1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expected value of observable over time

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    We have a linear combination of eigenstates of observable A [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] with eigenstates a and -a. The expected value of the energy for both states is 0, while [tex](\Phi_+,H\Phi_-)=E[/tex], with E real. Calculate the expected value of A for eigenstates [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] over time.

    2. Relevant equations

    I guess

    [tex](\Phi_+,H\Phi_+)=(\Phi_-,H\Phi_-)=0[/tex]
    [tex](\Phi_+,H\Phi_-)=E[/tex]
    [tex]\varphi=C_+\Phi_++C_-\Phi_-[/tex]

    3. The attempt at a solution

    I guess that for the given equations I have to obtain <H>

    [tex]<H>=\varphi^*H\varphi=(C_+\Phi_++C_-\Phi_-)^*H(C_+\Phi_++C_-\Phi_-)=C_+^*C_+(\Phi_+^*,H\Phi_+)+C_-^*C_-(\Phi_-^*,H\Phi_-)+C_+^*C_-(\Phi_+^*,H\Phi_-)+C_-^*C_+(\Phi_-^*,H\Phi_+)[/tex]

    Then I assume C's and [tex]\Phi[/tex]'s are real so
    [tex]<H>=2C_+C_-E[/tex]

    Now I have to compute this

    [tex]<A>=(\varphi^*,A\varphi)[/tex]


    How <H> plugs into the calculation of <A>?
     
  2. jcsd
  3. May 23, 2009 #2
    Express your states in terms of energy eigenstates. That way you can find out their time dependence from the Schrödinger equation.
     
  4. May 23, 2009 #3
    [tex]
    \varphi=C_+\Phi_+e^{-iE_+t/h}+C_-\Phi_-e^{-iE_-t/h}
    [/tex]

    Aren't [tex]E_+,E_-[/tex] suposed to be 0?
     
  5. May 23, 2009 #4
    No, [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] aren't eigenstates of energy so you can't use the simple exponential time dependence. You have to find what the eigenstates of energy are and express [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] in terms of those states.

    The way I would go about this is to use matrix representation for the operators in the basis of the eigenstates of A. Choose:

    [tex]\Phi_+ = \left( \begin{array}{c} 1 \\0 \end{array}\right) ; \Phi_- = \left(\begin{array}{c} 0 \\1 \end{array}\right)[/tex]

    [tex]\Rightarrow A = \left( \begin{array}{cc} a & 0 \\ 0 & -a \end{array}\right) ; H = \left( \begin{array}{cc} 0 & E \\ E & 0 \end{array}\right)[/tex]

    You can find the energy eigenstates (whose time development is just the exponential) by diagonalizing H and express [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] in terms of those states thus obtaining the time dependence of [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex].
     
  6. May 23, 2009 #5
    I see, the problem is that H and A are not represented in the same basis of eigenstates
     
  7. May 23, 2009 #6
    Well, actually they are. They are both represented in the basis of eigenstates of A. The problem is that time development is determined by the hamiltonian which means that you need eigenstates of H in order to obtain the time development of [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex]
     
  8. May 24, 2009 #7
    I found the diagonal matrix of H to be:

    [tex]
    \Rightarrow D = \left( \begin{array}{cc} E & 0 \\ 0 & -E \end{array}\right)
    [/tex]

    so the eigenvalues are E and -E and

    [tex]

    \varphi=C_+\Phi_+e^{-iEt/h}+C_-\Phi_-e^{iEt/h}

    [/tex] ????

    [tex]<A>_t=(\varphi,A\varphi)[/tex]

    but this is time independent ...
     
  9. May 24, 2009 #8
    Again you are treating [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] as if they were energy eigenstates which they are not. They don't have the exponential time development, energy eigenstates do.

    As you correctly calculated the eigenvalues of energy are ±E, so it follows that eigenvectors are

    [tex]\left|E_+\right> = \frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\1 \end{array}\right) ; \left|E_-\right> = \frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\-1 \end{array}\right)[/tex]

    From which it follows that

    [tex]\Phi_\pm = \frac{1}{\sqrt{2}}\left(\left|E_+\right> \pm \left|E_-\right>\right)[/tex]

    Now |E±> have the exponential time development:

    [tex]\left|E_\pm, t\right> = e^{\mp i Et/\hbar}\left|E_\pm, t=0\right>[/tex]

    From this you can deduce the time dependence of [tex]\Phi_+[/tex] and [tex]\Phi_-[/tex] and calculate the expectation value.

