Showing that Position and Momentum Operators are Hermitian

Question

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

Answer 1

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:

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

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:

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

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

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

Assume the wavefunctions go to zero at infinity then:

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

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

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

Thus $\hat{p}$ is Hermitian.

Answer 2

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

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

Answer 3

Ali's answer is highly rated. However, when I look to [1] I see that I must show something different than what Ali shows. My learning style is one where I adhere closely to definitions. As such, I would benefit from an additional proof. Maybe others would benefit too.

Proposition

The linear momentum operator $\widehat{p} : L^2([-\infty,\infty],dx)\to L^2([-\infty,\infty],dx)$, which is the rule that assigns to any $\psi(r)$ in the domain the one vector $$\psi'(r) = \left[\widehat{p}\psi\right](r) = \left[-i\hbar \frac{d}{dx}\psi\right](r)$$ in the codomain is, in fact, a Hermitian operator.

Proof

From [1] I know that $\widehat{p}$ is Hermitian if, and only if, $$ \langle \widehat{p}\psi,\phi\rangle =\langle \psi,\widehat{p}\phi\rangle ,\quad\text{for all}~\psi,\phi\in L^2([-\infty,\infty],dx). \tag{1} $$ From [2] I know that the inner product of functions $\psi\in L^2([-\infty,\infty],dx)$ and $\phi\in L^2([-\infty,\infty],dx)$ is defined in one of two ways. Here, I take it that $$ \langle \psi,\phi\rangle = \int_{\mathbb{R}} \psi(x) \operatorname{conj}\left(\phi\right)(x)\,dx. $$

Thus, the left hand side of Eq. (1) maybe rewritten as \begin{align} \langle \widehat{p}\psi,\phi\rangle &= \int_{\mathbb{R}} \left[\widehat{p}\psi\right](x) \operatorname{conj}\left(\phi\right)(x)\,dx \\ &= \int_{\mathbb{R}} \left[-i\hbar\frac{d}{dx} \psi\right](x) \operatorname{conj}\left(\phi\right)(x)\,dx \\ &= \int_{\mathbb{R}} \underbrace { \operatorname{conj}\left(\phi\right)(x) } _ {v} \underbrace{ \left[-i\hbar\frac{d}{dx} \psi\right](x) ~dx} _{du} \\ &= \left[ \underbrace { \operatorname{conj}\left(\phi\right)(x) } _ {v} \underbrace{ \left[-i\hbar \psi (x)\right] } _{ u} \right]_{-\infty}^\infty - \int_{\mathbb{R}} \underbrace{ \left[-i\hbar \psi (x)\right] } _{ u} \underbrace { \left[ \frac{d}{dx} \operatorname{conj}\left(\phi\right)\right](x) ~dx} _ {dv} \\ &= -i\hbar \left[ \psi (x)\operatorname{conj}\left(\phi\right)(x) \right]_{-\infty}^\infty + \int_{\mathbb{R}} \psi (x) \left[+i\hbar \frac{d}{dx} \operatorname{conj}\left(\phi\right)\right](x) ~dx \\ &= -i\hbar \left[ \psi (x)\operatorname{conj}\left(\phi\right)(x) \right]_{-\infty}^\infty + \int_{\mathbb{R}} \psi (x) \left[ \operatorname{conj}\left(-i\hbar \frac{d}{dx}\phi\right)\right](x) ~dx \\ &= -i\hbar \left[ \psi (x)\operatorname{conj}\left(\phi\right)(x) \right]_{-\infty}^\infty + \int_{\mathbb{R}} \psi (x) \left[\operatorname{conj}\left( \widehat{p}\phi\right)\right](x) ~dx \\ &= -i\hbar \left[ \psi (x)\operatorname{conj}\left(\phi\right)(x) \right]_{-\infty}^\infty + \langle \psi,\widehat{p}\phi\rangle . \end{align} Since $\phi,\psi\in L^2([-\infty,\infty],dx)$, therefore $\left[ \psi (x)\operatorname{conj}\left(\phi\right)(x) \right]_{-\infty}^\infty =0$. Thus, by the transitive law, I find that $$\langle \widehat{p}\psi,\phi\rangle =\langle \psi,\widehat{p}\phi\rangle . $$ This being so, $\widehat{p}$ is Hermitian.

Q.E.D.

Bibliography

[1] https://en.wikipedia.org/wiki/Self-adjoint_operator.

[2] https://en.wikipedia.org/wiki/Hilbert_space