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Time evolution problem.

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data


    A box containing a particle is divided into a left and right compartment
    by a thin foil. The two orthonormal base kets |L> and |R> stand for the
    particle being in either the left or the right compartment, respectively.
    Hence, any state ket in our system can be decomposed as
    |psi> = |R> <R|psi> + |L> <L|psi>
    Tunneling between them, through the foil is characterised by a quantity
    , with units of energy, such that the corresponding Hamiltonian reads
    H = V [|L> <R| + |R> <L|] :

    (a) Write the Hamiltonian in matrix form.


    (b) Find the normalised energy eigenkets and the corresponding energy
    eigenvalues.

    (c) Find the state ket |psi(t)> = psiL(t)|L> + psiR(t)|R> in the Schrodinger
    picture, if Psi L, R(t = 0) = psi L, R are known.


    2. Relevant equations

    Eigenvalue equations Av = yv
    Time evolution operator U(t,t0) = exp(iH(t-t0)E/hbar)


    3. The attempt at a solution

    I don't have a problem with part a) or b). I simply constructed my 2x2 matrix for the hamiltonian by using scalar products. This gave a matrix that looked like:

    0 V
    V 0

    I then calculated the eigenvalues of the matrix and associated eigenvectors, which I then normalised for part b). I ended up with eigenvalues of +V and -V, and eigen vectors of 1/sqrt(2) (1 1) and 1/sqrt(2) (1 -1).

    Now for part C I know I need to use the time evolution operator, but it won't work very well for a matrix that isn't diagonal and H isn't in the original basis. So consequently I understand that I need to perform the time independent time evolution operator (I'm assuming the operator itself is time independent since we are in the Schrodinger picture) on the basis of eigen vectors just calculated which will yield a diagonal matrix of eigenvalues:

    V 0
    0 -V

    this is about as far as I have got. I really don't quite understand how to represent the wavefunction in terms of this new basis of eigenvectors. I think the notation is a bit of a problem for me, I don't really know how to deal with the |psi> vs psi notation.
     
  2. jcsd
  3. May 7, 2014 #2

    Simon Bridge

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    How would you normally represent a vector in a new basis?

    ... basically you want to revise linear algebra where functions are treated as vectors. Look at the definitions of the inner and outer products etc.
    Since courses usually start with the Schrodinger picture, you should have some notes back from when you first learned Dirac notation about translating Schrodinger to Dirac.
     
  4. May 7, 2014 #3
    I would do a linear transformation on my vector to change the basis set from one to the other. It all seems fairly straight forward in a linear algebra course. But here I'm struggling to get my head round it, I know V is the diagonal matrix that I want and D = S*HS. For an arbitrary vector x, S*x would push the vector x from the old representation to the new representation.

    I don't think we went into any great detail about the differences between the function picture and the dirac picture. I think you can do something like <x|X> where you extract a position wave-function x from the state function X.
     
  5. May 7, 2014 #4

    Simon Bridge

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    You have the old representation - which you can write out in the dirac notation.
    You know the new basis vectors ... so you need the old basis vectors in terms of the new ones.
    http://www.physics.ohio-state.edu/~perry/p632_wi07/lectures/DiracNotation.pdf

    Courses don't normally go into a lot of detail about the conversion - but they do make the connection.
    Revise. Don't use your memory - actually go and look at what you wrote down and what the text says.
     
  6. May 8, 2014 #5
    I honestly don't have much about this, it was glossed over so quickly. But like I say there was a bit where the position wavefunction was extracted from the state function like psi(x) = <x|X> or in the case of momentum psi(p) = <p|X>
     
  7. May 8, 2014 #6

    Simon Bridge

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    Well that is why I gave you a link ;)
     
  8. May 8, 2014 #7
    I understand what dirac space is, and much of what is on that link I am okay with. The only point I can find that links the two notations is line 25 which is what I said in the last post. I'm not sure how to use this idea to solve the problem.
     
  9. May 8, 2014 #8

    Simon Bridge

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    I don't understand your problem then - if you have a vector representation in one basis, how would you normally represent it in another basis. Actually do it and see.
    How would you change basis using dirac notation?

    Maybe you need something more explicit:
    http://electron6.phys.utk.edu/qm1/modules/m3/change_rep.htm [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. May 8, 2014 #9

    OK so if I apply S* to psiL(t)|L> + psiR(t)|R> this will change psi into the basis where V is diagonal which is what I want. I know S* =

    1/sqrt(2) 1 1
    1 -1

    but how would I be able to apply this matrix to psiL(t)|L> + psiR(t)|R>?
     
    Last edited by a moderator: May 6, 2017
  11. May 8, 2014 #10

    Simon Bridge

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    ##\renewcommand{\bra}[1]{\left\langle #1\right|} \renewcommand{\xpt}[1]{\left\langle #1\right\rangle} \renewcommand{\ket}[1]{\left| #1\right\rangle} \renewcommand{\braket}[2]{\left\langle #1\right|\left. #2 \right\rangle}
    \renewcommand{\A}{\text{A}}\renewcommand{\B}{\text{B}}\renewcommand{\H}{\text{H}}\renewcommand{\C}{\text{C}} ##
    I just want to go back to basics for a bit to make sure I am reading you right...

