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Schrödinger and Heisenberg picture

  1. Jul 3, 2012 #1
    Im sorry, I accidently edited my opening post instead of posting a new one.. The question was regarding the statement that the state ket is stationary in the Heisenberg picture when the basis kets are transforming in time. Because the state ket is a superposition of the base kets it should the be evolving to, unless the coefficients change.
     
    Last edited: Jul 3, 2012
  2. jcsd
  3. Jul 3, 2012 #2
    Usually in S. picture you write the evolution of your state as:

    $$|\Psi(t)\rangle =U(t)|\Psi(t=0)\rangle $$

    where:

    $$U(t)=e^{-\frac{i}{\hbar}\hat{H}t}$$

    When you have a scalar product (which are the only physical quantities in quantum mechanics), you can always write (O is an observable):

    $$\langle \Psi(t)|O|\Psi(t)\rangle=\langle \Psi (t)|U(t)U^{ \dagger }(t)OU(t)U^{\dagger}(t)|\Psi(t)\rangle$$

    In H. picture the state vectors and the operators are redefined as:

    $$|\Psi(t)\rangle _H = U^{ \dagger }|\Psi(t)\rangle = U^{\dagger}(t) U(t) |\Psi(t=0)\rangle \equiv |\Psi(t=0)\rangle$$

    and:

    $$O_H(t)=U^{ \dagger }OU(t)$$

    So, as you can see, while the scalar product remain the same, now the state vectors are constant (their are fixed at their initial value) while the operators become time dependent.
     
  4. Jul 3, 2012 #3
    Ah thanks for the answer.
    I think my confusion comes from the fact that Sakurai writes that the basis kets evolves in the Heisenberg picture:

    [tex]\mid a' \rangle_H = \mid a'(t) \rangle_H, [/tex]
    while the state ket stays stationary. If the state ket should be stationary then the coefficients for a basis (which evolves with time) also has to evolve with time.

    Since:
    [tex] \mid \phi \rangle_H = \sum_{i=1}^n c_i \mid a' \rangle_H. [/tex]
    If the coefficient doesnt evolve with time.. doesnt this mean that:

    [tex]\mid \phi \rangle_H = \mid \phi(t) \rangle_H ?[/tex]
     
    Last edited: Jul 3, 2012
  5. Jul 3, 2012 #4
    Your reasoning is correct but where have you found that the coefficients of the expantion of the state ket are time indipendent? Maybe I'm missing something but, reading the Sakurai itself, at page 88 eq. (2.2.44b) it says that the coefficients are the same in both representation so in H. representation they are time dependent too...or at least so it seems to me :tongue:
     
  6. Jul 3, 2012 #5

    vanhees71

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    In the Heisenberg picture the full time dependence is put to the operators which describe observables while the state, i.e., the Statistical Operator (or in the case of pure states a Hilbert-space vector as its representant) is time independent.

    Mathematically that means that in the Heisenberg picture the time evolution of a (not explicitly time dependent) observable operator is given by
    [tex]\hat{O}(t)=\hat{U}(t,t_0) \hat{O}(t_0) \hat{U}^{\dagger}(t,t_0), \qquad (1)[/tex]
    where [itex]\hat{U}(t,t_0)[/itex] is the unitary time evolution oparator in the Heisenberg picture. For a non explicityly timedependent Hamiltonian you have
    [tex]\hat{U}(t,t_0)=\exp[+\mathrm{i} (t-t_0) \hat{H}].[/tex]
    A basis can be defined by an arbitrary complete set of observables, i.e., let [itex]\hat{O}_k(t)[/itex], [itex]k \in \{1,\ldots,N \}[/itex] be a set of pairwise commuting self-adjoint operators whose joint eigenspaces are non-degenerate, i.e., for each possible set of eigenvalues [itex](o_1,o_2,\ldots,o_N)[/itex] there is (up to a constant factor) one and only one eigenvector [itex]|o_1,\ldots,o_N;t \rangle[/itex]. The time evolution of these eigenvectors is, according to Eq. (1)
    [tex]|o_1,\ldots,o_N;t \rangle=\hat{U}(t,t_0) |o_1,\ldots o_N; t \rangle.[/tex]
    Note that the set of eigenvalues doesn't change with time since the unitary time evolution (1) for observable operators doesn't change the spectra of these operators. The state kets, however, evolve according to this very unitary time-evolution operator, as you can prove within one line immediately.

    Now, according to Born's Rule, the physical meaning of the formalism is that, if a quantum system is prepared at time, [itex]t[/itex] in the pure state represented by the state ket [itex]|\psi,t \rangle=|\psi,t_0 \rangle=:|\psi \rangle=\text{const}[/itex], the probability to find the outcome [itex](o_1,o_2,\ldots,o_N[/itex] for a simultaneous (and by assumption compatible!) measurements of the observables [itex]O_1,\ldots,O_N[/itex] is given with help of the wave function,
    [tex]\psi(q_1,\ldots,q_N;t)=\langle o_1,\ldots,o_N;t |\psi \rangle[/tex]
    via
    [tex]P_{\psi}(q_1,\ldots,q_N;t)=|\psi(q_1,\ldots,q_N;t)|^2.[/tex]
    This physical outcome is independent of the choice of the picture of time evolution by construction, i.e., you obtain the very same probability (distribution) also within the Schrödinger picture since it is connected to the Heisenberg picture by a unitary transformation. The same holds true for any general picture (which formalism has been developed by Dirac; particularly in connection with the socalled interaction picture, which is important in time-dependent perturbation theory).
     
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