Box normalization for the free particle-Position operator

[Joker93]

Hello,
When we normalize the free particle by putting it in a box with periodic boundary conditions, we avoid the "pathological" nature of the momentum representation that take place in the normal problem of a particle in a box with the usual boundary conditions of Ψ=0 at the two borders. Thus, the momentum operator is self-adjoint; something that does not hold for the normal problem of the infinite well.
My question is, what happens to the position representation in the case of the box normalization with periodic boundary conditions? Is the position operator self-adjoint and/or hermitian? How do we prove these things?
Thanks in advance!

Note: For information on some things I mentioned, you can go to https://www.physicsforums.com/threads/particle-in-a-box-in-momentum-basis.694158/page-2

 

[vanhees71]

Well, with periodic boundary conditions you have trouble with the position operator. Take as Hilbert space $\mathcal{H}$ the functions that are square integrable at the interval $[0,L]$ with periodic boundary conditions $\psi(0)=\psi(L)$. Now the position operator acts as $\hat{x} \psi(x)=x \psi(x)$, and $x \psi(x)|_{x=0}=0$ while $L \psi(L) \neq 0$ in general. Thus again, your position operator is only Hermitean but not self-adjoint in this case. It's obviously Hermitean, because
$$\langle \psi_1|\hat{x} \psi_2 \rangle=\int_0^L \mathrm{d} x \psi_1^*(x) x \psi_2(x) = \int_0^L \mathrm{d} x (x \psi_1)^* \psi_2(x)=\langle \hat{x} \psi_1|\psi_2 \rangle.$$
But since with $|\psi_1 \rangle \in \mathcal{H}$ the image under the position operator $\hat{x} |\psi_1 \rangle$ doesn't need to be in $\mathcal{H}$, it's not self-adjoint.

The momentum operator $\hat{p}=-\mathrm{i} \partial_x$, however, is self-adjoint. You can easily solve the eigenvalue problem. You get again
$$u_p(x)=N \exp(\mathrm{i} p x).$$
The periodic boundary conditions tell you that
$$p \in \frac{2 \pi}{L} \mathbb{Z},$$
and the eigenfunctions are a complete set of orthonormal states since
$$\langle u_p|u_{p'} \rangle=N^* N' \int_0^L \mathrm{d} x \exp[\mathrm{i}(p'-p) x].$$
Now for $p=p'$ you get
$$\langle u_p|u_{p'} \rangle = |N|^2 \int_0^L \mathrm{d} x=|N|^2 L \stackrel{!}{=} 1 \; \Rightarrow \; N=\sqrt{L^{-1}}.$$
For $p \neq p'$ you get
$$\langle u_p|u_{p'} \rangle=0$$
due to the periodic boundary conditions since for $p,p' \in 2 \pi/L \mathbb{Z}$ also $p-p' \in \pi/L \mathbb{Z}$.

The transformation from "position representation" to "momentum representation" and back is just the usual theory of Fourier series for periodic functions:
$$\tilde{\psi}(p) = \langle u_p|\psi \rangle=\int_0^{L} \frac{\mathrm{d} x}{\sqrt{L}} \exp(-\mathrm{i} p x) \psi(x)$$
and
$$\psi(x) = \sum_{p} \langle x|u_p \rangle \langle u_p|\psi \rangle=\sum_p \frac{1}{\sqrt{L}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$

 

[Joker93]

Well, with periodic boundary conditions you have trouble with the position operator. Take as Hilbert space $\mathcal{H}$ the functions that are square integrable at the interval $[0,L]$ with periodic boundary conditions $\psi(0)=\psi(L)$. Now the position operator acts as $\hat{x} \psi(x)=x \psi(x)$, and $x \psi(x)|_{x=0}=0$ while $L \psi(L) \neq 0$ in general. Thus again, your position operator is only Hermitean but not self-adjoint in this case. It's obviously Hermitean, because
$$\langle \psi_1|\hat{x} \psi_2 \rangle=\int_0^L \mathrm{d} x \psi_1^*(x) x \psi_2(x) = \int_0^L \mathrm{d} x (x \psi_1)^* \psi_2(x)=\langle \hat{x} \psi_1|\psi_2 \rangle.$$
But since with $|\psi_1 \rangle \in \mathcal{H}$ the image under the position operator $\hat{x} |\psi_1 \rangle$ doesn't need to be in $\mathcal{H}$, it's not self-adjoint.

