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Homework Help: QM: Changing wavefunctions after measurements

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all. My question is best illustrated with an example. Please, take a look:

    Let's say we have particle in a stationary state, so [tex]\Psi(x,0)=1\cdot \psi_{1,0}(x)[/tex] with energy E_{1,0}. Now at time t=0 the Hamiltonian of this particle changes, since the particle gains some energy. Thus the wavefunction [tex]\Psi(x,t)[/tex] changes, and it can be written as:


    where [tex]\psi_{2,n}(x)[/tex] are the new stationary states.

    Question: At time t=0 when the particle gains energy and hence [tex]\Psi(x,t)[/tex] changes, is the wavefunction given as:

    \Psi (x,0) = \psi _{1,0} (x) = \sum\limits_{n = 0}^\infty {c_n } \psi _{2,n} (x)

    So does the wavefunction change immediately or does it evovle in a slow fashion?
  2. jcsd
  3. Oct 16, 2008 #2
    The coefficients cn are time dependent here. You can write the wave-function as
  4. Oct 16, 2008 #3
    Does this follow from the fact that our wave function lives in Hilkbert space?
  5. Oct 16, 2008 #4
    Ok, I am having my doubts about why we are allowed to do this.

    When the Hamiltonian changes, the particle is not longer in the groundstate.

    So if it is not in the groundstate, then why are we even interested in using the following expression?

    \Psi (x,0) = \psi _{1,0} (x) = \sum\limits_{n = 0}^\infty {c_n } \psi _{2,n} (x)
    Last edited: Oct 16, 2008
  6. Oct 17, 2008 #5
    I don't mean to be impolite, but I really need to understand this. I still cannot see how [itex]\psi_{1,0}[/itex] comes into play, when the particle is not in that state anymore. Of course we can write it as a linearcombination of the new stationary states (as you said, nasu), but we could do that with any wavefunction.

    Why is [itex]\psi_{1,0}[/itex] so interesting in this case? Especially when our particle is not in this state anymore.
  7. Oct 17, 2008 #6
    At time t=0 the particle *is* still in the state [itex] \psi_{1,0} [/itex]. Now, instead of simply evolving as [itex] \psi_{1,0}\exp(-iE_{1,0}t/\hbar) [/itex] it evolves under the new Hamiltonian with new eigenenergies [itex] E'_n [/itex]. What you do is you first expand the initial state as
    \psi_{1,0}=\sum_n c_n \psi'_{n}
    in terms of the new eigenfunctions of the Hamiltonian. Then, once you have calculated those coefficients [itex] c_n [/itex] (they are given by the overlap between the initial state and the new eigenfunctions), the state evolves as
    \psi_{1,0}=\sum_n c_n \psi'_{n}\exp(-iE'_n t/\hbar)
  8. Oct 18, 2008 #7
    Hi borgwal

    First, thanks for answering me. The way I have understood it, [itex]\psi_{1,0}[/itex] is the stationary state with a definite energy. So if the particle at time t=0 is in state [itex]\psi_{1,0}[/itex], then doesn't this mean that it has the energy associated with the stationary state [itex]\psi_{1,0}[/itex]?

    In your last expression: [itex]\psi_{1,0}=\sum_n c_n \psi'_{n}\exp(-iE'_n t/\hbar)[/itex]: Isn't a timefactor [itex]\exp(-iE'_{1,0}t/\hbar)[/itex] missing on the left hand side?

    Again, thanks for answering. This forum is the only place I can get help at the moment, so I really appreciate it.
  9. Oct 18, 2008 #8
    You're changing the Hamiltonian at t=0: so, the *same* state that is stationary with respect to the old Hamiltonian is no longer stationary with respect to the new Hamiltonian.

    There is no timefactor [itex] \exp(-iE'_{1,0}t/\hbar)[/itex] missing in my equation: in fact there is no new energy [itex] E'_{1,0} [/itex] for the state [itex] \psi_{1,0} [/itex] anymore, because that wavefunction is no longer an energy eigenfunction.

    The concept of what a stationary state is, depends on what Hamiltonian you have!
    Similarly, the energy "associated with a state" depends on the Hamiltonian.
  10. Oct 20, 2008 #9
    Perfect borgwal, these are just the answers I needed. I have two final questions (REMARK: I can't get the LaTeX to work properly, so I'll just write the code. I guess it's just a temporary issue.):

    1) Prior to the measurement (i.e. the old Hamiltonian is still valid), our particle is in the stationary state \psi_{1, 0}. How does the time-dependent wavefunction look like for this state? Is it simply:

    \Psi(x, t) = \psi_{1, 0}?

    2) So have I understood it correctly when I say that the stationary state of the old Hamiltonian is now a time-dependent state after the Hamiltonian has changed?

    Thanks in advance. I really appreciate your effort.
    Last edited: Oct 20, 2008
  11. Oct 20, 2008 #10
    2) yes!

