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

[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 $1/n < x < 1 - 1/n$, 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.