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

[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?

 

[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.