does not have a clear understanding of quantum mechanics. Could you elucidate the quantum-mechanical notions of momentum and movement?

Does quantum mechanics provide any physical picture on how particles in a box move and collide?

 

[PeroK]

It should be apparent that I'm asking these questions because I am a beginner, who does not have a clear understanding of quantum mechanics. Could you elucidate the quantum-mechanical notions of momentum and movement?

Does quantum mechanics provide any physical picture on how particles in a box move and collide?

Let me answer a different question about the ground state of the hydrogen atom. The classical picture is a single electron "moving in an orbit" about a single proton.

But, if we look at the gorund state we find that the expected value of the Kinetic Energy is:

$\langle T \rangle = |E_0|$

And, that the expected value of the angular monentum squared is:

$\langle L^2 \rangle = 0$

That means that every measurement of total angular momentum gives the answer $0$.

What do you make of that? The electron is moving, but always "radially"? What sort of orbit is that?

There is, in fact, a long discussion about this here:

http://physics.stackexchange.com/qu...ly-radial-motion-in-the-hydrogen-ground-state

You might also like this:

https://www.physicsforums.com/threads/electrons-orbiting-the-nucleus-angular-momentum.863964/

The moral I took from this is that you cannot relate QM to classical notions of motion.

 

[Jilang]

Let me answer a different question about the ground state of the hydrogen atom. The classical picture is a single electron "moving in an orbit" about a single proton.

But, if we look at the gorund state we find that the expected value of the Kinetic Energy is:

$\langle T \rangle = |E_0|$

...
That means that every measurement of total angular momentum gives the answer $0$.
.

No, it doesn't mean that at all! It is an expectation value, an average. It requires many measurements to average to zero.

 

[stevendaryl]

No, it doesn't mean that at all! It is an expectation value, an average. It requires many measurements to average to zero.

If a quantity is never negative, then the only way for the average to be zero is if almost every measurement returns 0 (or since measurements have a finite accuracy, almost every measurement must be consistent with zero--less than the measurement accuracy).

 

[PeroK]

No, it doesn't mean that at all! It is an expectation value, an average. It requires many measurements to average to zero.

$L^2$ is non-negative, so it ain't that easy to get an average of 0 unless every measurement is 0.

 

[stevendaryl]

$L^2$ is non-negative, so it ain't that easy to get an average of 0 unless every measurement is 0.

But if your memory is wiped after each measurement...Sorry, wrong thread!

 

[Jilang]

How would one go about measuring it?

 

[Nugatory]

No, it doesn't mean that at all! It is an expectation value, an average. It requires many measurements to average to zero.

You are forgetting that he ground state is also an eigenstate of $L^2$ with eigenvalue zero, so every measurement must yield exactly zero. What you said could be correct for an observable that does not commute with $H$ and $L^2$ so does not have the ground state as an eigenstate. In that case repeated measurements of that observable when the electron is in the ground state wil yield different results that have to be averaged to get the expectation value; and if these are uniformly distributed on both sides of zero, then the expectation value will come out zero.

 

[stevendaryl]

You are forgetting that he ground state is also an eigenstate of $L^2$ with eigenvalue zero, so every measurement must yield exactly zero. What you said could be correct for an observable that does not commute with $H$ and $L^2$ so does not have the ground state as an eigenstate. In that case repeated measurements of that observable when the electron is in the ground state wil yield different results that have to be averaged to get the expectation value; and if these are uniformly distributed on both sides of zero, then the expectation value will come out zero.

But whether or not the state is an eigenstate of $L^2$, a measurement of $L^2$ has to result in a nonnegative quantity. So you can't have $\langle L^2 \rangle = 0$ except when the state is an eigenstate of $L^2$ with eigenvalue 0.

 

[Nugatory]

But whether or not the state is an eigenstate of $L^2$, a measurement of $L^2$ has to result in a nonnegative quantity. So you can't have $\langle L^2 \rangle = 0$ except when the state is an eigenstate of $L^2$ with eigenvalue 0.

That's also true, and pretty conclusively proves that the statement about a statistical spread averaging to zero cannot be possibly be correct.

I was trying to guess identify the error that had led Jilang to that easily disproven statement, and suspect that he was thinking about it the way we think about the position observable of a simple harmonic oscillator: expectation value zero, statistical spread on each side of zero when the oscillator is in any energy eigenstate.

 

[PeroK]

That's also true, and pretty conclusively proves that the statement about a statistical spread averaging to zero cannot be possibly be correct.

I was trying to guess identify the error that had led Jilang to that easily disproven statement, and suspect that he was thinking about it the way we think about the position observable of a simple harmonic oscillator: expectation value zero, statistical spread on each side of zero when the oscillator is in any energy eigenstate.