The momentum operator $\hat{p}=-\mathrm{i} \partial_x$, however, is self-adjoint. You can easily solve the eigenvalue problem. You get again
$$u_p(x)=N \exp(\mathrm{i} p x).$$
The periodic boundary conditions tell you that
$$p \in \frac{2 \pi}{L} \mathbb{Z},$$
and the eigenfunctions are a complete set of orthonormal states since
$$\langle u_p|u_{p'} \rangle=N^* N' \int_0^L \mathrm{d} x \exp[\mathrm{i}(p'-p) x].$$
Now for $p=p'$ you get
$$\langle u_p|u_{p'} \rangle = |N|^2 \int_0^L \mathrm{d} x=|N|^2 L \stackrel{!}{=} 1 \; \Rightarrow \; N=\sqrt{L^{-1}}.$$
For $p \neq p'$ you get
$$\langle u_p|u_{p'} \rangle=0$$
due to the periodic boundary conditions since for $p,p' \in 2 \pi/L \mathbb{Z}$ also $p-p' \in \pi/L \mathbb{Z}$.

The transformation from "position representation" to "momentum representation" and back is just the usual theory of Fourier series for periodic functions:
$$\tilde{\psi}(p) = \langle u_p|\psi \rangle=\int_0^{L} \frac{\mathrm{d} x}{\sqrt{L}} \exp(-\mathrm{i} p x) \psi(x)$$
and
$$\psi(x) = \sum_{p} \langle x|u_p \rangle \langle u_p|\psi \rangle=\sum_p \frac{1}{\sqrt{L}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$

Thanks for the detailed answer.
So, why use the box normalization to treat the non self-adjointness of the momentum operator if in this case it's the position operator which becomes non self-adjoint?

 

[vanhees71]

The preriodic box is usually used for cases, where you are not interested in the position observable. It's the standard way to make sense of some trouble with the continuous momentum spectrum in relativistic QFT. It's, e.g., one way to get rid of the trouble squaring the S-matrix elements in scattering theory, because in the infinite-volume limit you have $S_{fi} \propto \delta^{(4)}(p_{\text{final}}-p_{\text{initial}})$
if you use plane-wave (momentum eigenstates) for the asymptotic free states (external legs of Feynman diagrams) and it's not a priori clear what to make out of this. The reason for this trouble is clear: momentum eigenstates in the infinite-volume limit are not true eigenstates but generalized ones, and a true asymptotic free state is a free wave packet rather than a plane wave. You can indeed tame the difficulty also by using wave packets in the initial state, but it's a bit of a cumbersome calculation to get the limit for states sharply peaked in momentum space to describe the scattering processes.

Instead you can use a large finite volume (like a cube) with the periodic boundary conditions for the fields to have a well defined momentum operator. Then the energy-momentum conserving $\delta$ distribution becomes a harmless Kronecker $\delta$, which can be squared with no problems. Then you take the infinite volume (and time) limit for the corresponding transition-probability rates, leading to the same result for the cross sections as with the wave-packet approach. While the latter treatment is more physical the finite-volume regularization ("box quantization") is much more easy mathematically.

 

[Joker93]

The preriodic box is usually used for cases, where you are not interested in the position observable. It's the standard way to make sense of some trouble with the continuous momentum spectrum in relativistic QFT. It's, e.g., one way to get rid of the trouble squaring the S-matrix elements in scattering theory, because in the infinite-volume limit you have $S_{fi} \propto \delta^{(4)}(p_{\text{final}}-p_{\text{initial}})$
if you use plane-wave (momentum eigenstates) for the asymptotic free states (external legs of Feynman diagrams) and it's not a priori clear what to make out of this. The reason for this trouble is clear: momentum eigenstates in the infinite-volume limit are not true eigenstates but generalized ones, and a true asymptotic free state is a free wave packet rather than a plane wave. You can indeed tame the difficulty also by using wave packets in the initial state, but it's a bit of a cumbersome calculation to get the limit for states sharply peaked in momentum space to describe the scattering processes.

