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Eigenstate problem in QM

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data

    The Hamiltonian of a quantum-mechanical system has only two energy eigenstates, namely |1> and |2>. The system has three other properties, denoted by the observables A, B, and C, respectively. The normalized eigenstates |1> and |2> may or may not be eigenstates of A, B, or C. Find, as many as possible, the eigenvalues of the observables A, B, and C by making use of the following known matrix elements:

    (a) [itex]<2|A|2> = 1/3, <2|A^{2}|2> = 1/9 [/itex]

    (b) [itex]<1|B|1> = 1/2, <1|B^{2}|1> = 5/4, <2|B|2> = 0 [/itex]

    (c) [itex]<1|C|1> = 1 , <1|C^{2}|1> = 5/4, <1|C^{3}|1> = 7/4 [/itex]

    Comment briefly on whether the observables commute with the Hamiltonian.

    2. The attempt at a solution

    If a state is an eigenstate I can write it as [itex] a|a> [/itex]. So |2> is an eigenstate of A with eigenvalue 1/3. The states |1> and |2> are not eigenstates of B and C. Is this part correct?

    From this, we know that B and C don't commute with the Hamiltonian but A might (depending on whether |1> is also an eigenstate).

    That's about all I can get from this problem. I get the feeling that I'm not using a lot of the information so can someone help me figure out more eigenvalues for A, B and C and whether A commutes with H. Thank you
     
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  3. Jun 23, 2011 #2

    vela

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    Use the matrix representations of the operators. For example, from the information given, you know that A can be represented by the matrix
    [tex]A=\begin{pmatrix} a & b \\ c & 1/3 \end{pmatrix}[/tex]
    in the [itex]\{\vert 1 \rangle, \vert 2 \rangle\}[/itex] basis. Then square the matrix and use the fact that [itex]\langle 2 \vert A^2 \vert 2 \rangle = 1/9[/itex] to solve for the other matrix entries, if you can. The fact that A is an observable also places constraints on the allowed values of the entries.
     
  4. Jun 24, 2011 #3
    vela, how do you know that the matrices are 2x2? Is it because the Hamiltonian is 2x2 (since it has two eigenstates) and so every other observable for this system must be of the same dimension?

    Using your argument, I got the following results:

    [tex]A=\begin{pmatrix} a & 0 \\ 0 & 1/3 \end{pmatrix}[/tex]

    [tex]B=\begin{pmatrix} 1/2 & b \\ 1/b & 0 \end{pmatrix}[/tex]

    [tex]C=\begin{pmatrix} 1 & c \\ 1/(4c) & 1/2 \end{pmatrix}[/tex]

    The matrices must be Hermitian but I'm unable to use that condition to get rid of the variables unless the entries are real, and I don't see why that must be the case.
     
  5. Jun 24, 2011 #4

    vela

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    Yup.
    Looks good so far, though you should recheck <2|C|2>. I got that it was equal to 1. For matrix B, you found bc=1, and because the matrix is Hermitian, you must also have c=b*. So you get bc=bb*=|b|2=1; therefore, b can be written as e for some θ. You can do a similar thing for matrix C.

    With these matrices, you should be able to find all the eigenvalues, and with the matrix H for the Hamiltonian, you should be able to show whether the operators A, B, and C commute with the Hamiltonian. Note that B and C can commute with H if a certain condition holds.
     
  6. Jun 24, 2011 #5
    vela, how do you know <2|C|2> is 1?

    Anyway, as of now, I can find the eigenvalues for B and C by the usual way but I cannot do it for A because the entry a is unknown. So is this as far as we can go for eigenvalues or is there a way to find out the second eigenvalue of A?

    Now, regarding the commutation with H, I thought B and C definitely don't commute with H? If they did, |1> and |2> would be eigenstates of B and C but they are not. With A, I realize it does commute.

    Thank you for your help.
     
  7. Jun 24, 2011 #6

    vela

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    I calculated it. How did you get <2|C|2>=1/2?
    That's the best you can do.
    That's not necessarily true. Suppose, for example, the matrix for H was the identity matrix. Clearly, H would then commute with B and C.

    H usually won't commute with B and C, but it will under one condition. You should be able to deduce what that condition is.
     
  8. Jun 24, 2011 #7
    Ok I see my C matrix was wrong.

    Regarding the condition, well now that you pointed it out, any multiple of the identity matrix would allow them to commute i.e. the two states have the same eigenvalue for H. Or is there a stronger condition?
     
    Last edited: Jun 24, 2011
  9. Jun 24, 2011 #8

    vela

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    You can deduce what the exact condition is mathematically. Just calculate HB and BH and figure out under what conditions they are equal.
     
  10. Jun 25, 2011 #9
    Ok here is what I did.

    [tex]H=\begin{pmatrix} p & 0 \\ 0 & q \end{pmatrix}[/tex]

    [tex]B=\begin{pmatrix} 1/2 & b \\ 1/b & 0 \end{pmatrix}[/tex]

    [tex]HB=\begin{pmatrix} p/2 & pb \\ q/b & 0 \end{pmatrix}[/tex]

    [tex]BH=\begin{pmatrix} p/2 & qb \\ p/b & 0 \end{pmatrix}[/tex]

    So, they commute if p=q which is what I got previously.
     
  11. Jun 25, 2011 #10

    vela

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    Right, and you can see there's no other condition that allows H and B to commute.
     
  12. Jun 25, 2011 #11
    Thanks very much vela.
     
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