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Question about evolution of a wave function

  1. Jan 21, 2012 #1

    andrewkirk

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    As I understand it, where a system’s Hamiltonian is not time-dependent, the wave function of a system that is in state psi(0) at time t=0 evolves as:

    psi(t) = sum, over all eigenvalues E of operator H, of exp(-i*E*t / hbar) * <E|psi(0)> * | E>

    If the eigenvalues are continuous it is an integral rather than a sum but let’s assume they are discrete.

    In Shankar’s ‘Principles of Quantum Mechanics’, he claims on p118 (2nd edition) that multiplying a ket by a complex number of modulus one does not change the physical state. The first factor in the above formula has modulus one so, we can re-write the equation as:

    psi(t) = sum, over all eigenvalues E of operator H, of <E|psi(0)> * f(t,|E>)

    where f(t,|E>)=exp(-i*E*t / hbar) * | E> is a ket representing a state that is physically identical to |E>.

    Since f(t,|E>) is physically identical to |E> and the first postulate of QM tells us that the ket represents the physical state, it should make no physical difference if we replace f(t,|E>) by |E> in the formula.

    That then gives us:
    psi(t) = sum, over all eigenvalues E of operator H, of <E|psi(0)> * |E>

    But now we have a formula from which t has disappeared, which implies that the state does not change over time.

    Where did I go wrong?
     
  2. jcsd
  3. Jan 21, 2012 #2

    tom.stoer

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    very hard to read; please try LaTeX

    [tex]|\psi,t\rangle = e^{-iHt}\,|\psi,0\rangle[/tex]
     
  4. Jan 21, 2012 #3

    andrewkirk

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    OK here is a Tex version of the post. I hope somebody can help explain this puzzle. :confused:

    As I understand it, where a system’s Hamiltonian is time-independent and has discrete eigenvalues, the wave function of a system that is in state [itex]|\psi(0)\rangle[/itex] at time [itex]t=0[/itex] evolves as:

    [itex]|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ |E\rangle\ \ e^{-iEt/h}\ \ \ \ \ \ [/itex] Formula 1

    where [itex]J[/itex] is the set of eigenvalues of the Hamiltonian operator.

    In Shankar’s ‘Principles of Quantum Mechanics’, he claims on p118 (2nd edition) that multiplying a ket by a complex number of unit modulus does not change the physical state. The last factor [itex]e^{-iEt/h}[/itex] in the above formula has unit modulus, because E is an eigenvalue of a Hermitian operator and hence real. So so we can re-write the equation as:

    [itex]|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ f(t,|E\rangle)\ \ \ \ \ \ [/itex] Formula 2

    where [itex]f(t,|E\rangle) = \ |E\rangle\ e^{-iEt/h}[/itex] is a ket representing a state that is, according to Shankar, physically identical to [itex]|E\rangle[/itex].


    Since [itex]f(t,|E\rangle)[/itex] is physically identical to [itex]|E\rangle[/itex] and the first postulate of QM tells us that the ket represents the physical state, it should make no physical difference if we replace [itex]f(t,|E\rangle)[/itex] by [itex]|E\rangle[/itex] in formula 2.

    That then gives us:

    [itex]|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ |E\rangle\ \ \ \ \ \ [/itex] Formula 3


    But now we have a formula from which [itex]t[/itex] has disappeared, which implies that the state does not change over time, which is clearly not the case.

    Where did I go wrong?
     
  5. Jan 21, 2012 #4
    No, it isn't. You did not multiply the ket by a complex number, you multiplied it by a complex function (unlike a complex number, it depends on t). And I don't think this is the only problem.
     
  6. Jan 21, 2012 #5
    If you multiply the overall state vector by a phase, the physical state does not change. The *overall* phase is not physically significant. But time evolution does not do this, it multiplies different components of the overall state vector by different phases. If you write the state as a superposition of energy eigenstates, the *relative* phases between these eigenstates *are* physically significant.

    I suggest taking an explicit example, like

    |1> + exp(i alpha) |2>

    where |1> and |2> are the first two energy eigenstates of the infinite square well and alpha is a number between 0 and 2*pi. Try computing the expectation value of the position operator (this is clearly physically meaningful) in this state and you will see that it depends on alpha.

