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Qubit system, time dependent states.

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    See attachment

    2. Relevant equations



    3. The attempt at a solution

    (i) [tex]|\Psi(t)_{1}>=e^{{-itE_{1}/\hbar}}\frac{1}{\sqrt{2}}(|z^{+}>+|z^{-}>)[/tex]
    [tex]|\Psi(t)_{2}>=e^{{-itE_{2}/\hbar}}\frac{1}{\sqrt{2}}(|z^{+}>-|z^{-}>)[/tex]

    where [tex]E_{1}=\lambda\frac{\hbar^{2}}{4}+\mu\frac{\hbar}{2}[/tex]
    [tex]E_{2}=\lambda\frac{\hbar^{2}}{4}-\mu\frac{\hbar}{2}[/tex]

    (ii) [tex]|\Psi(t)>=\sum_{i}c_{i}(0)e^{{-itE_{i}/\hbar}}|\Psi_{i}>[/tex]

    (iii) [tex]|<\psi(t)_{2}|\psi(0)_{1}>|^{2}=[/tex]
    [tex]|(e^{it(\lambda\hbar^{2}/4-\mu\hbar/2)/\hbar}<z^{-}|)^{*}(|z^{+}>)|^{2}=0[/tex]

    I feel like i'm doing something really wrong could someone check over my attempt please and guide me in the right direction if i'm doing it wrong.
     

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  3. Dec 9, 2013 #2

    TSny

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    Part (i) looks good. The stationary states are clearly ##|x^+\rangle## and ##|x^-\rangle## and these can be expressed in terms of ##|z^+\rangle## and ##|z^-\rangle## as you have done.

    Some people would not include the time dependent exponential factors when writing the stationary states. The stationary states are often defined as solutions of the time independent Schrodinger equation, and so the stationary states are also time independent. The corresponding solutions to the full time dependent Schrodinger equation are then the stationary states multiplied by the exponential time dependent factors. Not a big deal in my opinion.

    Part (ii) looks good. I would change the notation to [tex] |\Psi(t)>=\sum_{i}c_{i}\; e^{{-itE_{i}/\hbar}}|\Psi(0)_{i}>[/tex] This is the same as just writing ##|\Psi(t)\rangle = c_1 e^{{-itE_{1}/\hbar}}|x^+\rangle + c_2 e^{{-itE_{2}/\hbar}}|x^-\rangle##

    You might want to write this out explicitly in terms of ##|z^+\rangle## and ##|z^-\rangle##.

    This is not correct. You want "the probability that the state ##|\Psi(t) \rangle## at time ##t## is in the state ##|z^- \rangle##".

    For practice, suppose you had a state that at time ##t## was given by ##|\phi(t)\rangle = .6|z^+\rangle + .8|z^-\rangle##.

    What is the probability that ##|\phi(t)\rangle## is in the state ##|z^-\rangle##?
     
    Last edited: Dec 9, 2013
  4. Dec 9, 2013 #3
    Thanks! In part iii) with your practice state the way you have worded it makes sense its just, [tex]|<\phi(t)|z^{-}>|^{2}[/tex]

    however in the actual question part iii) even if i got it mixed up and it should be
    [tex]|<\psi(t)|z^{-}>|^{2} [/tex]

    [tex]<z^{+}|z^{-}>[/tex] is still going to be = 0, unless i'm still missing something here. From what you've said, i'm interpreting that all I have done is mix up which state is to be found inside of which, so I just need to switch the |Z>'s round.

    Ultimately though the probability will still be equal to zero?
     
    Last edited: Dec 9, 2013
  5. Dec 9, 2013 #4

    TSny

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    Can you write the state ##|\Psi(t)\rangle## as

    ##|\Psi(t)\rangle = a_1(t)|z^+\rangle + a_2(t)|z^-\rangle##

    where ##a_1(t)## and ##a_2(t)## are certain time dependent coefficients?
     
  6. Dec 9, 2013 #5
    Yeh but if i'm measuring the probability of Z+ in Z- doesnt the state collapse?
     
  7. Dec 9, 2013 #6
    By any chance is this what your thinking of?
    [tex]|\Psi(t)\rangle=e^{{-itE_{1}/\hbar}}|z^{+}\rangle+e^{{-itE_{2}/\hbar}}|z^{+}\rangle[/tex]

    Like if I take the general state and then collapse it to [tex]|z^{+}\rangle[/tex] as this is what I am measuring the probability when the state is in [tex]|z^{-}\rangle[/tex], but then the z's still = 0
     
    Last edited: Dec 9, 2013
  8. Dec 9, 2013 #7

    TSny

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    Alternately, you can leave ##|\Psi(t)\rangle## expressed in terms of ##|x^+\rangle## and ##|x^-\rangle## with simple time dependent coefficients. And then construct the probability ##|\langle z^-|\Psi(t)\rangle|^2## which will involve evaluating ##\langle z^-|x^+\rangle## and ##\langle z^-|x^-\rangle##.
     
  9. Dec 9, 2013 #8
    dont you mean [tex]|\langle \psi(t)|z^-\rangle|^2[/tex]?

    As Quote:

    I think i'm starting to understand, the fact that it says the state at t=0 is in Z+ is irelevant right? once time t has passed the state will abide by the superposition of states originally found in Q1)?
     
    Last edited: Dec 9, 2013
  10. Dec 9, 2013 #9

    TSny

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    No, this isn't correct. The states that evolve with simple exponential time factors are the eigenstates of the Hamitonian: ##|x^+\rangle## and ##|x^- \rangle##. So, the correct expression is

    ##|\Psi(t)\rangle = c_1 e^{{-itE_{1}/\hbar}}|x^+\rangle + c_2 e^{{-itE_{2}/\hbar}}|x^-\rangle##.

    One approach is to express the x-states in this expression in terms of the z-states (which I think is what you did in your first post when you had expressions for ##|\Psi_1\rangle## and ##|\Psi_2\rangle##) . Then, if you collect together all of the ##|z^- \rangle## terms, you can just read off the probability amplitude for ##|\Psi(t)\rangle## to be ##|z^-\rangle##.

    Another approach is to leave ##|\Psi(t)\rangle## in terms of the x-sates and form the probability ##| \langle z^-|\Psi(t)\rangle|^2##
     
  11. Dec 9, 2013 #10

    TSny

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    ##\langle\phi|\psi \rangle = \langle\psi|\phi \rangle^*##,

    so ##|\langle\phi|\psi \rangle |^2 = | \langle\psi|\phi \rangle|^2##
     
  12. Dec 9, 2013 #11
    Ok you have helped plenty and I think I can now do this but could you just explain the significance in question iii) where it says "If the system is in state |z+> at t=0", surely this is irelevant to answering the question as i'm just trying to find the probability of finding the time evolved state (which is written in Q1/Q2) within |z->
     
    Last edited: Dec 9, 2013
  13. Dec 9, 2013 #12

    TSny

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    When you write the expression

    ##|\Psi(t)\rangle = c_1 e^{{-itE_{1}/\hbar}}|x^+\rangle + c_2 e^{{-itE_{2}/\hbar}}|x^-\rangle##.

    You will need to choose the constants ##c_1## and ##c_2## so that the state will be |z+> at time t = 0.
     
  14. Dec 9, 2013 #13
    Ahhh ok, well thanks very much for all your help! Much appreciated.
     
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