# Is the eigenvalue of Hamiltonian invariant under linear transformation of momentum operator?

It is given

The dynamics of a particle moving one-dimensionally in a potential V(x) is governed by the Hamiltonian $H_0 = p^2 /2m + V(x)$, where $p = -i\hbar d/dx$ is the momentum operator. Let $E_{n} ^{(0)}$ , $n = 1,2,3...$ be the eigenvalues of $H_0$. Now consider a new Hamiltonian $H = H_0 + \lambda p/m$ , where $\lambda$ is a given parameter. Given $\lambda, m$ and $E_{n}^{(0)}$, find the eigenvalues of H.

Now, the solution was conveniently given where the $H$ was transformed into a form $p' = p+ \lambda$ such that,

$$H = (p+\lambda)^2/2m + V(x) - \lambda^2/2m$$

Where another Hamiltonian $H'$ was defined such that $$H' = p'^2/2m + V(x)$$

But, it says the eigenvalues of $H'$ are the same with those of $H_0$, i.e., $E_n^{(0)}, n=1,2,3...$

Thus, $H + \lambda^2/2m= H'$ and $$(H + \lambda^2/2m)\psi= H'\psi$$

$$H\psi + (\lambda^2/2m) \psi = E_n^{(0)}\psi$$

$$H\psi = [E_n^{(0)}-\lambda^2/2m]\psi$$

Therefore, the eigenvalues of $H$ are,

$$E_n^{(0)}-\lambda^2/2m, n = 1,2,3...$$

But why is it that the eigenvalues remain invariant under such transformation of momentum operator? I've searched Translation operator, and learned that Hamiltonian and Translation commute, but translation operator linearly changes $x$, not $p$.

Also, would it be possible to explain it without the use of Translation operator?

This is because $H'=UHU^{-1}$ for a certain unitary operator $U$, therefore $\psi$ is an eigenvector of $H$ with an eigenvalue if an only if $U\psi$ is eigenvector of $H'$ with the same eigenvalue. Thus the two operators have the same point spectrum. $U = e^{i \lambda X/\hbar}$.
From $[X,P]= i \hbar I$ one finds $$e^{-i \lambda X/\hbar}Pe^{i \lambda X/\hbar}= P + \lambda I\:.$$ On the other hand $$e^{-i \lambda X/\hbar}V(x)e^{i \lambda X/\hbar}=V(x)$$ As a consequence (with $\hbar=1$) $$UHU^{-1}= U \left(\frac{1}{2m}P^2 + V(x)\right) U^{-1}= \frac{1}{2m} UPU^{-1}UP U^{-1}+ UV(x)U^{-1}= \frac{1}{2m} (P+\lambda I)^2 + V(x) = H'\:.$$
• I have carried out the transformation calculation given $U = e^{i\lambda X/\hbar}$ but failed to get the transformed $H'$, could you elaborate? Apr 23, 2016 at 14:16
• It is given $e^{A} B e^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + ...$ where, $B= P$ and $A =i\lambda X/ \hbar$ Therefore, the transformation of $P$ is $P + [i\lambda X/\hbar,P] +\frac{1}{2!}[i\lambda X/\hbar,[i\lambda X/\hbar,P]] ...$ but because $[i\lambda X/\hbar,P] = i \lambda / \bar [x,p] = -\lambda$ thus, further commutation relations will yield zero. Apr 23, 2016 at 15:13
From the equation $H$ is a linear function of $H_0$. In that case, the eigen values of $H_0$ are eigen values of $H$ also. That's why the eigen values remain invariant under such a transformation.