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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

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    $\begingroup$ 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: $$\langle \phi | \psi \rangle = \int_{\mathbb{R}} \phi^{*}(x) \psi(x) \, dx$$ $\endgroup$
    – Heath
    Nov 12, 2020 at 1:05
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    $\begingroup$ @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

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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.

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

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

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