# Peeter Joot's (OLD) Blog.

• ## Recent Comments

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 294,114

## PHY456H1F: Quantum Mechanics II. Lecture 14 (Taught by Prof J.E. Sipe). Representation of two state kets and Pauli spin matrices.

Posted by peeterjoot on October 26, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Representation of kets.

Reading: section 5.1 – section 5.9 and section 26 in [1].

We found the representations of the spin operators

\begin{aligned}S_x &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ S_y &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ S_z &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.1)

How about kets? For example for ${\left\lvert {\chi} \right\rangle} \in H_s$

\begin{aligned}{\left\lvert {\chi} \right\rangle} \rightarrow \begin{bmatrix}\left\langle{{+}} \vert {{\chi}}\right\rangle \\ \left\langle{{-}} \vert {{\chi}}\right\rangle\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.4)

and

\begin{aligned}{\left\lvert {+} \right\rangle} &\rightarrow \begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\left\lvert {0} \right\rangle} &\rightarrow \begin{bmatrix}0 \\ 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

So, for example

\begin{aligned}S_y{\left\lvert {+} \right\rangle} \rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix} =\frac{i\hbar}{2}\begin{bmatrix}0 \\ 1\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(2.7)

Kets in $H_o \otimes H_s$

\begin{aligned}{\left\lvert {\psi} \right\rangle} \rightarrow \begin{bmatrix}\left\langle{{\mathbf{r}+}} \vert {{\psi}}\right\rangle \\ \left\langle{{\mathbf{r}-}} \vert {{\psi}}\right\rangle\end{bmatrix}=\begin{bmatrix}\psi_{+}(\mathbf{r}) \\ \psi_{-}(\mathbf{r})\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.8)

This is a “spinor”

Put

\begin{aligned}\begin{aligned}\left\langle{{\mathbf{r} \pm}} \vert {{\psi}}\right\rangle&= \psi_{\pm}(\mathbf{r}) \\ &= \psi_{+} \begin{bmatrix}1 \\ 0\end{bmatrix}+\psi_{-} \begin{bmatrix}0 \\ 1 \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.9)

with

\begin{aligned}\left\langle{{\psi}} \vert {{\psi}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.10)

Use

\begin{aligned}\begin{aligned}I &= I_o \otimes I_s \\ &= \int d^3 \mathbf{r} {\left\lvert {\mathbf{r}} \right\rangle}{\left\langle {\mathbf{r}} \right\rvert} \otimes \left( {\left\lvert {{+}} \right\rangle}{\left\langle {{+}} \right\rvert}+{\left\lvert {{-}} \right\rangle}{\left\langle {{-}} \right\rvert}\right) \\ &=\int d^3 \mathbf{r} {\left\lvert {\mathbf{r}} \right\rangle}{\left\langle {\mathbf{r}} \right\rvert} \otimes \sum_{\sigma=\pm} {\left\lvert {{\sigma}} \right\rangle}{\left\langle {{\sigma}} \right\rvert} \\ &=\sum_{\sigma = \pm} \int d^3 \mathbf{r} {\left\lvert {{\mathbf{r} \sigma}} \right\rangle}{\left\langle {{\mathbf{r} \sigma}} \right\rvert} \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.11)

So

\begin{aligned}\begin{aligned}{\left\langle {\psi} \right\rvert} I {\left\lvert {\psi} \right\rangle} &= \sum_{\sigma = \pm} \int d^3 \mathbf{r} \left\langle{{\psi}} \vert {{\mathbf{r} \sigma}}\right\rangle \left\langle{{\mathbf{r} \sigma}} \vert {{\psi}}\right\rangle \\ &= \int d^3 \mathbf{r} \left( {\left\lvert{\psi_{+}(\mathbf{r})}\right\rvert}^2+{\left\lvert{\psi_{-}(\mathbf{r})}\right\rvert}^2\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.12)

Alternatively

\begin{aligned}\begin{aligned}{\left\lvert {\psi} \right\rangle} &= I {\left\lvert {\psi} \right\rangle} \\ &=\int d^3 \mathbf{r} \sum_{\sigma = \pm} {\left\lvert {\mathbf{r} \sigma} \right\rangle}\left\langle{{\mathbf{r} \sigma}} \vert {{\psi}}\right\rangle \\ &=\sum_{\sigma = \pm} \left(\int d^3 \mathbf{r} \psi_\sigma(\mathbf{r})\right){\left\lvert {\mathbf{r} \sigma} \right\rangle} \\ &=\sum_{\sigma = \pm} \left(\int d^3 \mathbf{r} \psi_\sigma(\mathbf{r}) {\left\lvert {\mathbf{r}} \right\rangle}\right)\otimes {\left\lvert {\sigma} \right\rangle} \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.13)