In the state $\psi_{210}$, say, $\langle L_z \rangle = 0$ but not every measurement of $L_z$ would be $0$.

 

[stevendaryl]

That's also true, and pretty conclusively proves that the statement about a statistical spread averaging to zero cannot be possibly be correct.

Jilang's question is appropriate, though: How can you measure $L^2$? Since that's $(L_x)^2 + (L_y)^2 + (L_z)^2$, let's ask: how do measure $(L_x)^2$? Is there a distinction between:

1. Directly measuring $(L_x)^2$
2. Measuring $L_x$ and squaring the result?

 

[PeroK]

Jilang's question is appropriate, though: How can you measure $L^2$? Since that's $(L_x)^2 + (L_y)^2 + (L_z)^2$, let's ask: how do measure $(L_x)^2$? Is there a distinction between:

1. Directly measuring $(L_x)^2$
2. Measuring $L_x$ and squaring the result?

There's something about it here:

https://physics.stackexchange.com/q...ntum-measured-in-experiments-in-practice?rq=1

 

[vanhees71]

Maybe it should be a separate thread (or perhaps it already is), but I still am not clear about the reasons that $\hat{p}$ is not self-adjoint. I understand that the spectral theorem tells us that if $\hat{p}$ is self-adjoint, then its eigenfunctions form a complete basis. But since $\hat{p}$ has no eigenfunctions (it has functions $\psi(x)$ satisfying $\hat{p} \psi(x) = \lambda \psi(x)$, but such functions are not elements of the Hilbert space), it can't be self-adjoint.

But the definition of self-adjoint is (if I have this right):

1. $\hat{p} |\psi \rangle = \hat{p}^\dagger |\psi \rangle$ whenever both sides are defined (that is, it's symmetric)
2. $dom(\hat{p}) = dom(\hat{p}^\dagger)$ where $dom$ means the domain (which I take to mean that $dom(A)$ = those $|\psi\rangle$ such that $A |\psi\rangle$ is an element of the Hilbert space)

The first is easily proved using integration by parts. So if $\hat{p}$ is not self-adjoint, it must be that $dom(\hat{p}^\dagger) \supset dom(\hat{p})$. That in turn means that there is some element $|\psi\rangle$ that is in the domain of $\hat{p}^\dagger$ but not in the domain of $\hat{p}$. So what's an example of such a $|\psi\rangle$?

Ok, once more. If $\hat{p}=-\mathrm{i} \partial_x$ on $\mathrm{L}^2([0,1])$ with rigid boundary conditions $\psi(0)=\psi(1)=0$ were self-adjoint, then it would have a complete set of orthonormal eigenvectors. The putative eigenfunctions are of course $u_p(x)=N_p \exp(\mathrm{i} p x)$, but these are not in the Hilbert space, i.e., there are no eigenvalues and eigenvectors at all.

For the same reason it doesn't make any sense to talk about the momentum representation and to think about position-momentum uncertainty relations. The "particle in the infinite square well" is a very bad example for beginners in QM 1. It should be avoided in teaching the subject! There's no harm (may be even less harm) done when simply ignoring it.

 

[Jilang]

There's something about it here:

https://physics.stackexchange.com/q...ntum-measured-in-experiments-in-practice?rq=1

Thank you, but that is about how you measure L.

 

[bob012345]

Of

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?

I think there is a contradiction since the Kinetic theory of gasses is a Classical theory and not a Quantum theory. My suggestion for this and the original question is to approach it with numbers from the classical regime down and the quantum regime up and study the boundary regions where the two models diverge. You will be fleshing out Bohr's Correspndence Principle.

 

[Prathyush]

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?

Interesting observation, I potentially have a method to resolve this issue, but it would require work. Instead of a sharp wall. we can work with a potential of this nature
V(x) = 0 for $x \in (-a,a)$
and for
$V(x) = -\alpha(x+a)$ for x<-a
$V(x) = \alpha(x-a)$ for x>a
To my knowledge Schrodinger equation can be exactly solved for a linear potential.
So the walls of the well become infinitely sharp at alpha tends to infinity. In this approximation we can calculate the wave functions, energies, eigenstates etc. I think that It would agree with the standard answer, however it has to be shown just to be sure.

 

[Akwinder Singh Mander]

m

how do we derive this?

 

[stevendaryl]

Ok, once more. If $\hat{p}=-\mathrm{i} \partial_x$ on $\mathrm{L}^2([0,1])$ with rigid boundary conditions $\psi(0)=\psi(1)=0$ were self-adjoint, then it would have a complete set of orthonormal eigenvectors. The putative eigenfunctions are of course $u_p(x)=N_p \exp(\mathrm{i} p x)$, but these are not in the Hilbert space, i.e., there are no eigenvalues and eigenvectors at all.

