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Time evolution, Hamiltonian

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A magnetic field pointing in ##\hat{x}##. The Hamiltonian for this is:

    ##H= \frac{eB}{mc}\begin{pmatrix}
    0 & \frac{1}{2}\\
    \frac{1}{2} & 0
    \end{pmatrix}##
    where the columns and rows represent ##{|u_z\rangle, |d_z\rangle}##.

    (a) Write this out in Dirac notation.

    (b) Compute ##e^{itH}## using 3'rd order taylor series.

    (c) Compute ##e^{itH}## exactly using wolfram.

    (d) If a state starts in ##|u_z\rangle##, write the state at a time t.

    (e) Describe in terms of the bloch sphere, what the state is doing.

    (f) If you do a measurement in the ##{|u_z\rangle, |d_z\rangle}## basis at time t, what is the probability of get "uz"?

    (g) If you do a measurement in the ##{|u_x\rangle, |d_x\rangle}## basis at time t, what is the probability of get "ux"?

    2. Relevant equations



    3. The attempt at a solution

    (a) I need to multiply out the matrix, I believe. I'll set ##\frac{eB}{mc}=w##.

    ##H= \begin{pmatrix}
    0 & \frac{w}{2}\\
    \frac{w}{2} & 0
    \end{pmatrix}##

    I'm not entirely sure how to expand this into Dirac. We were shown a similar problem in class but we didn't show the method, only the result.

    (b-g) I'll come back to this once I understand (a). I doubt I'll need help with all of them, though.
     
  2. jcsd
  3. Sep 6, 2014 #2

    Matterwave

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    So it seems that you don't know what operators look like in Dirac notation? Have you seen any examples of operators being written in Dirac notation before?

    What you have written there is a state, and not an operator.
     
  4. Sep 6, 2014 #3
    Not explicitly. They only ever gave a matrix example as above and then "multiplied" it against a state and gave the result. So, as far as I can confidently say, I only know what an operator multiplied against a state looks like.

    The example I got was:

    ##H= \begin{pmatrix}
    \frac{w}{2} & 0\\
    0 & -\frac{w}{2}
    \end{pmatrix}##

    ##|\Psi\rangle=\frac{1}{√2}(|u_x\rangle+e^{iΘ}|d_z\rangle)##

    ##H|\Psi\rangle= \frac{w}{2}(\frac{1}{√2}(|u_x\rangle-e^{iΘ}|d_z\rangle)##

    I don't know how to get the information I need out of that to put H in dirac notation. Perhaps the above could be used as an example to show me?
     
  5. Sep 6, 2014 #4

    Matterwave

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    In a Dirac notation, an operator is represented by a ket-bra combination. In other words, operators look something like this:

    $$W=\sum_n c_n \left|\psi_n\right>\left<\psi_n\right|$$

    This operator will take a state, for example ##\left|\phi\right>##, and map it to another state, namely:

    $$W\left|\phi\right>=\sum_n c_n \left|\psi_n\right>\left<\psi_n|\phi\right>$$

    The simplest type of operator is a projection operator which simply projects one state onto another, e.g. ##P=\left|\Psi\right>\left<\Psi\right|##. This will take an arbitrary state and project it onto the state ##\left|\Psi\right>##.

    It doesn't seem fair to ask you part (a) without showing you what a Dirac notation operator looks like...I would say definitely your textbook should have a section going over this. Luckily this is just a notational thing, and the later parts of the problem will not really require you to be able to write operators using Dirac notation rather than matrix notation.
     
  6. Sep 6, 2014 #5
    Okay, so something like:

    ##H= \begin{pmatrix}
    0 & \frac{w}{2}\\
    \frac{w}{2} & 0
    \end{pmatrix} =\frac{w}{2}[(0)(|u_z\rangle \langle u_z|+(|u_z\rangle \langle u_z|)+(|d_z\rangle \langle d_z|+(|d_z\rangle\langle d_z|)(0)]##

    ##=\frac{w}{2}[|u_z\rangle \langle u_z|+|d_z\rangle \langle d_z|]##

    I'm suspect of this because it is basically the same as the example but without the negative sign when multiplied by ##|\Psi \rangle ##.
     
    Last edited: Sep 6, 2014
  7. Sep 6, 2014 #6

    Matterwave

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    In your expansion you have two ##\left|u_z\right>\left<u_z\right|## terms. I assume one is a typo. Fix that one and I think you're good. :)
     
  8. Sep 6, 2014 #7
    I think I fixed it. All of the copying and pasting latex messed me up.
     
  9. Sep 6, 2014 #8

    Matterwave

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    You actually fixed the wrong terms haha, and now it's worse than when you started. Maybe it's easier if I expand a little bit on a what a matrix element is.

    A matrix element ##H_{ij}## is given by ##H_{ij}=\left<i\right|H\left|j\right>##. So say you had an operator ##H=\left|i\right>\left<i\right|+\left|j\right>\left<j\right|## what do you think this operator looks like in terms of the matrix ##H_{ij}##? What if ##H=\left|i\right>\left<j\right|+\left|j\right>\left<i\right|##?
     
    Last edited: Sep 6, 2014
  10. Sep 6, 2014 #9
    if ##H=\left|i\right>\left<i\right|+\left|j\right>\left<j\right|##
    ##H= \begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix} ##

    and if ##H=\left|i\right>\left<j\right|+\left|j\right>\left<i\right|##

    ##H= \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} ##

    I think that if I had a matrix:

    ##Y=\begin{pmatrix}
    a & b\\
    c & d
    \end{pmatrix} ##

    then 'a' and 'b' ##= \langle u_z |## and 'c' and 'd' ##=\langle d_z |##
    So if I multiply ##| u_z \rangle + |d_z \rangle## against Y I'd get the above two dirac equations added together.

    Have I got it now?
     
  11. Sep 6, 2014 #10

    Matterwave

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    I think you are making this more complicated than it needs to be (for that last line). You are correct in the previous 2 cases. We just showed how to generate the diagonal elements, and then the off diagonal elements. Now you just have to combine the two. The matrix

    ##Y=\begin{pmatrix}
    a & b\\
    c & d
    \end{pmatrix} ##

    Can be represented easily as ##Y=a\left|i\right>\left<i\right|+b\left|i\right>\left<j\right|+c\left|j\right>\left<i\right|+d\left|j\right>\left<j\right|##.
     
  12. Sep 6, 2014 #11
    Okay, it all makes sense to me now. I'll work on the rest of this. I'll probably just make a new post if I need additional help since it won't likely be until tomorrow

    Thank you.
     
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