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I understand that we can apply the Schrödinger equation to any wavefunction. Now, my question is, can we apply it to states that are being operated upon? Because, when we apply an operator on a state, we get another state which might/might not be a linear multiple of the original one, but still should satisfy the Schrödinger equation, right?

Basically, can I apply Schrödinger equation on states like $\hat{A}\psi $

Meaning, can I write:

$$i\hbar \frac{\partial (\hat{A}\psi) }{\partial t} = \hat{H}(\hat{A}\psi)$$

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If you start with $\psi$ that solves the Schrödinger equation: $$ i\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi $$ And acts $\hat{A}$ on $\psi$: $$ i\hbar \frac{\partial (\hat{A}\psi) }{\partial t} = \hat{H}(\hat{A}\psi) $$ Then because we work in the Schrödinger picture and if we assume that $\hat{A}$ doesn't explicitly depends on time:

$$ i\hbar \hat{A}\frac{\partial \psi }{\partial t} = \hat{A}\hat{H}\psi + [\hat{H}, \hat{A}] \psi $$

So $\hat{A}\psi$ will solve the Schrödinger equation if $[\hat{H}, \hat{A}]=0$, i.e. if $\hat{A}$ is a symmetry generator.

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    $\begingroup$ Sorry but I downvoted your answer because it is misleading. The Schrödinger equation holds for any state, including those that are some operator acting on another state. In the Schrödinger picture $\hat{A}\psi$ evolve with an evolution operator in front like everybody else, not with an evolution operator between $\hat{A}$ and $\psi$. $\endgroup$ – MannyC Feb 25 at 0:02
  • $\begingroup$ I exactly wrote "If you start with ψ that solves the Schrödinger equation", because question about new $\psi^\prime = \hat{A}\psi$ is obvious and empty:( As for me, it's just renaming of variable in Shrodinger equation $\endgroup$ – Nikita Feb 25 at 0:23
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Yes.

In the Schrödinger equation $\psi$ is arbitrary. Therefore there is nothing that makes $\hat{A}\psi$ special with respect to $\psi$. If you are bothered by this, expand $\hat{A}\psi$ in the complete basis of energy eigenstates and use the linearity of the Schrödinger equation.

The observation raised by @Nikita is incorrect because $$ \frac{\partial (\hat{A}\psi)}{\partial t} \neq \hat{A} \frac{\partial \psi}{\partial t}\,,\tag{1}\label{1} $$ regardless of the picture used. Actually the equality would hold if $\hat{A}$ were a symmetry of the Hamiltonian, thus the vanishing of the commutator of $\hat{A}$ with $H$ is consistent with it. I'm ignoring the situation where $\hat{A}$ depends explicitly on $t$ because that would be trivial.

The truth is that, in the Schrödinger picture, $\hat{A}\psi$ just evolves like a regular state $$ \hat{A}\psi \to e^{-iHt/\hbar}\hat{A}\psi\,, $$ and not like $$ \hat{A}\psi \not\!\to \hat{A}e^{-iHt/\hbar}\psi\,, $$ as \eqref{1} would imply. This follows from the fact that we demand that the norm of the state $\hat{A}\psi$ (or any state for that matter) must be constant with time . But if we evolved in the way above one would have $$ \|\hat{A}\psi\|^2 \to \|e^{iHt/\hbar}\hat{A}e^{-iHt/\hbar}\psi\|^2 \neq \|\hat{A}\psi\|^2\,, $$ again, unless $\hat{A}$ commutes with $H$. Another motivation is that is simply does not make sense to pick out a set of states that evolve à la Schrödinger and then make all the states obtained by acting with some operator on them evolve differently.

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  • $\begingroup$ I am sorry, maybe I am measleding, but why there is not equality in (1) if we concretely use Shrodinger picture?? $\endgroup$ – Nikita Feb 25 at 0:27
  • $\begingroup$ @Nikita because you can have explicitly time dependent operators in the Schrödinger picture as well. Say a magnetic field varying with time. $\endgroup$ – Superfast Jellyfish Feb 25 at 3:39
  • $\begingroup$ @Fellow Not necessarily. Even if $\hat{A}$ is independent of time (1) is true. The Schrödinger picture evolves all states in the same fashion. And $\hat{A}\psi$ is a state. $\endgroup$ – MannyC Feb 25 at 7:13
  • $\begingroup$ @MannyC if $A$ is independent of time then the equality holds. You can see this by expanding $\psi$ in energy eigenstates. $\endgroup$ – Superfast Jellyfish Feb 25 at 7:29
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    $\begingroup$ By symmetry do you mean that $[A,H]=0$? If so, then I think the actual condition for the equality to hold is both the commutation to be zero and there to be no explicit time dependence. Unless I’m missing out on something trivial. $\endgroup$ – Superfast Jellyfish Feb 25 at 10:12

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