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We know that states are time independent in Heisenberg picture. However, if I apply an operator to a state in Heisenberg picture, $$|\beta\rangle= A(t)|\alpha\rangle \equiv \exp(itH) A(0)\exp(−itH)|\alpha\rangle$$ then the state $|\beta\rangle$ will be time dependent. However, states are not time dependent in Heisenberg picture. So is $|\beta\rangle$ no longer in Heisenberg picture?

In schrodinger picture $A(t)$ would be $A$ but $|\alpha\rangle$ now would be $\exp (-itH)|\alpha\rangle$ thus $|\beta\rangle$ would be $A\exp (-itH)|\alpha\rangle$

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    $\begingroup$ You have to be careful about whether $A(t)$ is also in the Heisenberg picture. $\endgroup$ – DanielSank Jan 31 '18 at 21:26
  • $\begingroup$ You hit a time-independent state α with a time-dependent operator A(t) and merely renamed it as β(t) neglecting to acknowledge its time-dependence. How would you write this in the Schroedinger picture? Is your remark in alignment with the mainstream definitions? $\endgroup$ – Cosmas Zachos Jan 31 '18 at 21:46
  • $\begingroup$ @CosmasZachos see the edit. $\endgroup$ – physshyp Jan 31 '18 at 21:52
  • $\begingroup$ @CosmasZachos MY edit provides an equivalent inner product of $\langle\alpha|\beta\rangle$ but my question asks something else. $\endgroup$ – physshyp Jan 31 '18 at 22:00
  • $\begingroup$ You are tying yourself up in logical knots, mostly because you refuse to write expectation values. The dual of $|\beta(t)\rangle$, would be $\langle \alpha| \exp(iHt) A$, for a hermitian A, so the expectation values would match. $\endgroup$ – Cosmas Zachos Jan 31 '18 at 22:14
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The statement "states are time-independent in the Heisenberg picture" can be a bit misleading, for exactly the reason that you point out. States are not measurable in quantum mechanics; only inner products (or equivalently, matrix elements) are. The Schrodinger and Heisenberg pictures are just two ways of mentally grouping the terms in a matrix element with parentheses - they aren't completely well defined when thinking about bras and kets that haven't been contracted all the way down to scalars.

So the answer to your question is largely a matter of semantics. I personally would probably say that in the Heisenberg picture, the time-dependent ket $A(t) | \psi_0 \rangle$ should not be thought of as a "state", but as a piece of an "incomplete" matrix element. But it really depends on your exact definition of the word "state", and I wouldn't argue with someone who would prefer to use different words.

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In Schrodinger picture, let $U$ be the unitary evolution operator that corresponds to suddenly hitting the particle; explicitly it would be $$U \sim e^{i F t_0 \hat{x}}$$ which applies a large force $F$ for a very short time $t_0$. Given a state $|\psi \rangle$, $U |\psi\rangle$ is just what that state would be right after being hit.

In Heisenberg picture, the physical meaning of $U(t')$ is hitting a particle at time $t'$. That is, if $U(t') |\psi\rangle = |\phi\rangle$ in Heisenberg picture, then in Schrodinger picture, $$|\phi(t) \rangle = |\psi(t) \rangle \text{ after } |\psi\rangle \text{ is hit at time } t', \quad |\psi\rangle = |\psi(0) \rangle, \quad |\phi\rangle = |\phi(0) \rangle.$$ So $|\phi \rangle$ does depend on the parameter $t'$, which indicates the time when the hit occurred. But $t'$ is just a parameter, which I could have called $Q$ or $\mathfrak{X}$ or $\mathcal{R}$. The state $|\phi \rangle$ still doesn't change in the time $t$, since it's in Heisenberg picture. (You could have instead considered $U(t) |\psi\rangle$, but this is an unnatural quantity that is not in Heisenberg picture.)

To give an everyday analogy, how much money do you have to put in your bank account in 2018 so that you'll be as rich as me in the long run, provided that I start with some amount of money, and plan to deposit more in $t'$ years? The answer clearly depends on $t'$ because that changes how much interest I get. But "the amount of money you need to deposit in 2018" clearly can't depend on time.

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  • $\begingroup$ I don't agree with your answer because, in that case, the expectation values in Heisenberg picture would be expectation values evaluated in some specific time and fixed. But it is not true, expectation values are rotating in time, in every picture. $\endgroup$ – physshyp Jan 31 '18 at 22:07
  • $\begingroup$ @physshyp Can you specify which expectation values you're talking about? $\endgroup$ – knzhou Jan 31 '18 at 22:08
  • $\begingroup$ $\langle A\rangle_\alpha$ is time-dependent and rotates in time and should be same in both pictures. But if we think this as your answer, this expectation value would be fixed in time and evaluated in some $t'$. $\endgroup$ – physshyp Jan 31 '18 at 22:12
  • $\begingroup$ @physshyp That's a different situation. $\langle A(t) \rangle$ of course depends on $t$. What my answer is saying is that you don't want to consider $A(t) | \alpha \rangle$, because that's an unnatural quantity and not in Heisenberg picture; the thing you actually want 99% of the time is $A(t') |\alpha \rangle$ where $t'$ is a parameter; that quantity is both in Heisenberg picture, and has a reasonable physical interpretation. $\endgroup$ – knzhou Jan 31 '18 at 22:15
  • $\begingroup$ writing $\langle A(t)\rangle$ is not very correct because whether $A$ is time dependent or not the expectation value of it is time-dependent. since in other case states will be time dependent. Secondly, I see your point on $A(t)|\alpha\rangle$ but this does not answer my question and if I apply your method to a particle creation operators, I would get strange results. $\endgroup$ – physshyp Jan 31 '18 at 22:27

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