    1) no, before t=0 there is still the trivial time-dependence
    \Psi=\psi_{1,0}\exp(-i E_{1,0}t/\hbar),
    but I'm sure you knew that already. [And I see I had written that in my previous post, too]
  12. Oct 20, 2008 #11
    Yes, you actually did. Sorry for not seeing that. But it still confuses me for two reasons:

    1) We are in a stationary state. So why would it evolve?

    2) The square of the constants in the linearcombination at an arbitrary time t does not sum up to one. I mean, let's for example take t = 2 - then the sum of the squares are not 1?
  13. Oct 20, 2008 #12
    1) That's just a matter of terminology: expectation values like those of position and momentum do not depend on time. The overall (time-dependent) phase factor is irrelevant for any physical variable.

    2) The *absolute* values squared of all coefficients should add up to 1, at any time t.
  14. Oct 21, 2008 #13
    Ok, my last issues are these:

    1) So \psi_{1,0} is no longer a solution to the time-independent Schrödinger equation, because the Hamiltonian has changed. But \psi_{1,0}(x) is not time-dependent (it is only x-dependent), so when the Hamiltonian has changed and we get a linear combination of the new states, the linear combination is somehow constant in time, although it evolves? How is this to be understood?

    2) When we calculate the probability of a particle's wavefunction to collapse to an eigenstate, we use the absolute value of the coefficient in front of that eigenstate in the linear combination. At an arbitrary time t, is the exponential term also part of that constant? If yes, it will vanish, right?

    Again, I really appreciate it. These are my last questions, I promise.
    Last edited: Oct 21, 2008
  15. Oct 21, 2008 #14


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    The coefficients in that linear combination contain the factors

    exp(i E2,n t / hbar)

    You can think of this as causing interference among the various state components, sometimes constructive and sometimes destructive. But the key thing is, the amount of interference changes with time at any given location, so the total wavefunction amplitude at a given x will change in magnitude.

    That's why the wavefunction as a whole changes after t=0.

    I'm not understanding this question (it has been 20 years since I had quantum mechanics, sorry) but perhaps somebody else can address it.
  16. Oct 21, 2008 #15

    George Jones

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    I wouldn't, say that it vanishes, I would say that it drops out. Why? How, in general, is this probability calculated?
  17. Oct 21, 2008 #16
    By taking the absolute value, and since the term is complex, it drops out.

    This is just a general question: If the Hamiltonian has changed, then \psi_{1,0} is no longer a solution to the Schrödinger equation. Then how can the particle still be in this state?

  18. Oct 21, 2008 #17
    You're confusing stationary solutions of the S.E., which correspond to solutions to the time-independent S.E. with a fixed energy, with general solutions to the time-dependent S.E. A particle is always in a state that is a solution to the time-dependent S.E., but not necessarily in a stationary state.
  19. Oct 25, 2008 #18
    Ahh yes, you are correct. Thanks for mentioning that.

    Say we are looking at a particle in the infinite square well, where the potential is zero in the interval from 0 to L. The groundstate is given by:
    \psi (x) = \sqrt {\frac{2}{L}} \sin \left( {\frac{{n\pi }}{L}x} \right)

    Now length of the well increases from 0<x<L to 0<x<2L, and the new eigenstates are given by:

    \psi_{\text{new}} (x) = \sqrt {\frac{1}{L}} \sin \left( {\frac{{n\pi }}{{2L}}x} \right)

    From our preceding discussion, the particle is still in the state [tex]\psi(x)[/tex], but it is now a linear combination of the new eigenstates [tex]\psi_{\text{new}}(x)[/tex].

    So using the facts from our discussion, can we conclude that the probability of finding the particle outside 0<x<L is zero, since the particle is still in the state [tex]\psi(x)[/tex]?

    Thanks in advance.

    Best regards,
  20. Oct 25, 2008 #19
    Correct, at time t=0 the particle will not be found outside 0<x<L. But then the wavefunction starts evolving away from the original state, and the particle can be found in the region x>L.
  21. Oct 25, 2008 #20
    Hmm, this is what I do not understand. The new states are:

    \Psi(x,t) = \psi(x)=\sum_n c_n \psi_{new,\,\,n}(x)\exp(-iE_{new,\,\,n}t/\hbar)

    1) Regardless of what time t it is, [tex]\psi(x)[/tex] is zero for x>L and x=L. But am I supposed to understand it like this: That when we make a new measurent, then the possible states are [tex]\psi_{\text{new,\,\,n}}(x)[/tex], which are not 0 for x greater than or equal to L, and hence the particle can be found in 0 <x< 2L for times t>0?

    2) By the way, does the particle's wavefunction collapse to one of these new eigenstates immediately when the well becomes wider? Or does it only collapse upon a new measurement?
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