Instead you can use a large finite volume (like a cube) with the periodic boundary conditions for the fields to have a well defined momentum operator. Then the energy-momentum conserving $\delta$ distribution becomes a harmless Kronecker $\delta$, which can be squared with no problems. Then you take the infinite volume (and time) limit for the corresponding transition-probability rates, leading to the same result for the cross sections as with the wave-packet approach. While the latter treatment is more physical the finite-volume regularization ("box quantization") is much more easy mathematically.

Wow, this is great, thanks!
If I could ask you one last thing:
If the eigenfunctions of an operator don't satisfy the boundary conditions of a given problem, then we can immediately say that the operator is not self-adjoint? Is this condition sufficient to conclude if an operators is self-adjoint or not?
Also, when textbooks write that for a Hermitian operator, its eigenvalues are real, eigenfunctions orthogonal, etc, they really mean that these are valid for self-adjoint operators rather than Hermitian? If so, why an operator being Hermitian is so important? Why don't the textbooks write from the beginning these properties for self-adjoint operators(why do they even consider Hermitian operators if it's the self-adjoint operators that are of use)?

 

[vanhees71]

Indeed, if you cannot find any eigenfunctions fulfilling the boundary conditions, there are none, but if an operator is self-adjoint you know that there must be even a complete orthonormal set of eigenfunctions. So if there are none eigenfunctions at all the operator cannot be self-adjoint.

 

[Joker93]

Indeed, if you cannot find any eigenfunctions fulfilling the boundary conditions, there are none, but if an operator is self-adjoint you know that there must be even a complete orthonormal set of eigenfunctions. So if there are none eigenfunctions at all the operator cannot be self-adjoint.

So, why do textbooks bother with Hermitian operators rather with self-adjoint ones?

 

[Ravi Mohan]

So, why do textbooks bother with Hermitian operators rather with self-adjoint ones?

For finite dimensional Hilbert spaces $\mathcal{H} \simeq \mathbb{C}^n$, the domain of operators is automatically entire $\mathcal{H}$ (you can see this from the matrix representation). Hence self-adjoint and hermitian become synonymous for such cases. Many authors are not careful enough to mention this explicitly. Ballentine does mention it.

 

[Joker93]

Well, with periodic boundary conditions you have trouble with the position operator. Take as Hilbert space $\mathcal{H}$ the functions that are square integrable at the interval $[0,L]$ with periodic boundary conditions $\psi(0)=\psi(L)$. Now the position operator acts as $\hat{x} \psi(x)=x \psi(x)$, and $x \psi(x)|_{x=0}=0$ while $L \psi(L) \neq 0$ in general. Thus again, your position operator is only Hermitean but not self-adjoint in this case. It's obviously Hermitean, because
$$\langle \psi_1|\hat{x} \psi_2 \rangle=\int_0^L \mathrm{d} x \psi_1^*(x) x \psi_2(x) = \int_0^L \mathrm{d} x (x \psi_1)^* \psi_2(x)=\langle \hat{x} \psi_1|\psi_2 \rangle.$$
But since with $|\psi_1 \rangle \in \mathcal{H}$ the image under the position operator $\hat{x} |\psi_1 \rangle$ doesn't need to be in $\mathcal{H}$, it's not self-adjoint.

