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Transition between two states probability

  1. Apr 19, 2014 #1
    1. The problem statement, all variables and given/known data

    i6witl.png

    Part (a): Show probability to transit from state i to j is given by:
    Part (b)i: Use answer in part (a) to find probability
    Part (b)ii: Use time evolution to find probability

    2. Relevant equations



    3. The attempt at a solution

    Part (a) was alright, bookwork question on time dependent perturbation theory.

    What's interesting was, I got different answers for part (b)i and (b)ii.

    Part (b)i

    Amplitude of finding particle in state j:

    [tex]a_j = \frac{1}{i\hbar}\int V_{ji} exp(\frac{i\Delta E}{\hbar}t) dt[/tex]

    In this case, ##\Delta E = 2\mu B##, and since the additional hamiltonian is ##-\mu b \sigma_x##, hence ##V_{ji} = \mu b##.

    [tex] a_j = \frac{1}{i\hbar} V_{ij}^{*} \int_0^T exp(\frac{i2\mu B}{\hbar}t) dt [/tex]

    [tex] |a_j|^2 = (\frac{b}{B})^2 sin^2(\frac{\mu B l}{\hbar v}) [/tex]

    Part (b)ii

    The time evolution operator is given by ##U_{(t)} = e^{-\frac{iH}{\hbar}t}## such that ##|\psi_t> = U_t |\psi_0>##.

    To find the time evolution operator:
    [tex]e^{-\frac{iH}{\hbar}t} = exp(\frac{i\mu B \sigma_z}{\hbar}t) exp(\frac{i\mu b \sigma_x}{\hbar}t)[/tex]

    For each of the exponentials, I found them in matrix form:

    2884jnb.png

    Now to overlap the desired |-,B> = (0 1) state with evolved state ##|\psi_t>##:

    [tex] <-,B|\psi_t> = i sin (\frac{\mu b}{\hbar}t) exp(\frac{-i\mu B}{\hbar}t)[/tex]

    Thus, probability is ##sin^2 (\frac{\mu b l}{\hbar v})##
     
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  3. Apr 19, 2014 #2

    TSny

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    Keeping in mind that ##\sigma_z## and ##\sigma_x## do not commute, is this correct?
     
  4. Apr 20, 2014 #3
    I was thinking about that, as the original operator was: ##U = e^{-\frac{iH}{\hbar}t} = exp(\frac{i\mu B \sigma_z}{\hbar}t + \frac{i\mu b \sigma_x}{\hbar}t)##

    Do I use ##\sigma_x \sigma_z## or ## \sigma_z \sigma_x##?

    And is the answer to (b)i correct?
     
  5. Apr 20, 2014 #4

    TSny

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    Edit: I think your answer to (b)i is correct.

    For (b)ii, can you find the exact eigenvalues and eigenvectors for ##H=-\mu( B \sigma_z+ b \sigma_x)##?
     
    Last edited: Apr 20, 2014
  6. Apr 21, 2014 #5
    The eigenvalues are ##\lambda = \pm \mu \sqrt{b^2 + B^2}##. I'm having trouble finding the eigenstates. What's the point of finding them anyway?
     
  7. Apr 21, 2014 #6

    TSny

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    OK.

    How have you gone about it?

    The eigenstates of H have a simple time evolution and they can be used as basis states for expressing the spin state of the particle.
     
  8. Apr 21, 2014 #7
    For example, consider ##\lambda = \mu \sqrt{b^2 + B^2}##, then I get these two simultaneous equations:

    [tex]Be_1 + be_2 = \sqrt{b^2 + B^2}(e_1)[/tex]
    [tex]be_1 - Be_2 = \sqrt{b^2 + B^2}(e_2)[/tex]

    By the way, what's wrong with my method with the time evolution operator ##U_t## above?
     
  9. Apr 21, 2014 #8

    TSny

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    Looks like you might have forgotten the minus sign in the expression for the Hamiltonian, so some signs are incorrect here. The two equations are not independent, but you can use normalization to get a second condition.

    There is another way to get the eigenstates if you note that the particle will be in a net magnetic field that lies in the xz plane at an angle of θ = tan-1(b/B) to the z-axis. So, the eigenstates will be spin "up" and "down" along the direction of the net magnetic field. So, if you know how to rotate spin 1/2 states, you could rotate a spin up along the z-axis to produce a spin up or spin down along the net magnetic field direction.

    I don't think the method is wrong, but you can't factor the time evolution operator as you did.
     
  10. Apr 21, 2014 #9
    ok I ended up with this pair of simultaneous equations for the positive eigenvalue:

    [tex]-Be_1 - be_2 = \sqrt{b^2+B^2}e_1[/tex]
    [tex]-be_1+Be_2 = \sqrt{b^2+B^2}e_2[/tex]

    Still can't solve it simply, i'm trying to get something simple like ##e_1=\pm e_2##.

    For a spin orientation (θ,ø) where θ = tan-1(b/B) to z-axis, the spin of the particle can be characterized as:

    [tex]|+,t\rangle = e^{\frac{-i\phi}{2}} cos \left(\frac{\theta}{2}\right) |+,z\rangle + e^{\frac{i\theta}{2}} sin \left(\frac{\theta}{2}\right)|-,\rangle[/tex]

    Since ##\phi = 0##:

    [tex]|+,t=0\rangle = cos \left(\frac{\theta}{2}\right) |+,z\rangle + sin \left(\frac{\theta}{2}\right)|-,z\rangle[/tex]

    This is the particle's spin state at t=0.

