Commutators, translations, and the virial theorem

For $n=1$, this is the canonical relation. Suppose the formula holds at rank $n$. Writing $\hat P^{n+1}=\hat P^n\hat P$ and using the stated rule in the form $[\hat C,\hat A\hat B]=\hat A[\hat C,\hat B]+[\hat C,\hat A]\hat B$ gives

$$[\hat X,\hat P^{n+1}]=\hat P^n[\hat X,\hat P]+[\hat X,\hat P^n]\hat P=i\hbar\hat P^n+i\hbar n\hat P^{n-1}\hat P=i\hbar(n+1)\hat P^n,$$

which establishes the formula at rank $n+1$. Since the commutator is linear, it follows for a polynomial $f(p)=\sum_na_np^n$ that

$$[\hat X,f(\hat P)]=\sum_na_n[\hat X,\hat P^n]=i\hbar\sum_nna_n\hat P^{n-1}=i\hbar f'(\hat P).$$

The commutator with $\hat X$ therefore acts as differentiation with respect to $\hat P$. The same argument, starting from $[\hat P,\hat X]=-i\hbar\mathbf{1}$, gives $[\hat P,g(\hat X)]=-i\hbar\,g'(\hat X)$, a result that can also be verified directly in the position representation, where $\hat P=-i\hbar\,d/dx$. These formulae extend to functions admitting a power-series expansion.

From the previous question, with $f(p)=e^{-iap/\hbar}$, whose derivative is $f'(p)=-\frac{ia}{\hbar}f(p)$,

$$[\hat X,T(a)]=i\hbar\Bigl(-\frac{ia}{\hbar}\Bigr)T(a)=a\,T(a).$$

Thus $\hat XT(a)=T(a)\hat X+aT(a)$. The operator $T(a)$ is unitary, since $T(a)^\dagger=e^{ia\hat P/\hbar}=T(-a)=T(a)^{-1}$, as $\hat P$ is self-adjoint. Multiplying on the left by $T(a)^\dagger$ gives

$$T(a)^\dagger\hat X\,T(a)=\hat X+a\mathbf{1}.$$

For the translated state $\ket{\psi_a}=T(a)\ket\psi$, the expectation value of position is therefore

$$\bra{\psi_a}\hat X\ket{\psi_a}=\bra\psi T(a)^\dagger\hat XT(a)\ket\psi=\langle\hat X\rangle_\psi+a .$$

The operator $T(a)$ does indeed translate the state a distance $a$ to the right, consistently with its action $(T(a)\psi)(x)=\psi(x-a)$ in the position representation.

Using the stated rule,

$$[\hat H,\hat X\hat P]=[\hat H,\hat X]\,\hat P+\hat X\,[\hat H,\hat P].$$

On the one hand, only the kinetic energy contributes to the first commutator and, from the first question, $[\hat P^2,\hat X]=-[\hat X,\hat P^2]=-2i\hbar\hat P$, so $[\hat H,\hat X]=-\frac{i\hbar}{m}\hat P$. On the other hand, only the potential contributes to the second, and $[\hat H,\hat P]=[V(\hat X),\hat P]=i\hbar V'(\hat X)$. Thus

$$[\hat H,\hat X\hat P]=-\frac{i\hbar}{m}\hat P^2+i\hbar\,\hat XV'(\hat X).$$

The operator $\hat X\hat P$ has no explicit time dependence, and Ehrenfest's theorem gives

$$\frac{d}{dt}\langle\hat X\hat P\rangle=\frac{i}{\hbar}\langle[\hat H,\hat X\hat P]\rangle=\frac{\langle\hat P^2\rangle}{m}-\langle\hat XV'(\hat X)\rangle .$$

Note that $\hat X\hat P$ is not self-adjoint, but Ehrenfest's theorem, which relies only on the Schrödinger equation, still applies to its expectation value.

In a stationary state, all expectation values of time-independent observables are constant, since the state changes only by a global phase. The left-hand side of the preceding relation therefore vanishes, and $\langle\hat P^2\rangle/m=\langle\hat XV'(\hat X)\rangle$, that is, $2\langle\hat T\rangle=\langle\hat XV'(\hat X)\rangle$.

For the harmonic oscillator, $xV'(x)=m\omega^2x^2=2V(x)$, and the theorem gives $\langle\hat T\rangle=\langle V\rangle$: in every eigenstate, the energy is divided equally between kinetic and potential energy, each equal to $E_n/2$. For the potential $\lambda|x|$, we have $xV'(x)=\lambda x\,\mathrm{sgn}(x)=\lambda|x|=V(x)$, hence $2\langle\hat T\rangle=\langle V\rangle$: the expectation value of the potential energy is twice that of the kinetic energy, and therefore $\langle V\rangle=\frac23E$ and $\langle\hat T\rangle=\frac13E$. More generally, for $V(x)\propto|x|^s$, we find $2\langle\hat T\rangle=s\langle V\rangle$. The case of the three-dimensional Coulomb potential, $s=-1$, gives $2\langle\hat T\rangle=-\langle V\rangle$, a result that can be verified for the ground state of the hydrogen atom.