Physics Asked by Allod on March 13, 2021
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 $langle phi |X psi rangle = langle Xphi | psi rangle$ where $phi, psi in H$, and the same for $hat P$.
I know this can be demonstrated easily in the specific case $langle psi | Xpsi rangle = langle X psi | psi rangle$ 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
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:
$$langle{phi|hat{x}psi}rangle=int{phi^*(xpsi)dx}$$
RHS:
$$langle{psi|hat{x}phi}rangle^*=(int{psi^*(xphi)dx})^*$$
Eigenvalues of $hat{x}$ are real, $x=x^*$:
$$langle{psi|hat{x}phi}rangle^*=int{psi(xphi^*)dx}$$
$$=int{phi^*(xpsi)dx}$$
$$thereforelangle{phi|hat{x}psi}rangle=langle{psi|hat{x}phi}rangle^*$$
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:
$$langle{phi|hat{p}psi}rangle=int{phi^*(-ihbar frac{partialpsi}{partial x})dx}$$
RHS: $$langle{psi|hat{p}phi}rangle^*=(int{psi^*(-ihbar frac{partialphi}{partial x})dx})^*$$ $$=int{psi(ihbar frac{partialphi^*}{partial x})dx}$$ $$=ihbar int{psi(frac{partialphi^*}{partial x})dx}$$
Using integration by parts gives: $$langle{psi|hat{p}phi}rangle^*=[phi^*psi]_{-infty}^{infty}-ihbar int{phi^*(frac{partialpsi}{partial x})dx}$$
Assume the wavefunctions go to zero at infinity then:
$$langle{psi|hat{p}phi}rangle^*= -ihbar int{phi^*(frac{partialpsi}{partial x})dx}$$
$$= int{phi^*(-ihbarfrac{partialpsi}{partial x})dx}$$
$$therefore langle{phi|hat{p}psi}rangle=langle{psi|hat{p}phi}rangle^*$$
Thus $hat{p}$ is Hermitian.
Correct answer by Ali on March 13, 2021
(1) Just commute $x$. You’re in the privileged basis for it.
(2) Integrate by parts, raising $psi$ and lowering $phi.$
Answered by CR Drost on March 13, 2021
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