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Instantaneous eigenstates in the Heisenberg picture

  1. Dec 22, 2015 #1
    I'm a bit confused as to what is meant by instantaneous eigenstates in the Heisenberg picture. Does it simply mean that if vectors in the corresponding Hilbert space are eigenstates of some operator, then they won't necessarily be so for all times ##t##, the eigenstates of the operator will change as it evolves in time?
    As far as I understand, in the Heisenberg picture the states are time-independent and instead it is the operators that evolve in time - does this mean that their eigenstates also evolve in time or are how I described above?

    The reason I ask is that I've been reading a text in which they consider the position operator ##\hat{x}## in both the Schrödinger picture and the Heisenberg picture. They start off in the Heisenberg picture and state that an instantaneous eigenstate of the position operator ##\hat{x}(t)## at some time ##t_{i}## is defined by $$\hat{x}(t_{i})\lvert x,t_{i}\rangle =x_{i}\lvert x,t_{i}\rangle$$ and take care to point out that the eigenstate ##\lvert x,t\rangle## is simply parametrised by time ##t## and is not dynamical - it simply labels the time at which the position is measured. They then move on to consider the same operator in the Schrödinger picture, denoting it as ##\hat{x}=\hat{x}(0)## to explicitly show that it is not time dependent in this picture. Clearly the eigenvalues of the operator should be the same in both pictures, and so $$\hat{x}\lvert x\rangle =x\lvert x\rangle$$ where ##\lvert x\rangle## is a corresponding eigenstate of the position operator in the Schrödinger picture. They then note that the Heisenberg position operator is related to its Schrödinger counterpart by $$\hat{x}(t_{i})=e^{i\hat{H}t_{i}}\hat{x}e^{-i\hat{H}t_{i}}$$ and proceeds as follows $$ \hat{x}(t_{i})\lvert x,t_{i}\rangle =e^{i\hat{H}t_{i}}\hat{x}e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle \quad\Rightarrow\quad \hat{x}\left(e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\right)=x_{i}\left(e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\right)$$ where we have operated from the left on both sides by ##e^{-i\hat{H}t_{i}}##. Upon comparing this with $$\hat{x}\lvert x\rangle =x\lvert x\rangle$$, the above implies that the Schrödinger and Heisenberg eigenstates are related by $$\lvert x_{i}\rangle =e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\quad\Rightarrow\quad \lvert x,t_{i}\rangle =e^{i\hat{H}t_{i}}\lvert x_{i}\rangle$$

    I get that the eigenstates of an operator in the Schrödinger picture should be time independent since the operator itself is time independent and so it's eigenstates should remain constant in time. However, I'm unsure as to how this concept translates in to the Heisenberg picture?
     
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  3. Dec 22, 2015 #2

    blue_leaf77

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    The time dependency of eigenstates in Heisenberg picture actually applies as a general rule, not only to the position eigenstates.
    To start, let's write the eigenvalue equation for an operator ##A^{(S)}## (Schroedinger picture operator):
    $$
    A^{(S)} |a\rangle = a |a\rangle
    $$
    Multiply both sides with the inverse-time evolution operator ##U^\dagger## to get
    $$
    U^\dagger A^{(S)} |a\rangle = a U^\dagger|a\rangle
    $$
    Then insert an identity operator ##1 = UU^\dagger## in between ##A^{(S)}## and ## |a\rangle## in the left hand side.
    $$
    (U^\dagger A^{(S)} U)U^\dagger |a\rangle = a U^\dagger|a\rangle
    $$
    The terms in the parentheses in the LHS constitute Heisenberg version of ##A^{(S)}##, so in the end we get the eigenvalue equation in Heisenberg picture ## A^{(H)} U^\dagger |a\rangle = a U^\dagger|a\rangle##, with the eigenstates ##U^\dagger|a\rangle = e^{iHt} |a\rangle##.
     
  4. Dec 22, 2015 #3
    Yes, sorry, I realise that. What confuses me is that in the Heisenberg picture states are meant to be time independent and in the Schrödinger picture they are time dependent, right? How does this tally up with what we've been discussing here?
     
  5. Dec 22, 2015 #4

    blue_leaf77

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    Are you seeking a way to reconcile the two pictures? Basically, Schroedinger and Heisenberg pictures differ in the way the dynamics is treated, in S picture the evolutionary history of the system is encoded in the state of that system while in H picture, it's the observables which carries the time evolution. Despite this difference, the consequence to measurement is unchanged between the two pictures. If you measure the expectation value of certain observable by treating the system in S and H pictures separately, the answer should be identical.
     
  6. Dec 22, 2015 #5
    I think my confusion is over whether the eigenstates in the Heisenberg picture are time dependent or not and whether they are time dependent in the Schrödinger picture as well? And why?
     
  7. Dec 22, 2015 #6

    blue_leaf77

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    As is shown in post #2, the eigenstate in Heisenberg pictures evolves with time, it does not in Schroedinger picture.
    Because in H picture, the operators are time dependent.
     
  8. Dec 22, 2015 #7
    But what if we express the state vector ##\lvert\psi\rangle## in terms of an eigenbasis. In the Heisenberg picture, states are time independent, i.e. ##\lvert\psi\rangle_{H}=\lvert\psi (0)\rangle_{S}##, so how can it's decomposition onto a particular basis shouldn't be time dependent, right?!
     
