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How Can I Write an Operator in the Coupled and Uncoupled Basis?

  1. Aug 1, 2014 #1
    Given a system of two identical particles (let's say electrons), of (max) spin 1/2 (which means the magnetic quantum number of each of the electrons can be either 1/2 or -1/2), how can we write the operators (total angular momentum, z-component of the total angular momentum etc.) (a) in the uncoupled basis and (b) in the uncoupled basis?

    Please be specific and give numerical examples.

    First, my problem is that I don't know how to write an operator in either the coupled or the uncoupled basis, and I really searched the internet and a few well-known quantum mechanics books (e.g. Griffiths). Please help.
     
  2. jcsd
  3. Aug 1, 2014 #2
    It doesn't matter if your two particles are identical or not, your total operator is the sum of the operators acting on each particle:
    $$\vec{L}_{tot}=\vec{L}_1+\vec{L}_2$$

    The square of this is:

    $$\vec{L}^2_{tot}=\vec{L}^2_1+\vec{L}^2_2+2\vec{L}_1 \cdot \vec{L}_2$$

    One way to calculate the matrix elements is to rewrite

    $$\vec{L}^2_{tot}=\vec{L}^2_1 I_2+I_1\vec{L}^2_2+2\vec{L}_1 \cdot \vec{L}_2$$

    Now consider the first term [itex]\vec{L}^2_1 I_2[/itex]. First write the matrix element acting on the 1st particle:

    $$\begin{pmatrix}3\hbar^2/4 & 0 \\
    0 &3\hbar^2/4 \end{pmatrix}$$

    Them multiply each element of that matrix with the matrix element acting on the 2nd particle:

    $$\begin{pmatrix}3\hbar^2/4 \begin{pmatrix}1 & 0 \\ 0&1\end{pmatrix} & 0 \begin{pmatrix}1 & 0 \\ 0&1\end{pmatrix} \\
    0\begin{pmatrix}1 & 0 \\ 0&1\end{pmatrix} &3\hbar^2/4\begin{pmatrix}1 & 0 \\ 0&1\end{pmatrix} \end{pmatrix}=
    \begin{pmatrix}3\hbar^2/4 & 0 & 0 & 0\\
    0 & 3\hbar^2/4 & 0 & 0\\
    0 & 0 & 3\hbar^2/4 & 0 \\
    0 &0&0& 3\hbar^2/4
    \end{pmatrix}
    $$

    Do this for each of the terms, and add them all up.
     
  4. Aug 1, 2014 #3
    addendum: I didn't read your question carefully. For the total angular momentum operator, you'd write:

    $$\vec{J}_{tot}^2=(\vec{L_1}+\vec{S_1}+\vec{L_2}+\vec{S_2})^2$$

    Expand it out. Then you'd do the same thing, except you'll have 4 nested matrices instead of 2. The basis would be the uncoupled basis.
     
  5. Aug 2, 2014 #4
    Hi geoduck and thank you very much for taking the time to explain.

    Could you explain what ## I_2 ## stands for? (and also what is ## I_1 ## )

    Also, I did not understand how you start from 2 X 2 matrices and end up with 4 X 4 matrices, as well as *why* do you need to do this. Could you explain in more detail?
     
  6. Aug 2, 2014 #5
    Also, what is the reason behind using the unitary matrix and not some other matrix in your calculation above?

    Again, sorry if this seem like a dumb question but I really need to understand.
     
  7. Aug 2, 2014 #6
    If you only consider spin and you have a single electron, it can be described by |+> or |->. Therefore a matrix operating on these states is 2x2. If you have two electrons, you need to specify the state of the first electron and the state of the second electron, so you need 4 vectors: |++>, |+->, |-+>, and |-->. A matrix operating on these states would then be 4x4. If you include orbital angular momentum of the first and second particles also, you'd have states |l1 m1s1l2 m2 s2>. So your matrices keep on getting larger.

    The fact that the electrons are identical can reduce the dimensionality because you only need the action of the operator on antisymmetric states.

    I1 and I2 are the identity operators on particle 1, and particle 2, respectively.
     
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