    If you want you could do the same thing for the state

    exp(i alpha) |1> + exp(i beta) |2>

    and you will find that physically meaningful numbers depend only on the relative phase (alpha - beta).
     
  7. Jan 21, 2012 #6

    andrewkirk

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    Why do you think that observation would change things? If it were a function of position or momentum, I can see this would be an issue, but it is not - it's only a function of t. At any given future time t', [itex]e^{-iEt'/h}[/itex] is a simple complex number, not a function of position or momentum, so [itex]f(t',|E\rangle) = \ |E\rangle\ e^{-iEt'/h}[/itex] is just [itex]|E\rangle[/itex] multiplied by a complex number of unit modulus.

    In fact, Shankar even uses the term "stationary state". This is what he says:

    "The normal modes [itex]|E(t)\rangle\ =\ |E\rangle\ \ e^{-iEt/h}[/itex] are also called stationary states for the following reason: the probability distribution P(ω) for any variable Ω is time-independent in such a state."

    This makes me think the time-dependence must somehow be introduced via the summation over different eigenvalues, but I can't see how that would work.
     
  8. Jan 22, 2012 #7
    With all due respect, we are talking about physics and mathematics, not about abstract painting, so we have to deal with some precise statements and take them seriously. What's happening, actually? You quote Shankar, who discusses multiplying by a complex number, then you multiply by a time-dependent function and get some nonsense as a result. I would say this is your problem, not Shankar's, not mine. This is pretty much the same as if you read in a book that the sum of angles of a planar triangle always equals 180 degrees and then complained that the sum of angles of a triangle on a sphere is less than 180 degrees. I don't know how to explain that this is wrong. As I said, this is not the only problematic thing about your reasoning, but until you remove this problem, further discussion of your reasoning does not make much sense.
     
  9. Jan 22, 2012 #8

    tom.stoer

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    As said physical states are rays in an Hilbert space; as such physics doesn't change if all state vectors [itex]|\psi\rangle[/itex] of a given system are multiplied by a constant number z with modulus one.

    But b/c the dynamics (time dependence) of a stationary state is encoded in it's phase it's nonsense to use a time dependent phase instead.
     
  10. Jan 22, 2012 #9

    andrewkirk

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    Thanks Duck and Tom, your comments really helped me to get a clearer view of this. That example you suggested Duck clarified things a lot.

    What follows is how I understand it now, including where the problem was, really written just for my own benefit to help solidify my thoughts, and in case anybody else with this query comes across this discussion:

    A ket in the Hilbert space completely describes a state. Any ket that is a complex scalar multiple of that ket (ie in the same ray) describes the same state. However that does not mean they are the same ket. They are not. It is a many-to-one relationship from kets to states.

    When we express the evolved state of a system in terms of the energy basis, we write the system's representative ket as a sum of the basis kets. Note this is a sum of kets, not a sum of states. We can multiply the whole sum by a phase factor exp(iBt) without changing the system state, as that is just multiplying the state's ket by a complex scalar. That is why a state corresponding to an energy eigenket is called stationary, because it physically does not change as time goes by.

    For a state with nonzero projection on more than one of the energy eigenkets, its expansion in the energy basis is a sum of more than one basis ket. These kets evolve over time by changing phase with different frequencies. While the state to which each ket in the sum can be mapped does not change as the phase changes, that is irrelevant, as the overall state is specified via a ket that is a sum of kets, not states. That ket sum changes over time because of the differing phases of the components, and that change involves a change of state.

    I think the source of the confusion is that people sometimes talk of a quantum state as being a superposition of states which, in my mind at least, conjures up images of some kind of a sum of states, which is an error. Indeed I wonder whether the concept of superposition is actually intrinsically misleading and best avoided.
     
  11. Jan 22, 2012 #10

    Ken G

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    I think the problem is actually that you are associating the "superposition" with the coefficients of the kets. The problem is not that the superposition is an invalid concept because states not unambiguously associated with kets, the problem is that the superposition is not the list of ket coefficients. The kets are acting like a basis here, and the coefficients of the kets are acting like coordinates, and like any coordinates, the same state can be associated with many different coordinates if the basis is different. All the same, there is a physical connection between the superposition state and the physical states that make up that superposition. There is just some ambiguity in the coordinates of that superposition.
     
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