The momentum operator $\hat{p}=-\mathrm{i} \partial_x$, however, is self-adjoint. You can easily solve the eigenvalue problem. You get again
$$u_p(x)=N \exp(\mathrm{i} p x).$$
The periodic boundary conditions tell you that
$$p \in \frac{2 \pi}{L} \mathbb{Z},$$
and the eigenfunctions are a complete set of orthonormal states since
$$\langle u_p|u_{p'} \rangle=N^* N' \int_0^L \mathrm{d} x \exp[\mathrm{i}(p'-p) x].$$
Now for $p=p'$ you get
$$\langle u_p|u_{p'} \rangle = |N|^2 \int_0^L \mathrm{d} x=|N|^2 L \stackrel{!}{=} 1 \; \Rightarrow \; N=\sqrt{L^{-1}}.$$
For $p \neq p'$ you get
$$\langle u_p|u_{p'} \rangle=0$$
due to the periodic boundary conditions since for $p,p' \in 2 \pi/L \mathbb{Z}$ also $p-p' \in \pi/L \mathbb{Z}$.

The transformation from "position representation" to "momentum representation" and back is just the usual theory of Fourier series for periodic functions:
$$\tilde{\psi}(p) = \langle u_p|\psi \rangle=\int_0^{L} \frac{\mathrm{d} x}{\sqrt{L}} \exp(-\mathrm{i} p x) \psi(x)$$
and
$$\psi(x) = \sum_{p} \langle x|u_p \rangle \langle u_p|\psi \rangle=\sum_p \frac{1}{\sqrt{L}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$

Sorry for bothering again, but I just needed a final clarification.
In order to conclude that an operator Q is self-adjoint or not, we must figure out if its eigenfunctions satisfy the boundary conditions of the given problem or we should figure out if the image of $\hat{Q} \psi(x)$ (for $|\psi \rangle \in \mathcal{H}$) satisfies the boundary conditions? I am confused for which one of those conditions is the one that tells us if the operator Q is self-adjoint or not. So, obviously, I haven't understood it yet.
Thanks in advance.

 

[vanhees71]

You need to check both. First of all the operator should be Hermitean, i.e., for all vectors in the domain of the operator you should have
$$\langle \psi_1|\hat{Q} \psi_2 \rangle=\langle \hat{Q} \psi_1|\psi_2 \rangle.$$
Then you find the (generalized) eigenvectors and check whether their images under the operator are also in the Hilbert space. Of course, if this is not the case since for a eigenvector you must have
$$\hat{Q} |\psi \rangle=\lambda |\psi \rangle,$$
i.e., since $|\psi \rangle$ is in the Hilbert space also $\lambda |\psi \rangle$ and thus $\hat{Q} |\psi \rangle$ is in the Hilbert space.

In the case of $-\mathrm{i} \partial_x$ on the Hilbert space of wave functions defined in the interval $[0,L]$ with boundary conditions $\psi(0)=0, \quad \psi(L)=0$ there are no eigenvectors to begin with and thus, although the operator is Hermitean, it's however not self-adjoint since a self-adjoint operator always provides a complete set of (generalized) eigenfunctions.

 

[Joker93]

You need to check both. First of all the operator should be Hermitean, i.e., for all vectors in the domain of the operator you should have
$$\langle \psi_1|\hat{Q} \psi_2 \rangle=\langle \hat{Q} \psi_1|\psi_2 \rangle.$$
Then you find the (generalized) eigenvectors and check whether their images under the operator are also in the Hilbert space. Of course, if this is not the case since for a eigenvector you must have
$$\hat{Q} |\psi \rangle=\lambda |\psi \rangle,$$
i.e., since $|\psi \rangle$ is in the Hilbert space also $\lambda |\psi \rangle$ and thus $\hat{Q} |\psi \rangle$ is in the Hilbert space.

In the case of $-\mathrm{i} \partial_x$ on the Hilbert space of wave functions defined in the interval $[0,L]$ with boundary conditions $\psi(0)=0, \quad \psi(L)=0$ there are no eigenvectors to begin with and thus, although the operator is Hermitean, it's however not self-adjoint since a self-adjoint operator always provides a complete set of (generalized) eigenfunctions.

So, to conclude, given that an operator is Hermitian, if its eigenvectors satisfy the boundary conditions then this is sufficient to conclude that the operator is self-adjoint, right?