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

[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 [/color]

 

[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.

 

[stevendaryl]

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).

But what in the definition of a "Hilbert space" prevents you from making a Hilbert space that consists of only the functions satisfying the boundary conditions?

 

[dextercioby]

Completion.

 

[vanhees71]

Well, but on your space, i.e., simply $\mathrm{L}^2([0,a])$ without any boundary conditions, $-\mathrm{i} \mathrm{d}_x$ is not even Hermitean, because
$$\int_0^a \mathrm{d} x \psi_1^*(x) (-\mathrm{i}) \psi_2'(x)=\psi_1^*(a) \psi_2(a)-\psi_1^*(0) \psi_2(0) +\int_0^a \mathrm{d} x [(-\mathrm{i}) \psi_1'(x)]^* \psi_2(x).$$
So even for getting the operator Hermitean you need boundary conditions to ensure that the integral-free part vanishes.

 

[stevendaryl]

Completion.

Completion in this sense is something like:

If \psi_1, \psi_2, ... is a converging infinite sequence of elements, then the limit is also an element?

So a counterexample might be if \psi_n(x) is the function that is equal to 1 in the interval 1/n &lt; x &lt; 1 - 1/n, but is equal to 0 outside that interval. Each \psi_n is in the space (of functions that are 0 on x=0 and x=1) but the limit is the constant function 1, which is not in that space.

Okay. Everything is illuminated now.