You're repeating what was already said. But the definition of "self-adjoint" implies that the domain of $\hat{p}$ is unequal to the domain of $\hat{p}^\dagger$. So I was asking: what function is in the domain of one but not the other?

 

[stevendaryl]

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.

Maybe you can help me understand the last point in this case. If $p$ is not self-adjoint, then point 3 means that the domain of $p$ is not equal to the domain of $p^\dagger$. So what's an example of an element of the domain of $p^\dagger$ that is not in the domain of $p$?

 

[dextercioby]

Maybe you can help me understand the last point in this case. If $p$ is not self-adjoint, then point 3 means that the domain of $p$ is not equal to the domain of $p^\dagger$. So what's an example of an element of the domain of $p^\dagger$ that is not in the domain of $p$?

It is covered by Gieres in his article on pages 11 and 12 (by the pdf numbering that is 13 and 14).

 

[DrDu]

Maybe you can help me understand the last point in this case. If $p$ is not self-adjoint, then point 3 means that the domain of $p$ is not equal to the domain of $p^\dagger$. So what's an example of an element of the domain of $p^\dagger$ that is not in the domain of $p$?

So let $p=-i d/dx$. For p to be hermitian, $-i \int_0^a dx \psi_2^* \psi'_1 =i \int_0^a dx \psi'^*_2 \psi_1$, which leads to the condition that $\psi^*_2 (0) \psi_1(0) -\psi^*_2(a)\psi_1(a)=0$. If we chose the same condition for $\psi_1$ as for the operator $H=-d^2/dx^2$, namely $\psi_1(0)=\psi_1(a)=0$, then then there arise no restrictions on $\psi_2$. Hence the domain of $p^\dagger$ is larger than that of $p$. So if we want to write $H=p^2$, this is not possilbe with a self adjoint operator p.

 

[houlahound]

Why are we discussing this due to the fact the box has infinite walls, nobody can ever measure anything in it because there is no way in ...or out.

This box is more black than a black box.

I always assumed it was a pretend problem for displaying some simple conclusions to beginners.

 

[stevendaryl]

So let $p=-i d/dx$. For p to be hermitian, $-i \int_0^a dx \psi_2^* \psi'_1 =i \int_0^a dx \psi'^*_2 \psi_1$, which leads to the condition that $\psi^*_2 (0) \psi_1(0) -\psi^*_2(a)\psi_1(a)=0$. If we chose the same condition for $\psi_1$ as for the operator $H=-d^2/dx^2$, namely $\psi_1(0)=\psi_1(a)=0$, then then there arise no restrictions on $\psi_2$. Hence the domain of $p^\dagger$ is larger than that of $p$. So if we want to write $H=p^2$, this is not possilbe with a self adjoint operator p.

Thank you! So a concrete example is the ground state: $\psi(x) = sin(\frac{\pi x}{a})$. This function is in the domain of $p^\dagger$, but not in the domain of $p$ (since $p \psi(x) \propto cos(\frac{\pi x}{a})$ does not vanish at the walls.).

That helps a lot, except that it makes me wonder if this snippet from the Wikipedia article on self-adjoint operators is wrong:

By the Riesz representation theorem for linear functionals, if x is in the domain of $A^\dagger$, there is a unique vector z in H such that

$\langle x ∣ A y \rangle = \langle z ∣ y \rangle\ \forall y \in dom ⁡ A$​

This vector z is defined to be $A^\dagger x$.

It seems that it is not enough that $x \in domA^\dagger$; it seems that it has to be in $dom A$ as well.

 

[dextercioby]

Thank you! So a concrete example is the ground state: $\psi(x) = sin(\frac{\pi x}{a})$. This function is in the domain of $p^\dagger$, but not in the domain of $p$ (since $p \psi(x) \propto cos(\frac{\pi x}{a})$ does not vanish at the walls.).

That helps a lot, except that it makes me wonder if this snippet from the Wikipedia article on self-adjoint operators is wrong:
It seems that it is not enough that $x \in domA^\dagger$; it seems that it has to be in $dom A$ as well.

It is not wrong. An operator in which the domain is contained in the domain of its adjoint is called symmetric. But to require that $x\in D_A$ is not needed to define the adjoint.

 

[stevendaryl]

It is not wrong. An operator in which the domain is contained in the domain of its adjoint is called symmetric. But to require that $x\in D_A$ is not needed to define the adjoint.

