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We start with the familiar Schrodinger equation: $$ i\hbar \frac{\partial \left|\psi_S\right\rangle}{\partial t} = \hat{H}_S \left|\psi_S\right\rangle $$

As we switch to a different picture than Schrodinger picture with a unitary transformation $\hat U$:

$$ \left|\psi_S\right\rangle = \hat{U}\left|\psi_P\right\rangle $$($S$ indicating Schrodinger picture and $P$ indicating the arbitrary picture) If we plug in $\left|\psi_S\right\rangle = \hat{U}\left|\psi_P\right\rangle$ into the Schrodinger equation, we obtain:

$$ i\hbar \frac{\partial \left|\psi_P\right\rangle}{\partial t} = \hat{H}_P \left|\psi_P\right\rangle $$ where $$ \hat{H}_P = U^\dagger \hat{H_S} U - i\hbar U^\dagger \frac{\partial U}{\partial t} $$ is the Hamiltonian in this arbitrary picture.

So, the question is - if Hamiltonian is an observable, shouldn't it have the same expectation values in both pictures - yet the second term in $\hat H_P$ makes them unequal. Because:

$$ \left\langle\psi_P\right|\hat{H}_P\left|\psi_P\right\rangle = \left\langle\psi_P\right| U^\dagger \hat{H_S} U \left|\psi_P\right\rangle - i\hbar \left\langle\psi_P\right| U^\dagger \frac{\partial U}{\partial t}\left|\psi_P\right\rangle $$ this simplifies to: $$ \left\langle\psi_P\right|\hat{H}_P\left|\psi_P\right\rangle = \left\langle\psi_S\right| \hat{H_S} \left|\psi_S\right\rangle - i\hbar \left\langle\psi_P\right| U^\dagger \frac{\partial U}{\partial t}\left|\psi_P\right\rangle $$ telling us that the expectation values in two different pictures are not the same. I do not see a reason why that the last term should be zero. What is wrong here? Is Hamiltonian somewhat different from other observables?

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  • $\begingroup$ The definition of $H_p$ in op's post is not physically motivated for me. Take Heisenberg picture for instance, which the state vector is stationary. Thus $$ i \partial | \psi_p \rangle / \partial t = 0 $$, thus $H_p =0$.... We know the motivated (or common) definition is $$H_p := U^{\dagger} H_s U$$ $\endgroup$ – user26143 Apr 24 '14 at 7:37
  • $\begingroup$ Here is an explicit example of this kind of transformation, the questions therein(question 3 and 1) is very similar to this post, hope it can help someone and me to understand this kind of situation better. $\endgroup$ – an offer can't refuse Apr 24 '14 at 11:41
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It is only in the Schroedinger picture (and those connected to it by time-independent $U$) that the Hamiltonian can be read off from the dynamical equation for $\vert\psi\rangle$. In all other pictures you would find there a kind of "effective Hamiltonian" which is different from the actual Hamiltonian in this picture. So $$\hat{H}_P = U^\dagger \hat{H}_S U$$ is the Hamiltonian in P and $$\hat{H}_{P,eff} = U^\dagger \hat{H}_S U - i \hbar U^\dagger \partial_t U$$ is the one you read off from $$i\hbar \partial_t\vert\psi_P\rangle = \hat{H}_{P,eff}\vert\psi_P\rangle.$$

Now it becomes clear that $\langle\psi_P\vert\hat{H}_P\vert\psi_P\rangle=\langle\psi_S\vert\hat{H}_S\vert\psi_S\rangle\neq\langle\psi_P\vert\hat{H}_{P,eff}\vert\psi_P\rangle$ in general.

For instance, in the Heisenberg picture ($P\to H$) we have $$\hat{H}_H = \hat{H}_S$$ but $$\hat{H}_{H,eff} = 0.$$

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