Quantum Fundamentals: Winter-2026
HW 7 (SOLUTION): Due W4 D3
- Spin One Interferometer Brief
S1 5421S
Consider a spin 1 interferometer which prepares the state as \(| 1\rangle\), then sends this state through an \(S_x\) apparatus and then an \(S_z\) apparatus. For the four possible cases where a pair of beams or all three beams from the \(S_x\) Stern-Gernach analyzer are used, calculate the probabilities that a particle entering the last Stern-Gerlach device will be measured to have each possible value of \(S_z\). Compare your theoretical calculations to results of the simulation. Make sure that you explicitly discuss your choice of projection operators.
Note: You do not need to do the first case, as we have done it in class.
In the case of combining the \(\hbar\) and \(0\hbar\) beams, the relevant projection operator is:
\[P_{10}=\left|{1}\right\rangle _x{}_x\left\langle {1}\right|+\left|{0}\right\rangle _x{}_x\left\langle {0}\right|\]
Therefore, the state can be written as: \begin{eqnarray*} \left|{\psi_{10}}\right\rangle &=& \frac{P_{10} \left|{1}\right\rangle }{\sqrt{\left\langle {1}\right|P_{10}\left|{1}\right\rangle }}\\ &=& \frac{\left(\left|{1}\right\rangle _x{}_x\left\langle {1}\right|+\left|{0}\right\rangle _x{}_x\left\langle {0}\right|\right) \left|{1}\right\rangle }{\sqrt{\left\langle {1}\right|\left(\left|{1}\right\rangle _x{}_x\left\langle {1}\right|+\left|{0}\right\rangle _x{}_x\left\langle {0}\right|\right)\left|{1}\right\rangle }}\\ \end{eqnarray*} Distribute the \(\left\langle {1}\right|\) and \(\left|{1}\right\rangle \) through the parentheses:
\[ \left|{\psi_{10}}\right\rangle = \frac{\left|{1}\right\rangle _x{}_x\left\langle {1}\middle|{1}\right\rangle +\left|{0}\right\rangle _x{}_x\left\langle {0}\middle|{1}\right\rangle }{\sqrt{\left\langle {1}\middle|{1}\right\rangle _x{}_x\left\langle {1}\middle|{1}\right\rangle +\left\langle {1}\middle|{0}\right\rangle _x{}_x\left\langle {0}\middle|{1}\right\rangle }}\\ \]
Evaluate the inner products by using the Spins Reference Sheet:
\[ \left|{\psi_{10}}\right\rangle = \frac{\left|{1}\right\rangle _x\frac{1}{2}+\left|{0}\right\rangle _x\frac{1}{\sqrt{2}}}{\sqrt{\frac{1}{2}\frac{1}{2}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}}}\\ \]
Simplifying: \begin{eqnarray*} \left|{\psi_{10}}\right\rangle &=& \frac{\frac{1}{2}\left|{1}\right\rangle _x+\frac{1}{\sqrt{2}}\left|{0}\right\rangle _x}{\sqrt{\frac{3}{4}}}\\ &=& \frac{1}{\sqrt{3}}\left|{1}\right\rangle _x+\sqrt{\frac{2}{3}}\left|{0}\right\rangle _x \end{eqnarray*} The probabilities of measuring specific values of the z-component of spin in the 3rd analyzer are: \begin{eqnarray*} \mathcal{P}(S_z=\hbar) &=& |\left\langle {1}\middle|{\psi_{10}}\right\rangle |^2 \\ &=& \left|\frac{1}{\sqrt{3}}\left\langle {1}\middle|{1}\right\rangle _x+\sqrt{\frac{2}{3}}\left\langle {1}\middle|{0}\right\rangle _x\right|^2\\ &=& \left|\frac{1}{\sqrt{3}}\frac{1}{2}+\sqrt{\frac{2}{3}}\frac{1}{\sqrt{2}}\right|^2\\ &=& \left|\frac{1}{\sqrt{12}}+\frac{2}{\sqrt{12}}\right|^2\\ &=& \frac{3}{4}\\ \mathcal{P}(S_z=0\hbar) &=& |\left\langle {0}\middle|{\psi_{10}}\right\rangle |^2 \\ &=& \left|\frac{1}{\sqrt{3}}\left\langle {0}\middle|{1}\right\rangle _x+\sqrt{\frac{2}{3}}\left\langle {0}\middle|{0}\right\rangle _x\right|^2\\ &=& \left|\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}\right|^2\\ &=& \frac{1}{6}\\ \mathcal{P}(S_z=-\hbar) &=& |\left\langle {-1}\middle|{\psi_{10}}\right\rangle |^2 \\ &=& \left|\frac{1}{\sqrt{3}}\left\langle {-1}\middle|{1}\right\rangle _x+\sqrt{\frac{2}{3}}\left\langle {-1}\middle|{0}\right\rangle _x\right|^2\\ &=& \left|\frac{1}{\sqrt{3}}\frac{1}{2}+\sqrt{\frac{2}{3}}\frac{-1}{\sqrt{2}}\right|^2\\ &=& \frac{1}{12} \end{eqnarray*} For the other interferometers, the projection operators, \(\left|{\psi}\right\rangle \) and probabilities are: \begin{eqnarray*} P_{0-1} &=& \left|{0}\right\rangle _x{}_x\left\langle {0}\right|+\left|{-1}\right\rangle _x{}_x\left\langle {-1}\right| \\ \left|{\psi_{0-1}}\right\rangle &=& \sqrt{\frac{2}{3}}\left|{0}\right\rangle _x+\frac{1}{\sqrt{3}}\left|{-1}\right\rangle _x\\ \mathcal{P}(S_z=\hbar) &=& \frac{3}{4} \\ \mathcal{P}(S_z=0\hbar) &=& \frac{1}{6} \\ \mathcal{P}(S_z=-\hbar) &=& \frac{1}{12} \\ \end{eqnarray*} \begin{eqnarray*} P_{11} &=& \left|{1}\right\rangle _x{}_x\left\langle {1}\right|+\left|{-1}\right\rangle _x{}_x\left\langle {-1}\right| \\ \left|{\psi_{11}}\right\rangle &=& \frac{1}{\sqrt{2}}\left|{1}\right\rangle _x+\frac{1}{\sqrt{2}}\left|{-1}\right\rangle _x\\ \mathcal{P}(S_z=\hbar) &=& \frac{1}{2} \\ \mathcal{P}(S_z=0\hbar) &=& 0 \\ \mathcal{P}(S_z=-\hbar) &=& \frac{1}{2} \\ \end{eqnarray*} \begin{eqnarray*} P_{101} &=& \left|{1}\right\rangle _x{}_x\left\langle {1}\right|+\left|{0}\right\rangle _x{}_x\left\langle {0}\right| +\left|{-1}\right\rangle _x{}_x\left\langle {-1}\right| \\ \left|{\psi_{101}}\right\rangle &=& \left|{1}\right\rangle \\ \mathcal{P}(S_z=\hbar) &=& 1 \\ \mathcal{P}(S_z=0\hbar) &=& 0 \\ \mathcal{P}(S_z=-\hbar) &=& 0 \\ \end{eqnarray*}
- Spin One Eigenvectors
S1 5421S
The operator \(\hat{S}_x\) for spin-1 particles, can be written in matrix form in the \(S_z\) basis as:
\[\hat{S}_x=\frac{\hbar}{\sqrt{2}}
\begin{pmatrix}
0&1&0\\ 1&0&1 \\ 0&1&0 \\
\end{pmatrix}
\]
- Find the eigenvalues of this matrix.
To find the eigenvalues, evaluate the following determinant. \begin{align*} \det\left(\hat{S}_x-\lambda {\cal I}\right) &=\det\left( \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0&1&0\\ 1&0&1 \\ 0&1&0 \\ \end{pmatrix} -\lambda \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}\right)\\ \\ &=\det\begin{pmatrix} -\lambda&\frac{\hbar}{\sqrt{2}}&0\\ \frac{\hbar}{\sqrt{2}}&-\lambda&\frac{\hbar}{\sqrt{2}}\\ 0&\frac{\hbar}{\sqrt{2}}&-\lambda \end{pmatrix}\\ \\ &=-\lambda^3+0+0+\frac{\hbar^2}{2}\lambda+\frac{\hbar^2}{2}\lambda+0\\ \\ &=-\lambda\left(\lambda^2-\hbar^2\right)\\ \\ \Rightarrow\lambda&=\left\{0, \pm\hbar\right\} \end{align*}
- Find the eigenvectors corresponding to each eigenvalue.
The eigenvector corresponding to \(\lambda=0\) is found by solving: \begin{align*} \begin{pmatrix} 0\\0\\0 \end{pmatrix} &=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0&1&0\\ 1&0&1 \\ 0&1&0 \\ \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix}\\ \\ &=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} y\\ x+z \\ y \\ \end{pmatrix} \end{align*} From which we can find that we must have \(y=0\) and \(z=-x\), i.e. the (un-normalized) eigenvector is \[ \begin{pmatrix} 1\\0\\-1\end{pmatrix}\]
The eigenvectors corresponding to \(\lambda=\pm\hbar\) are found by solving: \begin{align*} \pm\hbar\begin{pmatrix} x\\y\\z \end{pmatrix} &=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0&1&0\\ 1&0&1 \\ 0&1&0 \\ \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix}\\ \\ &=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} y\\ x+z \\ y \\ \end{pmatrix} \end{align*} From which we can find that we must have \begin{align*} \pm x&=\frac{1}{\sqrt{2}}y\\ \pm z&=\frac{1}{\sqrt{2}}y\\ \pm y&=\frac{1}{\sqrt{2}}(x+z) \end{align*} If we choose \(y=\sqrt{2}\), then the first two equations say \(x=z=\pm 1\) and the third equation is automatically satisfied. Therefore the (un-normalized) eigenvectors are \[ \begin{pmatrix} \pm 1\\\sqrt{2}\\\pm 1\end{pmatrix}\]
- Find the eigenvalues of this matrix.
- Finding Matrix Elements
S1 5421S
Carry out the following matrix calculations.
\[ \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33}\cr\end{pmatrix} \begin{pmatrix}1\cr0\cr0\cr\end{pmatrix}\] and
\[\begin{pmatrix} 0 & 1 & 0\end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33}\end{pmatrix} \begin{pmatrix}0\cr1\cr0\cr\end{pmatrix} \]
\begin{align*} \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33}\cr\end{pmatrix} \begin{pmatrix}1\cr0\cr0\cr\end{pmatrix} &=\begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}a_{11}\\a_{21}\\a_{31}\end{pmatrix}\\ &=a_{21} \end{align*} \begin{align*} \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33}\cr\end{pmatrix} \begin{pmatrix}0\cr1\cr0\cr\end{pmatrix} &=\begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}a_{12}\\a_{22}\\a_{32}\end{pmatrix}\\ &=a_{22} \end{align*}
What matrix multiplication would you do if you wanted the answer to be \(a_{31}\)?
The combination (row)(matrix)(column) where the row and column are elements of the standard basis picks out one element of the matrix--called a “matrix element.” The row matrix at the left of the initial expression tells us which row we will pick out and the column matrix at the right of the initial expression tells us which column will be picked out. So, if we want to find \(a_{31}\), you need a row matrix with a 1 in the third place and a column matrix with a 1 in the first place, i.e. \[ a_{31} =\begin{pmatrix} 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33}\cr\end{pmatrix} \begin{pmatrix}1\cr0\cr0\cr\end{pmatrix} \]
In the first question above, the bra/ket representations for the calulations are:
\[\left\langle {2}\right| A\left|{1}\right\rangle = ? \quad \hbox{and} \quad \left\langle {2}\right| A\left|{2}\right\rangle = ?\]
Write the second question in bra/ket notation.
We are trying to find \(a_{31}\), so we need to pick out the third row and first column. So we need to use the third standard basis element for the “bra” and the first standard basis element for the “ket,” i.e. \[ \left\langle {3}\right| A \left|{1}\right\rangle =a_{31} \]
- Spin Three Halves Operators
S1 5421S
If a beam of spin-3/2 particles is input to a Stern-Gerlach analyzer, there are four output beams whose deflections are consistent with magnetic moments arising from spin angular momentum components of \(\frac{3}{2}\hbar\), \(\frac{1}{2}\hbar\), \(-\frac{1}{2}\hbar\), and \(-\frac{3}{2}\hbar\). For a spin-3/2 system:
- Write down the eigenvalue equations for the \(\hat S_z\) operator.
I have been told in the statement fot the problem that the possible measurements, and therefore the eigenvalues, are \(\frac{3}{2}\hbar\), \(\frac{1}{2}\hbar\), \(-\frac{1}{2}\hbar\), and \(-\frac{3}{2}\hbar\). Therefore, the eigenvalue equations are: \begin{eqnarray*} \hat S_z\vert \frac{3}{2}\rangle &=& \frac{3}{2}\hbar \vert \frac{3}{2}\rangle \\ \hat S_z\vert \frac{1}{2}\rangle &=& \frac{1}{2}\hbar \vert \frac{1}{2}\rangle \\ \hat S_z\vert -\frac{1}{2}\rangle &=& -\frac{1}{2}\hbar \vert -\frac{1}{2}\rangle \\ \hat S_z\vert -\frac{3}{2}\rangle &=& -\frac{3}{2}\hbar \vert -\frac{3}{2}\rangle \\ \end{eqnarray*}
- Write down the matrix representation of the \(\hat S_z\) eigenstates in the \(S_z\) basis.
There are four eigenvalues and four independent eigenvectors. Therefore all the matrices will have 4 rows and or columns. In the \(S_z\) basis, its own eigenstates are just the standard basis: \[\left|{\frac{3}{2}}\right\rangle \doteq \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \left|{\frac{1}{2}}\right\rangle \doteq \begin{pmatrix} 0\\ 1\\ 0\\ 0 \end{pmatrix}, \left|{-\frac{1}{2}}\right\rangle \doteq \begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix}, \left|{-\frac{3}{2}}\right\rangle \doteq \begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}, \]
- Write down the matrix representation of the \(\hat S_z\) operator in the \(S_z\) basis.
In its own basis, the \(\hat S_z\) operator is diagonal and its diagonal entries are its eigenvalues: \[\hat S_z\doteq \hbar \begin{pmatrix} \frac{3}{2}&0&0&0\\0&\frac{1}{2}&0&0\\0&0&-\frac{1}{2}&0\\0&0&0&-\frac{3}{2} \end{pmatrix} \]
- Write down the eigenvalue equations for the \(\hat {S^2}\) operator. (The eigenvalues of the \(S^2\) are \(\hbar^2s(s+1)\), where \(s\) is the spin quantum number. \(\hat {S^2}=(\hat S_x)^2+(\hat S_y)^2+(\hat S_z)^2\), which is proportional to the identify operator. For spin-3/2 system, \(s=\frac{3}{2}\))
Because the \(\hat {S^2}\) operator is proportional to the identify operator, any vector is an eigenvector of the \(\hat {S^2}\) operator. For this system the spin quantum number is 3/2. \[\hat {S^2}\vert \psi\rangle=s(s+1)\hbar^2 \vert \psi\rangle =\frac{15}{4}\hbar^2 \vert \psi\rangle \]
- Write down the matrix representation of the \(\hat {S^2}\) operator in the \(S_z\) basis. Check Beasts: Is your operator proportional to the identity operator?
In its own basis, the \(\hat {S^2}\) operator is diagonal and its diagonal entries are its eigenvalues. There is only one (degenerate) eigenvalue, so \(\hat {S^2}\) is a multiple of the identity: \[S^2\doteq\frac{15}{4}\hbar^2 \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{pmatrix} \]
- Write down the eigenvalue equations for the \(\hat S_z\) operator.