    So at time t, the particle evolves as:

    [tex]|+,t\rangle = cos (\frac{\theta}{2})*exp\left(\frac{-i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |+,z\rangle + sin (\frac{\theta}{2})*exp\left(\frac{i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |-,z\rangle[/tex]

    Thus the probability of it being in the ##|-,z\rangle## state at time t is

    [tex]sin^2\left(\frac{\theta}{2}\right) = \frac{1}{2}\left[1 - sin\theta\right][/tex]

    Using ##tan\theta = \frac{b}{B}##, ##sin\theta = \frac{b}{\sqrt{b^2+B^2}}##:


    Probability is ##\frac{1}{2}\left[1 - \frac{b}{\sqrt{b^2+B^2}}\right]##

    This is clearly a different answer as it is a constant and there should be a factor of ##sin^2(\theta)## according to perturbation theory..
     
    Last edited: Apr 21, 2014
  11. Apr 21, 2014 #10

    TSny

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    Looks good.
    I found it helpful to divide both sides of the first equation by ##\small \sqrt{b^2+B^2}## and then express the coefficients in terms of cos θ and sin θ where θ = tan-1(b/B).

    The right side of your equation represents a state with spin that is parallel to the net B-field.

    But this is not the spin state at t = 0. At t = 0 the particle has spin in the z-direction.

    No. The states ##|+,z\rangle## and ##|-,z\rangle## do not evolve with time this way. It's the spin states that are parallel or antiparallel to Bnet that have simple time evolution.
     
  12. Apr 21, 2014 #11
    Yeah I got these two eigenstates:
    551r2x.png

    At t = 0, the state of the particle is along z-axis, but we wish to express it in basis along net B field, B'.

    [tex]|\psi_0\rangle = |+,z\rangle = a|+,B'\rangle + c|-,B'\rangle[/tex]

    This implies that ##a = \langle+,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec \theta + 1)}}(1+sec\theta) = \sqrt{\frac{1+cos\theta}{2}}## and ##c = \langle-,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec\theta-1)}}(1-sec\theta) = -\sqrt{\frac{1-cos\theta}{2}}##

    That's true, because these are basis kets in the z-direction, and not in direction of tan θ. I suppose the above method can only be used.

    Following from above, the ket at time t is:
    [tex]|\psi_t\rangle = a*exp(\frac{-i\lambda_{+}}{\hbar}t)|+,B'\rangle + c*exp(\frac{-i\lambda_{-}}{\hbar}t)|-,B'\rangle[/tex]

    Thus probability amplitude is:

    [tex]\langle -,z|\psi_t\rangle = a*exp(\frac{-i\lambda_{+}}{\hbar}t)\langle-,z|+,B'\rangle + c*exp(\frac{-i\lambda_{-}}{\hbar}t)\langle -,z|-,B'\rangle[/tex]

    where ##\langle-,z|+,B'\rangle = \langle -,z|-,B'\rangle = \frac{1}{\sqrt{2sec\theta(sec \theta + 1)}} tan \theta##

    Separating real and imaginary parts:

    [tex]\langle -,z|\psi_t\rangle = cos \left(\frac{\lambda_+}{\hbar}t\right)\langle-,z|+,B'\rangle\left(a+c\right) + i sin \left( \frac{\lambda_-}{\hbar}t \right)\langle-,z|+,B'\rangle (a-c)[/tex]
    [tex] = cos \left(\frac{\lambda_+}{\hbar}t\right)\langle-,z|+,B'\rangle\left(a+c\right) - i sin \left( \frac{\lambda_+}{\hbar}t \right)\langle-,z|+,B'\rangle (a-c)[/tex]

    Finally, the probability is:
    [tex]|\langle -,z|\psi_t\rangle|^2 = \left(\langle-,z|+,B'\rangle\right)^2 \left[cos^2\left(\frac{\lambda_+}{\hbar}t\right)(a+c) + sin^2 \left(\frac{\lambda_+}{\hbar}t\right) (a-c)\right][/tex]

    [tex] = \left(\langle-,z|+,B'\rangle\right)^2 \left[a + c*cos\left(\frac{2\lambda_+}{\hbar}t\right)\right][/tex]

    [tex] = \left[\frac{sin^2\theta}{2(cos\theta+1)}\right]\left[\sqrt{\frac{1+cos\theta}{2}} - cos\left(\frac{2\lambda_+}{\hbar}t\right)\sqrt{\frac{1-cos\theta}{2}}\right][/tex]

    This looks crazy, compared to the answer in part (b)i using perturbation theory.
     
    Last edited: Apr 21, 2014
  13. Apr 21, 2014 #12

    TSny

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    If you forget normalization for a moment, notice that ##|+,B'\rangle \sim \left(
    \begin{array}{c}
    1+\sec\theta\\
    \tan\theta\\
    \end{array}
    \right)
    \sim \left(
    \begin{array}{c}
    \cos\theta +1\\
    \sin\theta\\
    \end{array}
    \right)
    ##

    Use half-angle identities to express ##1+\cos\theta## and ##\sin\theta## in terms of ##\theta/2## and simplify.

    Note that the eigenvalue ##\lambda_+## corresponds to the higher energy and therefore belongs to the spin state that is anti-parallel to B'. This is due to the overall negative sign in H.
     
  14. Apr 23, 2014 #13
    Yeah it gets pretty messy, at this stage its hard to believe that it will all turn out to be a simple sin2θ function. I will definitely work it out later today or tomorrow. But for now, is this concept/method right?

    1. Express initial state |+,z> in terms of basis along B'.
    2. Each ket evolves with time given by exp(-iEt/h)
    3. Overlap with <-,z| to find probability amplitude
     
  15. Apr 23, 2014 #14

    TSny

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    Yes. That will solve it.
     
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