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I Are Eigenstates of operators always stationary states?

  1. Jan 21, 2017 #1
    Hello everyone,

    I am wondering if the eigenstates of Hermitian operators, which represent possible wavefunctions representing the system, are always stationary wavefunctions, i.e. the deriving probability distribution function is always time invariant. I would think so since these eigenstates arise when the system is bound... Am I correct?

    Thanks!
     
  2. jcsd
  3. Jan 21, 2017 #2

    Orodruin

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    No. The eigenstates of an Hermitian operator are stationary only if that operator commutes with the Hamiltonian of the system.
     
  4. Jan 21, 2017 #3
    Ok. I hear and accept that.

    Let's just focus on the Hamiltonian (energy operator). Its eigenvalues can form a discrete or continuous set depending on the physical situation. To each eigenvalue correspond an energy eigenstate. Are the energy eigenstates always stationary states?
     
  5. Jan 21, 2017 #4

    PeroK

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    Does the Hamiltonian always commute with itself?
     
  6. Jan 21, 2017 #5

    vanhees71

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    Let's work in the Schrödinger picture. Then the state vectors carry the entire time evolution, while the essentially self-adjoint (Hermitean is not sufficient!) operators that represent the observables are time-independent (we leave out the somwhat more complicated case of explicitly time-dependent observables). Then the state vector obeys
    $$\mathrm{i} \hbar \partial_t |\psi(t) \rangle = \hat{H} |\psi(t) \rangle.$$
    Obviously the formal solution solution of this equation is
    $$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |\psi(0) \rangle.$$
    Now suppose at the initial time ##t=0## the system has been prepared in an energy eigenstate ##|\psi(0) \rangle=|u_E \rangle##. Then you have [corrected in view of #6]
    $$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |u_E \rangle =
    \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right ) |u_E \rangle,$$
    which means that the time evolution of the state is just given by multiplying ##|u_E \rangle## with a phase factor.

    Now since the physical content of the state ket is just given by Born's rule, i.e., if you have a complete set of compatible observables ##A_i## (##i \in \{1,\ldots,n \}##) and ##|a_1,\ldots,a_n \rangle## a complete set of common eigenvectors of the corresponding essentially self-adjoint operators ##\hat{A}_i##, then the probability (distribution) to measure at time ##t## the values ##(a_i)## is given by
    $$P_{\psi}(t,a_1,\ldots,a_n) = |\langle a_1,\ldots a_n |\psi(t) \rangle|^2,$$
    a phase factor doesn't play any role, i.e., for the above case that an energy eigenstate is the initial state of the system, the probability distribution is time-independent
    $$P_{\psi}(t,a_1,\ldots a_n)=|\langle a_1,\ldots a_n |u_E \rangle|^2.$$
    Thus, the eigenvectors of the Hamiltonian represent stationary states of the quantum system.
     
    Last edited: Jan 21, 2017
  7. Jan 21, 2017 #6

    PeroK

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    Perhaps you meant:

    $$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |u_E \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right ) |u_E \rangle,$$

    o_O
     
  8. Jan 21, 2017 #7
    Thanks.

    Wouldn't it depend on the type of Hamiltonian operator?
    There is a the free space Hamiltonian operator (if the potential is zero) and other Hamiltonian operator depending on the form of the potential operator involved....

    In the case of free particle represented by a wavepacket traveling in free space (zero potential), the probability distribution function expands (hence varies) in time. What does it say about the Hamiltonian for that particular system?
     
  9. Jan 21, 2017 #8

    PeroK

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    An operator ##H## commutes with itself if ##HH = HH##

    That doesn't leave a lot of room for any operator to get out of that one!
     
  10. Jan 21, 2017 #9
    I am familiar with what you are saying.

    Ok, but isn't the Hamiltonian a composite operator involving the potential energy operator and the kinetic energy operator? I would think that depending on the type of potential operator, the Hamiltonian may or may not commute with itself.
     
  11. Jan 21, 2017 #10

    PeroK

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    So, for a non-self-commuting operator, H, we have:

    ##HH \ne HH##?
     
  12. Jan 21, 2017 #11

    vanhees71

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    I don't see any difference between my formula and yours, or am I overlooking something?
     
  13. Jan 21, 2017 #12

    PeroK

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    You just missed out the ##\exp## from the last expression.
     
  14. Jan 21, 2017 #13

    vanhees71

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    Argh, yes, you are right. I'll correct it. Thanks.
     
  15. Jan 22, 2017 #14

    PeterDonis

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    Write ##H = T + V##, where ##T## is kinetic and ##V## is potential. Then write out the product of ##T + V## with itself. Then switch the order of the factors in each term. Does the product change?
     
  16. Jan 22, 2017 #15

    vanhees71

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    Well, trivially any operator commutes with itself. The commutator of two operators is defined by
    $$[\hat{A},\hat{B}]=\hat{A} \hat{B}-\hat{B} \hat{A}.$$
    Now set ##\hat{A}=\hat{B}##, and you immediately get 0!
     
  17. Jan 22, 2017 #16

    dextercioby

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    If the Hamiltonian is explicitely time dependent in the Schrödinger picture, then H(t1) will not strongly commute with H(t2). This is actually the root cause for Dyson's series.
     
  18. Jan 22, 2017 #17

    Orodruin

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    If the Hamiltonian is explicitly time-dependent it would generally not make any sense to talk about stationary states anyway.
     
  19. Jan 22, 2017 #18

    PeterDonis

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    That's because H(t1) and H(t2) in this case are different operators. The OP was talking about the time independent case, where you just have the single operator H that never changes.
     
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