I'm still not understanding this. You have a Hilbert space $H$: the square-integrable functions that vanish on $x=0$ and $x=a$. Let $H^*$ be the set of linear functionals on $H$. For every $\psi$ in $H$, we can associate a corresponding functional $F_\psi$ in $H^*$: $F_\psi(\phi) = \langle \psi | p \phi \rangle$. But only for certain values of $\psi$ will it be the case that there is a ket $|p^\dagger \psi\rangle$ in $H$ satisfying $\langle p^\dagger \psi | \phi \rangle = \langle \psi | p \phi \rangle$

Once again, consider the example $\psi(x) = sin(\frac{\pi x}{a})$. Then what is $p^\dagger \psi$? It can't be $-i \partial_x \psi(x) \propto cos(\frac{\pi x}{a})$, because that is not an element of $H$. The Wikipedia definition says that $p^\dagger \psi$ is that $z$ in $H$ such that $\langle z | \phi \rangle = \langle \psi | p \phi \rangle$.

 

[dextercioby]

No, actually, you're mistaking the Hilbert space for the subset of it which is the (maximal) domain of the momentum operator. More precisely:

Once again, consider the example $\psi(x) = sin(\frac{\pi x}{a})$. Then what is $p^\dagger \psi$? It can't yes, it can be $-i \partial_x \psi(x) \propto cos(\frac{\pi x}{a})$, because that is not an element of $H$ but your H is actually the domain of p, not the whole Hilbert space

 

[DrDu]

I'm still not understanding this. You have a Hilbert space $H$: the square-integrable functions that vanish on $x=0$ and $x=a$.

The boundary conditions don't belong to the definition of the hilbert space.

 

[stevendaryl]

No, actually, you're mistaking the Hilbert space for the subset of it which is the (maximal) domain of the momentum operator. More precisely:

I'm not mistaking those two. I'm just not sure which one is relevant. Let me just define a whole bunch of spaces:

• $H$ is the set of square-integrable functions that vanish when $x \leq 0$ or $x \geq a$.
• $H_2$ is the set of all functions $\psi$ such that $p \psi$ is an element of $H$.
• $H_3$ is the intersection of the two.
  
• $H^*$ is the space of "kets"; linear functionals on $H$.
• $H_4$ is the set of all functions $\psi$ such that $\langle \psi|$ is in $H^*$ (note: $H \subset H_4$, because there is no constraint that $\psi$ vanish at the walls.)
• $H_5$ is the intersection of $H_4$ and $H$.

My understanding was that

• $dom\ p = H_3$, those elements $\psi$ of $H$ such that $-i \partial_x \psi$ is also an element of $H$.
  
• $dom\ p^\dagger = H_5$, those elements $\psi$ of $H$ such that $-i \partial_x \psi$ is an element of $H_4$

So $dom\ p \subset dom\ p^\dagger$. But if $\psi$ is in $dom \ p^\dagger$ but not in $dom\ p$, then that means $-i \partial_x \psi$ is not an element of $H$ (because it doesn't have to vanish at the walls).

 

[vanhees71]

The boundary conditions don't belong to the definition of the hilbert space.

Well, in this case the boundary conditions are part of the definition of the Hilbert space. That's why $-\mathrm{i} \mathrm{d}_x$ is not self-adjoint in this particular Hilbert space $\mathrm{L}^2([0,a])$ with these "rigid boundary conditions", but it indeed is for the Hilbert space with "periodic boundary conditions" or for $\mathrm{L}^2(\mathbb{R})$.

 

[stevendaryl]

The boundary conditions don't belong to the definition of the hilbert space.

Ah! Well, that explains it. But why not? I thought a Hilbert space was an arbitrary set equipped with an inner product such that blah, blah, blah...

 

[dextercioby]

Ah! Well, that explains it. But why not? I thought a Hilbert space was an arbitrary set equipped with an inner product such that blah, blah, blah...

No, the Hilbert space is L^2 ([0,a], dx), that is the set of all (equivalence classes of) functions with finite norm, that is the Hilbert space is:

$H = \{\psi (x) | \int_{0}^{a} |\psi^2 (x)| dx <\infty \}$

The necessity of boundary conditions on the wavefunction stems from the fact that the Hamiltonian and the momentum are (in coordinates representation) derivative operators, hence their spectral equations are (Sturm-Liouville) differential equations whose unique solutions need boundary conditions (Dirichlet or Neumann).

 

[vanhees71]

Ok, then I guess $-\mathrm{i} \mathrm{d}_x$ could be made self-adjoint, because the subspace of wave functions with periodic boundary conditions is a dense subset in this larger Hilbert space. The more I think about it, the more I hate this apparently simple but unphysical example of the infinite-square well. To define this problem one way is to use the finite square well and take the limit. Then you clearly get the Hilbert space of functions with rigid boundary conditions, i.e., $\psi(0)=\psi(a)=0$, in which $-\mathrm{i} \mathrm{d}_x$ is not self-adjoint.