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QM: Questions on spin

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I am looking at the spin-state (i.e. we neglect other degrees of freedom for this system) of two particles 1 and 2 given by:

    [tex]
    \left| {\psi (t = 0)} \right\rangle = \frac{1}{{\sqrt 3 }}\left( {\left| \uparrow \right\rangle _1 \left| \uparrow \right\rangle _2 + \left| \uparrow \right\rangle _1 \left| \downarrow \right\rangle _2 + \left| \downarrow \right\rangle _1 \left| \downarrow \right\rangle _2 } \right)
    [/tex]

    where the subscript denotes the particle which has a spin-direction given by the arrow (up or down).

    Question #1: Say I conduct a measurement of the spin for particle 1, and I get spin up. This means that there are two possible states the system is now in, more specifically the first and second state. Now I measure the spin of particle 2, and I can get either up or down, but what is the probability of this?

    Attempt for #1: I believe it is simply ½, since the particles are now in a state, which is a linear combination of the two first states in [itex]\psi(t=0)[/itex]. Can you confirm this?

    ******

    Question #2: The two particles are in the "original" state [itex]\psi(t=0)[/itex]. We let them evovle by according to the Hamiltonian given by:

    [tex]
    \hat H = \omega_1 S_{1,z} + \omega_2 S_{2,z},
    [/tex]

    where the omega's are just positive constants and the operatores are spin in the z-direction for particle 1 and 2. I have to find [itex]\psi(t)[/itex] at some random time t.

    Attempt for #2: I find the eigenenergies of each of the three possible states at time t=0, and then I just multiply each of the three states in [itex]\psi(t=0)[/itex] with the time-constant exp(-iEt/hbar), with the respective energy for each state.

    If this is correct, then why can we just multiply by the exponential time factor? I mean, this came when the solved the S.E., but that was for a different Hamiltonian.

    Thanks in advance.

    Best regards,
    Niles.
     
  2. jcsd
  3. Mar 21, 2009 #2
    Your answers for both questions are correct.

    As for the second question, the idea is as follows. Any wavefunction [tex]\Psi(x,t)[/tex] can be expanded in terms of energy eigenstates [tex]\psi_n(x)[/tex]. At t=0, this expansion looks like:

    [tex]\Psi(x,0) = \sum_n c_n \psi_n(x)[/tex]

    Now you can prove, on very general grounds, that the evolution of an energy-eigenstate through time is quite simple, namely, we only have an energy and time-dependent phase factor: [tex]e^{iE_nt/\hbar}[/tex]. So the complete decomposition is:

    [tex]\Psi(x,t) = \sum_n c_n \psi_n(x)e^{iE_nt/\hbar}[/tex]

    This is precisely the answer you gave, only now for a specific Hamiltonian (i.e. set of energy states and energy eigenvalues). It is in fact a very general result.
     
  4. Mar 21, 2009 #3
    Thank you for reading through my post, and taking the time to reply. I really appreciate it.
     
  5. Mar 22, 2009 #4
    If the states [itex]\psi_n(x)[/itex] are not energy eigenstates (i.e. not eigenfunctions of the Hamiltonian), then how does the time-dependence look like?

    Is it still exp(-iEt/hbar)? Or is it not the energy E, but the relevant eigenvalue that should be used instead?
     
    Last edited: Mar 22, 2009
  6. Mar 23, 2009 #5
    No, the time-dependence is always given in terms of the energy-eigenstates. If you use basis states which are not eigenstates of the Hamiltonian, then their evolution gets a lot more complicated. Probably the only way to know what the exact expression is, is to write these basis states in terms of energy eigenstates.

    This is why energy-eigenstates are so important: their time evolution is quite simple.
     
  7. Mar 23, 2009 #6
    Ahh, ok, very interesting. Thanks!
     
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