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Seemingly simple quantum mechanics/linear algebra problem

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    For some reason, I am making a trivial mistake somewhere. I just need to find the eigenvalues and normalized eigenvectors of the following matrix:

    H = (1/sqrt(2))*Matrix(Row 1 Row 2 Row 3)

    Row 1 = [0 -i 0]
    Row 2 = [i 3 3]
    Row 3 = [0 3 0]

    (Sorry, I don't know the proper latex code for a matrix)

    If someone can work this out and just provide the solutions so I can compare, I would really appreciate it.





    2. Relevant equations

    ....



    3. The attempt at a solution

    I tried.
     
  2. jcsd
  3. Oct 22, 2009 #2
    Use the normal eigenvalue equation:

    [tex]
    \det(H-\lambda\mathbb{I})=0
    [/tex]

    Solve for [tex]\lambda[/tex] (you should have 3 values since you have a [tex]3\times3[/tex] matrix).

    For the eigenvectors,

    [tex]
    \left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\lambda\left(\begin{array}{c}x\\y\\z\end{array}\right)
    [/tex]

    where you have to do this as many times as you have [tex]\lambda[/tex] values. That should help you a lot

    PS: to write matrices in LaTeX, use the command \begin{array} & \end{array}. The following produced the matrix above:

    \left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)
     
  4. Oct 22, 2009 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Click on the [itex]\LaTeX[/itex] image below to see the code that generated it

    [tex]H=\frac{1}{\sqrt{2}}\begin{pmatrix}0 & -i & 0 \\ i & 3 & 3 \\ 0 & 3 & 0 \end{pmatrix}[/tex]

    You've been here long enough to know that isn't what we do here.

    Then show us!
     
  5. Oct 22, 2009 #4
    I forgot the [tex]\frac{1}{\sqrt{2}}[/tex] constant in front of my matrix for the eigenvectors, it should read:


    [tex]
    \frac{1}{\sqrt{2}}
    \left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)\left(\begin{array}{c}x\\y\ \z\end{array}\right)=\lambda\left(\begin{array}{c} x\\y\\z\end{array}\right)

    [/tex]
     
  6. Oct 22, 2009 #5
    Thanks guys.

    For my eigenvalues, I got [tex]\lambda = 0, -2, 5[/tex]

    For the last two eigenvalues, when I set up the linear equations to solve I get something of the sort [tex]cy = cy[/tex] for one of them. What does that mean for [tex]y[/tex]? When the authors of my textbook do an example like this and run into the same situation, they postulate that [tex]y = 1[/tex]. But how is that when the variable cancels out on both sides?

    Thanks again.
     
  7. Oct 22, 2009 #6

    gabbagabbahey

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    Homework Helper
    Gold Member

    Try it again without forgetting the factor of [itex]1/\sqrt{2}[/itex]....:wink:
     
  8. Oct 23, 2009 #7
    Doesn't that answer your question? If the authors are allowed to set a canceled variable to be 1, doesn't that mean you can too? Usually you let the canceled variable be 1 for simplicity, but you can make it a billion if you really wanted, it'll just make your later calculations a bit more difficult.

    But as Gabba suggested, you should add in that [tex]1/\sqrt{2}[/tex] factor to find your eigenvalues before you turn in your assignment.
     
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