I am trying to understand the following statement:

One sometimes summarizes this situation with the slogan: In the Schr¨odinger picture the basis vectors (provided by the observables) are fixed, while the state vector evolves in time. In the Heisenberg picture the state vectors are held fixed, but the basis vectors evolve in time (in the inverse manner). This is an instance of the “active vs. passive” representation of a transformation — in this case time evolution.

I want to mathematically show what is being described above but I am not sure whether I achieve in doing so or not. Also one thing that I find hard to understand, is how in the Heisenberg picture even though the basis kets are time-dependent, the state in this picture, which can be expressed as a linear combination of the basis kets (which are time-dependent) is not?

We have: $A|\phi_n\rangle=n|\phi_n\rangle$

In Schrodinger picture:

$|\psi(t)\rangle =\sum_n c_n(t_0)e^{-\frac{iE_n(t-t_0)}{\hbar}}|\phi_n\rangle$

$|\psi(t)\rangle =\sum_n c_n(t)|\phi_n\rangle$

Here we can see that the state vector is time-dependent while the basis kets are not. So I believe, that I correctly translate mathematically the initial part of the statement.

Than in the Heisenberg picture at some arbitrary time t:

$|\psi(t)\rangle_H=\sum_n c_n(t_0)e^{-\frac{iE_n(t-t_0)}{\hbar}}|\phi_n\rangle$

$|\psi(t)\rangle_H=\sum_n c_n(t_0)|\phi_n(t)\rangle$

But at the same time when we move from the Schroedinger picture to the Heisenberg one we do:

$|\psi(t)\rangle_H=U(t,t_0)^\dagger |\psi(t)\rangle=U^{\dagger} U |\psi(t_0)\rangle$.

So I don't know how to show the 2nd part of the statement.

  • $\begingroup$ You are unequivocally cool with WP, right? One rarely represents Heisenberg operators through ket spectral decomposition... $\endgroup$ Sep 9, 2023 at 19:50
  • $\begingroup$ I am ok with WP in general, but I just wanted to mathematically translate the above statement $\endgroup$
    – imbAF
    Sep 9, 2023 at 19:52
  • $\begingroup$ I'd start by working with the (time-independent) hamiltonian eigenkets, not your operator's eigenkets. $\endgroup$ Sep 9, 2023 at 19:58
  • $\begingroup$ I am assuming that the operator commutes with the Hamiltonian $\endgroup$
    – imbAF
    Sep 9, 2023 at 20:07
  • $\begingroup$ Bit if it did, the Heisenberg operator would coincide with the Schroedinger one, and the heisenberg equations of motion would trivialize. Did you read the WP article cited? $\endgroup$ Sep 9, 2023 at 20:08

1 Answer 1


But at the same time when we move from the Schröedinger picture to the Heisenberg one we do:


So I don't know how to show the 2nd part of the statement.

Careful here. Actually we have $|ϕ(t)⟩_H = U(t,t_0)^†|ϕ(t_0)⟩$. We are dealing with base kets, not with the state ket. Remember that the base kets become time dependent and that the whole state is now time independent.

The correct way to see the state in the Heisenberg picture would be

$|ψ(t)⟩ = ∑_n c_n(t) |ϕ_n⟩ = ∑_n c_n(t) e^{-i \frac{E_n (t−t_0)}{ℏ}} e^{i \frac{E_n (t−t_0)}{ℏ}} |ϕ_n⟩,$

$|ψ(t)⟩ = ∑_n c_n(t) e^{-i \frac{E_n (t−t_0)}{ℏ}} |ϕ_n (t)⟩_H,$

$|ψ(t)⟩ = ∑_n c_n(t) U\, |ϕ_n (t)⟩_H = U \, ∑_n c_n(t)\, |ϕ_n (t)⟩_H$.

Now, we have $|ψ⟩_H = U^† |ψ(t)⟩$. So $|ψ⟩_H = \, ∑_n c_n(t)\, |ϕ_n (t)⟩_H$. But remember that $|ϕ_n (t)⟩_H = e^{\mathbf{+} i \frac{E_n (t−t_0)}{ℏ}} |ϕ_n⟩ $ (with a POSITIVE sign). So there is no temporal dependence in $|ψ⟩_H$.

You might wonder what happened to the temporal dependece? This is actually present in the observables instead that in the state. Now the observable $A$ would be - in the Heisenberg picture - $A_H (t) = U^† A (t_0) U$.

Check that we retrieve $\langle ψ_H | A_H(t) |ψ⟩_H = \langle ψ (t) | U A_H(t) U^† |ψ(t)⟩ = \langle ψ (t) | A |ψ(t)⟩ = a(t)$.

  • $\begingroup$ Thanks for your answer. You are assuming that the Hamiltonian is time independent right? $\endgroup$
    – imbAF
    Sep 9, 2023 at 22:09
  • $\begingroup$ Yeah, but I don't think that much would change with a time dependent H(t). I am a bit rusty, so don't trust me 100%! $\endgroup$ Sep 9, 2023 at 22:20

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