  9. Dec 22, 2015 #8

    blue_leaf77

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    In Heisenberg picture, the expansion coefficients are time dependent (it must be the case as the expansion coefficients are the probability of finding the state to be in one of the eigenbasis of an evolving observable), which has the same expression with that in the Schroedinger picture. In Schroedinger picture you have ##c_a(t) = e^{-iE_a t} \langle a|\psi,0\rangle##. Using the expression for ##|a,t\rangle## in Heisenberg picture in post #2, you should find that the projection ##\langle a,t|\psi,0\rangle## is identical to that in Schroedinger picture.
     
  10. Dec 22, 2015 #9
    So do the coefficients and eigenstates evolve such that the overall state describing the system remains constant?
    What confuses me is that in the book I've been reading it makes a point at highlighting that the time ##t## in ##\lvert x,t\rangle## is parametrising the point at which the position is measured, it is not the same as the time dependent Schrödinger state that evolves according to the Schrödinger equation.
     
  11. Dec 24, 2015 #10

    vanhees71

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    This is often very confusing, because in the textbooks it's not clearly stated that the choice of the picture just means how you "distribute" the time-dependence among operators that represent operators and states. The physical observables are completely picture independent as it must be. Take you examples, the Schrödinger and the Heisenberg picture.

    In the Schrödinger picture the observable-operators are time independent, and the full time dependence is on the states. For a pure state, represented by a Hilbert-space vector you have
    $$|\psi,t \rangle_{\text{S}} = \exp(-\mathrm{i} \hat{H} t) |\psi,t=0 \rangle_{\text{S}}.$$
    The observable-operators (take the position operator ##\hat{\vec{x}}## as an example) are time-independent and so are their (generalized) eigenvectors, i.e.,
    $$\hat{\vec{x}}_{\text{S}}|\vec{x} \rangle_{\text{S}}=\vec{x} |\vec{x} \rangle_{\text{S}},$$
    The wave function is
    $$\psi(t,\vec{x})=_{\text{S}} \langle \vec{x}|\psi,t \rangle_{\text{S}}=_{\text{S}} \langle \vec{x} \exp(-\hat{H} t)|psi,t=0 \rangle_{\text{S}}.$$
    Now in the Heisenberg picture the choice of time dependence is the opposite, i.e., the full time dependence is on the observable-operators, and the states are time-independent. You get it by the unitary transformation of the Schrödinger-picture quantities given by the time-evolution operator. Assuming without loss of generality that the quantities of both pictures coincide at ##t=0## you get
    $$|\psi \rangle_{\text{H}}=\exp(+\mathrm{i} \hat{H} t) |\psi,t \rangle_{\text{S}}=|\psi,t=0 \rangle_{\text{S}}=\text{const}.$$
    Then all observable-operators must also be transformed with this unitary transformation, i.e., for the position operator
    $$\hat{\vec{x}}_{\text{H}}(t) = \exp(+\mathrm{i} \hat{H} t) \hat{\vec{x}}_{\text{S}} \exp(-\mathrm{i} \hat{H} t).$$
    Now the spectrum of the operators is of course the same in any picture, because a unitary transformation doesn't change the eigenvalues of the operators. In the Heisenberg picture the eigenvectors for a given fixed eigenvalue ##\vec{x}## of the position operator thus must be time dependent:
    $$\hat{\vec{x}}_{\text{H}}(t) |\vec{x},t \rangle_{\text{H}} = \vec{x} |\vec{x},t \rangle_{\text{H}}.$$
    Now the relation between the eigenvectors in the pictures must be given by the same unitary transformation as for the states. Indeed
    $$\hat{\vec{x}}_{\text{H}}(t) \exp(+\mathrm{i} t \hat{H}) |\vec{x} \rangle_{\text{S}} = \exp(+\mathrm{i} t \hat{H}) \hat{\vec{x}}_{\text{S}} \exp(-\mathrm{i} t \hat{H}) \exp(+\mathrm{i} t \hat{H}) |\vec{x} \rangle_{\text{S}} = \exp(+\mathrm{i} t \hat{H}) \hat{\vec{x}}_{\text{S}} |\vec{x} \rangle_{\text{S}}=\vec{x} \exp(-\mathrm{i} t \hat{H})|\vec{x} \rangle_{\text{S}},$$
    i.e., indeed ##\exp(+\mathrm{i} t \hat{H}) |\vec{x} \rangle_{\text{S}}## is an eigenvector of ##\hat{\vec{x}}_{\text{H}}(t)##, and thus you can set
    $$|\vec{x},t \rangle_{\text{H}} = \exp(+\mathrm{i} t \hat{H})|\vec{x} \rangle_{\text{S}}.$$
    Thus, the change from the Schrödinger to the Heisenberg picture is just a unitary transformation on all vectors and operators, and thus the measurable consequences are the same in both pictures. Particularly the wave function is the same in both pictures and thus the position-probability distribution, which is it's modulus squared, is the same:
    $$_{\text{H}}\langle \vec{x},t|\psi_{\text{H}} \rangle_{\text{H}} =_{\text{S}}\langle \exp(+\mathrm{i} t \hat{H}) \vec{x}|\exp(+\mathrm{i} t \hat{H})\psi,t \rangle_{\text{S}} = _{\text{S}}\langle \vec{x} |\exp(-\mathrm{i} t \hat{H}) \exp(+\mathrm{i} t \hat{H}) \psi,t \rangle_{\text{S}} = _{\text{S}}\langle \vec{x}|\psi,t \rangle_{\text{S}}=\psi(t,\vec{x}).$$
    As you might have realized, I've not distinguished ##\hat{H}_{\text{S}}## and ##\hat{H}_{\text{H}}##, but that's unimportant, because ##\hat{H}## commutes with the time-evolution operator and thus the Hamilton operator is the same in both pictures anyway.
     
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