    Btw, I don't get why you're using [tex]\varphi[/tex]. I think what you are asked to find is

    [tex](\Phi_+^*(t), A\Phi_+(t))[/tex] and [tex](\Phi_-^*(t), A\Phi_-(t))[/tex]
     
  10. May 24, 2009 #9
    yes I know I have a lot to learn, I don't even know what they ask for. Let's see

    I guess the [tex]\sqrt{2}[/tex] is to normalise the vectors

    [tex]
    (\Phi_+^*(t), A\Phi_+(t))=\frac{1}{2}(e^{-jEt/h}+e^{jEt/h})a(e^{jEt/h}+e^{-jEt/h})=\frac{a}{2}(2+e^{-2jEt/h}+e^{2jEt/h})=a(1+cos(2Et/h))=2acos^2(Et/h)
    [/tex]

    and I guess the other should look like [tex]2asin^2(Et/h)[/tex]
     
  11. May 24, 2009 #10
    Indeed [tex]\sqrt{2}[/tex] is for normalization. Also you can't just replace A with the eigenvalue because the state depends on time. At t=0 it is indeed an eigenstate of A but after that it evolves into something else (in fact the state oscillates between the two eigenstates). Use the the matrix representation of operators and states and calculate the expectation value in the traditional way:

    [tex]\Phi_+(t) = \frac{1}{\sqrt{2}}\left(e^{-iEt/\hbar}\left|E_+\right> + e^{iEt/\hbar}\left|E_-\right>\right) = \frac{1}{2}\left( \begin{array}{c} e^{-iEt/\hbar} + e^{iEt/\hbar}\\e^{-iEt/\hbar} - e^{iEt/\hbar} \end{array}\right) = \left( \begin{array}{c} \cos(Et/\hbar) \\ -i\sin(Et/\hbar) \end{array}\right)[/tex]

    [tex]\left<A(t)\right>_+ = \Phi_+^{\dagger}(t) A \Phi_+(t)[/tex]

    And the one with the minus sign correspondingly.

    EDIT:

    There is also an alternative way of doing this and that is the Heisenberg picture. Instead of having time dependent states (Sxhrödinger picture) one can take the time dependence into the operators. A(t) = eiHt/hAe-iHt/h. By using the fact that H2 = E2I2 (I2 being the identity matrix) one can deduce that

    [tex]e^{\pm iHt/\hbar} = \cos(Et/\hbar) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \pm i\sin(Et/\hbar) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/tex]

    and thus get the time dependence of the operator A. But if you haven't yet studied the Heisenberg picture then just disregard this piece of information. I think it's easier to do this in the Schrödinger picture anyway, but one should keep in mind that there are often several ways to solve the problem.
     
    Last edited: May 24, 2009
  12. May 24, 2009 #11
    is this right?

    [tex]
    \Phi_+^{\dagger}(t) A \Phi_+(t)= \left( \cos(Et/\hbar) \: i\sin(Et/\hbar) \right) \left( \begin{array}{cc} a & 0 \\ 0 & -a \end{array}\right) \left( \begin{array}{c} \cos(Et/\hbar) \\ -i\sin(Et/\hbar) \end{array}\right)=a(cos^2(Et/h)-sin^2(Et/h))
    [/tex]
     
  13. May 24, 2009 #12
    I think so. You can write it a little more compactly, though, recall that [tex]\cos^2(x) - \sin^2(x) = \cos(2x)[/tex]
     
  14. May 24, 2009 #13
    thanks a lot
     
  15. May 30, 2009 #14
    Hi,

    I have a similar question now. From the text I take this
    [tex]
    \varphi=+\frac{1}{2}\Phi_+\frac{1}{\sqrt{2}}\Phi_0+\frac{1}{2}\Phi_-
    [/tex]
    [tex]
    \Phi_+ = \left( \begin{array}{c} 1 \\0\\0 \end{array}\right) ; \Phi_- = \left(\begin{array}{c} 0\\0 \\1 \end{array}\right); \Phi_0 = \left(\begin{array}{c} 0\\1 \\0 \end{array}\right)
    [/tex]
    [tex]
    \Rightarrow A = \left( \begin{array}{ccc} 0 & 0 & a \\ 0 & a & 0 \\ a & 0 & 0 \end{array}\right) ; H = \left( \begin{array}{ccc} E & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E \end{array}\right)
    [/tex]

    Folowing the same methodology I diagonalise A and get the eigenvalues of A to be a (two times) and -a .

    The question is if A is a constant of movement. From what I know this is true if [A,H]=0. I guess this A refers to the matrix formed bythe eigenvalues of A

    [tex]
    \Rightarrow A = \left( \begin{array}{ccc} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & -a \end{array}\right)
    [/tex]

    If I do [A,H] = AH-HA = 0.
    Meaning A is a constant of movement and <A> is constant in time.

    The next question is to calculate <A>t . Is it then <A>t= <A> at t=0 ?

    [tex]
    <A>_0=+\frac{1}{4}a+\frac{1}{2}a+\frac{1}{4}(-a)
    [/tex]

    I guess this last part is wrong, from where I take the coefficients? from the eigenvectors that I calculate when I diagonalise A?
     
  16. May 30, 2009 #15
    When calculating the commutator [A,H] you have to do it in the same basis. If you use the basis you defined above (the basis of energy eigenstates) then both A and H have to be expressed in that basis, that is

    [tex]

    \Rightarrow A = \left( \begin{array}{ccc} 0 & 0 & a \\ 0 & a & 0 \\ a & 0 & 0 \end{array}\right) ; H = \left( \begin{array}{ccc} E & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E \end{array}\right)

    [/tex]

    The diagonalized A is expressed in a different basis, namely in the basis of its own eigenstates. You don't need to diagonalize anything, just work in the basis you defined. And since [tex]\varphi[/tex] is already expressed as a linear combination of eigenstates of H you know how it evolves in time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook