Activity: Raising and Lowering Operators for Spin

Central Forces 2023 (2 years)
What students learn This activity is the same as problem 942.

For \(\ell=1\), the operators that measure the three components of angular momentum in matrix notation are given by: \begin{align} L_x&=\frac{\hbar}{\sqrt{2}}\left( \begin{matrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{matrix} \right)\\ L_y&=\frac{\hbar}{\sqrt{2}}\left( \begin{matrix} 0&-i&0\\ i&0&-i\\ 0&i&0 \end{matrix} \right)\\ L_z&=\;\;\;\hbar\left( \begin{matrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{matrix} \right) \end{align}

Show that:

  1. Find the commutator of \(L_x\) and \(L_y\).
  2. Find the matrix representation of \(L^2=L_x^2+L_y^2+L_z^2\).
  3. Find the matrix representations of the raising and lowering operators \(L_{\pm}=L_x\pm iL_y\).
  4. Show that \([L_z, L_{\pm}]=\lambda L_{\pm}\). Find \(\lambda\). Interpret this expression as an eigenvalue equation. What is the operator?
  5. Let \(L_{+}\) act on the following three states given in matrix representation. \begin{equation} \left|{1,1}\right\rangle =\left( \begin{matrix} 1\\0\\0 \end{matrix} \right)\qquad \left|{1,0}\right\rangle =\left( \begin{matrix} 0\\1\\0 \end{matrix} \right)\qquad \left|{1,-1}\right\rangle =\left( \begin{matrix} 0\\0\\1 \end{matrix} \right) \end{equation} Why is \(L_{+}\) called a “raising operator”?

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    Writing an operator in matrix notation in its own basis is easy: it is diagonal with the eigenvalues on the diagonal.

    What if I want to calculate the matrix elements using a different basis??

    The eigenvalue equation tells me what happens when an operator acts on its own eigenstate. For example: \(\hat{S}_y\left|{\pm}\right\rangle _y=\pm\frac{\hbar}{2}\left|{\pm}\right\rangle _y\)

    In Dirac bra-ket notation, to know what an operator does to a ket, I needs to write the ket in the basis that is the eigenstates of the operator (in order to use the eigenvalue equation.)

    One way to do this to stick completeness relationships into the braket: \begin{eqnarray*} \left\langle {+}\right|\hat{S_y}\left|{+}\right\rangle = \left\langle {+}\right|(I)\hat{S_y}(I)\left|{+}\right\rangle \end{eqnarray*}

    where \(I\) is the identity operator: \(I=\color{blue}{\left|{+}\right\rangle _{yy}\left\langle {+}\right|}\;+\;\color{blue}{\left|{-}\right\rangle _{yy}\left\langle {-}\right|}\). This effectively rewrite the \(\left|{+}\right\rangle \) in the \(\left|{\pm}\right\rangle _y\) basis.

    Find the top row matrix elements of the operator \(\hat{S}_y\) in the \(S_z\) basis by inserting completeness relations into the brakets. (The answer is already on the Spins Reference Sheet, but I want you do demonstrate the calculation.)

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  • assignment Frequency

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  • assignment Phase

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      • \(z_1=i\),

      • \(z_2=2+2i\),
      • \(z_3=3-4i\).
    2. In quantum mechanics, it turns out that the overall phase for a state does not have any physical significance. Therefore, you will need to become quick at rearranging the phase of various states. For each of the vectors listed below, rewrite the vector as an overall complex phase times a new vector whose first component is real and positive. \[\left|D\right\rangle\doteq \begin{pmatrix} 7e^{i\frac{\pi}{6}}\\ 3e^{i\frac{\pi}{2}}\\ -1\\ \end{pmatrix}\\ \left|E\right\rangle\doteq \begin{pmatrix} i\\ 4\\ \end{pmatrix}\\ \left|F\right\rangle\doteq \begin{pmatrix} 2+2i\\ 3-4i\\ \end{pmatrix} \]
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Learning Outcomes