    Defining some terms:
    ##\A## is an operator with eigenstates ##\ket{a}## so that ##\A\ket a =a\ket a## ... the set of ##\ket a## form a basis, we can assert that they are orthonormal so ##\braket{a}{a}=1## and ##\sum_a \ket{a}\bra{a}=1##

    (Strictly: ##\braket{a_i}{a_j}=\delta_{ij}## and ##\sum_i \ket{a_i}\bra{a_i}=1##, but I hate subscripts - let me know if it gets confusing.)

    Similarly:
    ##\B## is an operator with eigenstates ##\ket{b}## so that ##\B\ket b =b\ket b## ... the set of ##\ket b## form a basis. We can assert that they are orthonormal so ##\braket{b}{b}=1## and ##\sum_b \ket{b}\bra{b}=1##

    What I think we are talking about...
    You want to know how to convert a representation of an arbitrary vector in one basis into the corresponding representation in the other basis. Is that correct?

    Still checking:
    You already have the expansion in ##\A##: ##\ket\psi = \sum_a c_a\ket a: c_a=\braket{a}{\psi}##

    To get: ##\ket\psi = \sum_b c_b\ket b##

    You need: ##c_b=\braket{b}{\psi}=\bra{b}\sum_a c_a\ket a##

    Isn't that how you'd normally approach it?

    You should be able to express that as a matrix operation between vectors of the coefficients.
    I have a feeling you are having trouble seeing how the matrix you have relates to the picture above. Is that correct?

    Meantime - I'll have a closer look at your problem statement. Please be aware that I am somewhat restricted in my answer since I am not allowed to do your work for you.
    All I can do is point you at a likely direction. The more I understand the way you are thinking about the problem, the better I can do this.
     
  12. May 8, 2014 #11

    Simon Bridge

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    ##\renewcommand{\bra}[1]{\left\langle #1\right|} \renewcommand{\xpt}[1]{\left\langle #1\right\rangle} \renewcommand{\ket}[1]{\left| #1\right\rangle} \renewcommand{\braket}[2]{\left\langle #1\right|\left. #2 \right\rangle}
    \renewcommand{\A}{\text{A}}\renewcommand{\B}{\text{B}}\renewcommand{\H}{\text{H}}\renewcommand{\C}{\text{C}} ##Maybe I'm mixing up two different things - i.e. change of basis and time evolution - in your question?

    Notation conversion example:$$\ket\psi = \ket{L}\braket{L}{\psi}+\ket{R}\braket{R}{\psi} = c_L\ket{L}+c_R\ket{R} \Leftrightarrow \psi = c_L\psi_L + c_R\psi_R$$

    Time-evolution:$$\ket{\psi(t)}=\text{U}(t-t_0)\ket{\psi(t_0)}$$
    http://ocw.mit.edu/courses/nuclear-...ng-2012/lecture-notes/MIT22_02S12_lec_ch6.pdf

    Could this be the issue:
    I'm not sure I'm reading that right ... is it:$$\ket{\psi(t)}=\psi_L(t)\ket{L}+\psi_R(t)\ket{R}$$...???

    I would write that as: $$\ket{\psi(t)}=\text{c}_L(t)\ket{L}+\text{c}_R(t)\ket{R}$$ ... which is identical but with more intuition-friendly names, considering the first line about notation conversion.

    Is this closer to what you are puzzled about?
    This is not a change in basis - this is a time-evolution within the same L and R eigenket basis as before.
    You do have another set of eigenkets (you should name them) corresponding to the Hamiltonian. To use those, you need a change in basis.
     
    Last edited: May 8, 2014
  13. May 9, 2014 #12
    Yes, so I do want to apply U to my <L,R| basis, but since H is not diagonal I have to perform a change of basis to my basis of eigenvectors where H is a diagonal matrix. Then I can apply my time evolution operator to that basis and then convert back into the <L,R| basis. At least that was my interpretation. I think my change of basis matrix is simply a matrix of eigenvectors, but I'm not really sure how to apply it.
     
  14. May 9, 2014 #13

    Simon Bridge

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    ##\renewcommand{\bra}[1]{\left\langle #1\right|} \renewcommand{\xpt}[1]{\left\langle #1\right\rangle} \renewcommand{\ket}[1]{\left| #1\right\rangle} \renewcommand{\braket}[2]{\left\langle #1\right|\left. #2 \right\rangle}
    \renewcommand{\A}{\text{A}}\renewcommand{\B}{\text{B}}\renewcommand{\H}{\text{H}}\renewcommand{\X}{\text{X}} ##Well... you already have the ##\H## eigenvectors (and H) in terms of the R-L basis.
    You have found a diagonal H (in the H basis).

    You should be able to do the inverse transformation - representing ##\ket R## and ##\ket L## in terms of the ##\H## basis. This means you can get ##\ket\psi## at ##t=0## in the ##\H## basis, all ready to do a time evolution.

    I think you should name the ##\H## eigenkets explictly BTW.
    ##\ket +## and ##\ket -## are usual.

    Note: "R-L basis" is the basis of an operator that just measures which side the particle is on.
    It may help you to name the operator ... i.e. call it ##\X## so that ##\X\ket R = R\ket R## and ##\X\ket L = L\ket L## where ##R## and ##L## just indicate the left or right side. Now you can call it the "X basis".

    Discussion:
    If the system were prepared so that ##\ket{\psi(t=0)}=\ket L## (result of a measurement of ##\X## yields ##L##) ... what would be the result of ##\X\ket{\psi(t>0)}## (result of repeating that measurement at a later time)?

    I think this is the core lesson they are after.
     
    Last edited: May 9, 2014
  15. May 19, 2014 #14

    Simon Bridge

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    How did you get on?
     
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