5
$\begingroup$

I'd like to show that the position operator $ X = x$ and momentum operator $ P = \frac \hbar i \frac \partial {\partial x}$ are Hermitian/Self Adjoint when acting in the Hilbert Space $H = L^2(R)$. I would like to show this in the general case $\langle \phi |X \psi \rangle = \langle X\phi | \psi \rangle$ where $\phi, \psi \in H$, and the same for $\hat P$.

I know this can be demonstrated easily in the specific case $\langle \psi | X\psi \rangle = \langle X \psi | \psi \rangle$ using: $$\langle \psi | \psi \rangle = \int_\infty^\infty \psi^*(x) \psi(x) \ dx = \int_\infty^\infty |\psi(x)|^2 dx$$ But I am not sure how to expand this to the general case for $\langle \phi| \psi \rangle$? I'd appreciate any help to get me on the right track

$\endgroup$
5
  • 1
    $\begingroup$ The position operator does not map $L^2$ to itself. $\endgroup$
    – fqq
    Nov 11, 2020 at 22:50
  • $\begingroup$ Is $\hat X |\psi \rangle \neq x |\psi \rangle$ in $L^2(R)$? $\endgroup$
    – Allod
    Nov 11, 2020 at 23:30
  • $\begingroup$ I am confused about the demonstration you have in mind for the case $\langle\psi|X\psi\rangle$ because it doesn't use in any way that the two $\psi$'s are the same. $\endgroup$
    – MannyC
    Nov 11, 2020 at 23:50
  • $\begingroup$ Small nitpick: the first $\psi$ in your inner product definition needs to be conjugated. The inner product for this space is: $$\langle \phi | \psi \rangle = \int_{\mathbb{R}} \phi^{*}(x) \psi(x) \, dx$$ $\endgroup$
    – Andrew
    Nov 12, 2020 at 1:05
  • 1
    $\begingroup$ @AndreasMastronikolis sorry I formatted that poorly, the * was supposed to signify the conjugate of $\psi$ but I realize now it looked like I was just multiplying them. $\endgroup$
    – Allod
    Nov 12, 2020 at 16:23

2 Answers 2

9
$\begingroup$

For all the following integrals, the limits are from $-\infty$ to $\infty$.

Assume we are working in the position representation.

For $\hat{x}$ to be Hermitian we must show that:

$$\langle{\phi|\hat{x}\psi}\rangle=\langle{\psi|\hat{x}\phi}\rangle^*$$

LHS:

$$\langle{\phi|\hat{x}\psi}\rangle=\int{\phi^*(x\psi)dx}$$

RHS:

$$\langle{\psi|\hat{x}\phi}\rangle^*=(\int{\psi^*(x\phi)dx})^*$$

Eigenvalues of $\hat{x}$ are real, $x=x^*$:

$$\langle{\psi|\hat{x}\phi}\rangle^*=\int{\psi(x\phi^*)dx}$$

$$=\int{\phi^*(x\psi)dx}$$

$$\therefore\langle{\phi|\hat{x}\psi}\rangle=\langle{\psi|\hat{x}\phi}\rangle^*$$

Thus, $\hat{x}$ is Hermitian.

For $\hat{p}$ to be hermitian we must show the following:

$$\langle{\phi|\hat{p}\psi}\rangle=\langle{\psi|\hat{p}\phi}\rangle^*$$

LHS:

$$\langle{\phi|\hat{p}\psi}\rangle=\int{\phi^*(-i\hbar \frac{\partial\psi}{\partial x})dx}$$

RHS: $$\langle{\psi|\hat{p}\phi}\rangle^*=(\int{\psi^*(-i\hbar \frac{\partial\phi}{\partial x})dx})^*$$ $$=\int{\psi(i\hbar \frac{\partial\phi^*}{\partial x})dx}$$ $$=i\hbar \int{\psi(\frac{\partial\phi^*}{\partial x})dx}$$

Using integration by parts gives: $$\langle{\psi|\hat{p}\phi}\rangle^*=[\phi^*\psi]_{-\infty}^{\infty}-i\hbar \int{\phi^*(\frac{\partial\psi}{\partial x})dx}$$

Assume the wavefunctions go to zero at infinity then:

$$\langle{\psi|\hat{p}\phi}\rangle^*= -i\hbar \int{\phi^*(\frac{\partial\psi}{\partial x})dx}$$

$$= \int{\phi^*(-i\hbar\frac{\partial\psi}{\partial x})dx}$$

$$\therefore \langle{\phi|\hat{p}\psi}\rangle=\langle{\psi|\hat{p}\phi}\rangle^*$$

Thus $\hat{p}$ is Hermitian.

$\endgroup$
5
  • $\begingroup$ Hi Ali, please use MathJax to typeset your equations. If you're not familiar with it, you can find a tutorial here. $\endgroup$
    – J. Murray
    Nov 12, 2020 at 3:48
  • $\begingroup$ Cheers. Will take a look. $\endgroup$
    – Ali
    Nov 12, 2020 at 3:50
  • $\begingroup$ Updated the answer! $\endgroup$
    – Ali
    Nov 12, 2020 at 4:32
  • $\begingroup$ Ah thank you, I see what was holding me back now. I didn't make the connection that $x = x^*$ and $[\phi^*\psi]_{-\infty}^{\infty} = 0$. Thank you for the help! $\endgroup$
    – Allod
    Nov 12, 2020 at 16:25
  • $\begingroup$ I spent a while puzzling over this one, I agree completely with your logic, but I want to avoid making the statment that x must be real. My plan was to say "let's show $\hat{x}$ is hermitian, then we'll know that x is real". Is it neccesary to assume x is real? $\endgroup$
    – Jack
    Sep 15, 2023 at 9:34
2
$\begingroup$

(1) Just commute $x$. You’re in the privileged basis for it.

(2) Integrate by parts, raising $\psi$ and lowering $\phi.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.