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Unitary equivalence of QM (not for energy?)

  1. Sep 3, 2012 #1
    Since all the observables in QM is on the form

    [tex]\langle \alpha |A| \beta \rangle[/tex]

    where A is an observable, one can transform the observables and states like

    [tex] A \to A' = UAU^{-1} \ \ \ |\beta \rangle \to |\beta '\rangle = U |\beta \rangle[/tex]
    where U is a unitary transformatioin. These descriptions of the theory is equivalent because

    [tex]\langle \alpha' |A'| \beta' \rangle = \langle \alpha |U^{-1} U A U U^{-1}| \beta \rangle = \langle \alpha |A| \beta \rangle.[/tex]

    However by using the Schrödinger equation one can show that the Hamiltonian transforms like

    [tex] H = H' = UHU^{-1} + i\hbar \frac{dU}{dt} U^{-1}[/tex]

    which means that the expectation value of H in the transformed representation is

    [tex] \langle \psi'| H'|\psi '\rangle = \langle \psi| H \psi \rangle + i\hbar \langle \psi |U^{-1}\frac{dU}{dt} |\psi \rangle \neq \langle \psi| H \psi \rangle.[/tex]

    What is the meaning of this inequivalence? Is not the expectation value of H supposed to be equal in two descriptions which differ by a unitary transformation?
  2. jcsd
  3. Sep 3, 2012 #2
    I think the assumption is usually that U≠U(t). This seems intuitively reasonable since time-dependent rotations (a subset of U(t)), such as the one required to transform classically from an inertial frame to a rotating frame, do not generally preserve the (classical) energy (they don't even preserve Newton's laws).
  4. Sep 3, 2012 #3
    Please show this.
  5. Sep 3, 2012 #4
    [tex] i\hbar \dot{|\psi '(t) \rangle } = i\hbar (\dot{U}(t) |\psi(t) \rangle + U \dot{|\psi(t)\rangle} = (UH + i\hbar \frac{dU}{dt} ) |\psi(t)\rangle = (UHU^{-1} ih \frac{dU}{dt}U^{-1})|\psi'(t)\rangle = H' |\psi'(t) \rangle [/tex]

    where I've used the Schrödinger equation in the second equality and in the third equality I've expressed the non-transformed state in terms of the transformed one.
  6. Sep 3, 2012 #5
    But then what about the Heisenberg picture? There certainly U=U(t), in fact it's the inverse of the time evolution operator. Is the energy of the system different in the Heisenberg picture?
  7. Sep 3, 2012 #6
    I was thinking of the usual simple cases, such as rotational transformations. You are correct that some time-dependent unitary transformations do not change the expectation of H.
    This is not quite correct see Wikipedia-Heisenberg picture.
  8. Sep 3, 2012 #7
    Ok, what you are doing reminds me of the derivation of the Berry phase. Basically, because you make a time-dependent similarity transformation, if the system described by the ket [itex]\vert \Psi(t) \rangle[/itex] is isolated (and its time-evolution is given by Schroedinger's equation through the Hamiltonian H), then the system described by [itex]\vert \psi'(t) \rangle \equiv U(t) \, \vert \psi(t) \rangle[/itex] is, in general, open, and its energy is not conserved (and it's evolution is described by another time-dependent Hamiltonian [itex]H' = U \, H \, U^{\dagger} + i \, \hbar \, \dot{U} \, U^{\dagger}[/itex]).
  9. Sep 4, 2012 #8
    wait, I think you might be mixing up some time dependence and the derivatives... maybe
    consider the following:
    |\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle
    Now if we re-do this derivation watching the [itex]t[/itex]s carefully, we get a different result
    i\hbar \frac{d}{dt}|\psi (t)\rangle &= i\hbar \frac{d}{dt}U(t,t_0)|\psi (t_0)\rangle =i\hbar \dot{U}(t,t_0)|\psi (t_0)\rangle = i\hbar \dot{U}(t,t_0)U^{-1}U|\psi (t_0)\rangle \\
    &= i\hbar \dot{U}U^{-1}|\psi (t)\rangle = H|\psi (t)\rangle
    because the time derivative doesn't hit both the [itex]U[/itex] and the state. then
    i\hbar \frac{d U}{dt} = HU
    which is normal and I imagine [itex]UHU^{-1}[/itex] still holds... but I might be mixed up.
  10. Sep 4, 2012 #9
    If I understand you correctly it seems like you've assumed that U is the time evolution operator, but I meant a general time dependent, unitary transformation.
  11. Sep 4, 2012 #10
    I think it is, I did use the fact that for a state in the Schrödinger picture
    [tex]i\hbar \dot{|\psi(t)\rangle }= H|\psi(t)\rangle.[/tex]

    I have to agree with your explanation though. Thanks :)
  12. Sep 4, 2012 #11
    Another way to derive what you did is:[tex][U,H]=i\hbar{}\frac{dU}{dt}\Rightarrow{}[U,H]U^{-1}=UHU^{-1}-HUU^{-1}=UHU^{-1}-H=i\hbar{}\frac{dU}{dt}U^{-1}\Rightarrow{}UHU^{-1}=i\hbar{}\frac{dU}{dt}U^{-1}+H[/tex]. However, this derivation is only correct for U that do not explicitly depend on time. Per the wikipedia page, in general [itex][H,A]=-i\hbar\frac{dA}{dt}+i\hbar\frac{\partial{A}}{ \partial{t}}[/itex]. The correspoding result being: [tex]UHU^{-1}=i\hbar{}(\frac{dU}{dt}-\frac{\partial{U}}{ \partial{t}})U^{-1}+H[/tex]. When U is the time-evolution operator it so happens that [itex]\frac{dU}{dt}=\frac{\partial{U}}{ \partial{t}}[/itex], so the equation reduces to UHU-1=H. This is simply a statement that H commutes with the time-evolution operator, something already known to be true.
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