# Showing that Position and Momentum Operators are Hermitian

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

• The position operator does not map $L^2$ to itself.
– fqq
Nov 11, 2020 at 22:50
• Is $\hat X |\psi \rangle \neq x |\psi \rangle$ in $L^2(R)$? Nov 11, 2020 at 23:30
• 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. Nov 11, 2020 at 23:50
• Small nitpick: the first $\psi$ in your inner product definition needs to be conjugated. The inner product for this space is: $$\langle \phi | \psi \rangle = \int_{\mathbb{R}} \phi^{*}(x) \psi(x) \, dx$$ Nov 12, 2020 at 1:05
• @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. Nov 12, 2020 at 16:23

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.

• Hi Ali, please use MathJax to typeset your equations. If you're not familiar with it, you can find a tutorial here. Nov 12, 2020 at 3:48
• Cheers. Will take a look.
– Ali
Nov 12, 2020 at 3:50
• 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! Nov 12, 2020 at 16:25
(1) Just commute $$x$$. You’re in the privileged basis for it.
(2) Integrate by parts, raising $$\psi$$ and lowering $$\phi.$$