In braces we have a ket in $H_o$, let’s call it

\begin{aligned}{\left\lvert {\psi_\sigma} \right\rangle} = \int d^3 \mathbf{r} \psi_\sigma(\mathbf{r}) {\left\lvert {\mathbf{r}} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.14)

then

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_{+}} \right\rangle} {\left\lvert {+} \right\rangle} + {\left\lvert {\psi_{-}} \right\rangle} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.15)

where the direct product $\otimes$ is implied.

We can form a ket in $H_s$ as

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi}}\right\rangle = \psi_{+}(\mathbf{r}) {\left\lvert {+} \right\rangle} + \psi_{-}(\mathbf{r}) {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.16)

An operator $O_o$ which acts on $H_o$ alone can be promoted to $O_o \otimes I_s$, which is now an operator that acts on $H_o \otimes H_s$. We are sometimes a little cavalier in notation and leave this off, but we should remember this.

\begin{aligned}O_o {\left\lvert {\psi} \right\rangle} = (O_o {\left\lvert {\psi+} \right\rangle}) {\left\lvert {+} \right\rangle}+ (O_o {\left\lvert {\psi+} \right\rangle}) {\left\lvert {+} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.17)

and likewise

\begin{aligned}O_s {\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi+} \right\rangle} (O_s {\left\lvert {+} \right\rangle})+{\left\lvert {\psi-} \right\rangle} (O_s {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.18)

and

\begin{aligned}O_o O_s {\left\lvert {\psi} \right\rangle} = (O_o {\left\lvert {\psi+} \right\rangle}) (O_s {\left\lvert {+} \right\rangle})+(O_o {\left\lvert {\psi-} \right\rangle}) (O_s {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.19)

Suppose we want to rotate a ket, we do this with a full angular momentum operator

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}=e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.20)

(recalling that $\mathbf{L}$ and $\mathbf{S}$ commute)

So

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}=(e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} {\left\lvert {\psi+} \right\rangle}) (e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {+} \right\rangle})+(e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} {\left\lvert {\psi-} \right\rangle}) (e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.21)

## A simple example.

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_+} \right\rangle} {\left\lvert {+} \right\rangle}+{\left\lvert {\psi_-} \right\rangle} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.22)

Suppose

\begin{aligned}{\left\lvert {\psi_+} \right\rangle} &= \alpha {\left\lvert {\psi_0} \right\rangle} \\ {\left\lvert {\psi_-} \right\rangle} &= \beta {\left\lvert {\psi_0} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.23)

where

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 + {\left\lvert{\beta}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(2.25)

Then

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_0} \right\rangle} {\left\lvert {\chi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.26)

where

\begin{aligned}{\left\lvert {\chi} \right\rangle} = \alpha {\left\lvert {+} \right\rangle} + \beta {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.27)

for

\begin{aligned}\left\langle{{\psi}} \vert {{\psi}}\right\rangle = 1,\end{aligned} \hspace{\stretch{1}}(2.28)

\begin{aligned}\left\langle{{\psi_0}} \vert {{\psi_0}}\right\rangle \left\langle{{\chi}} \vert {{\chi}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.29)

so

\begin{aligned}\left\langle{{\psi_0}} \vert {{\psi_0}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.30)

We are going to concentrate on the unentagled state of 2.26.

\begin{itemize}
\item
How about with

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 = 1, \beta = 0\end{aligned} \hspace{\stretch{1}}(2.31)

${\left\lvert {\chi} \right\rangle}$ is an eigenket of $S_z$ with eigenvalue $\hbar/2$.
\item

\begin{aligned}{\left\lvert{\beta}\right\rvert}^2 = 1, \alpha = 0\end{aligned} \hspace{\stretch{1}}(2.32)

${\left\lvert {\chi} \right\rangle}$ is an eigenket of $S_z$ with eigenvalue $-\hbar/2$.

\item
What is ${\left\lvert {\chi} \right\rangle}$ if it is an eigenket of $\hat{\mathbf{n}} \cdot \mathbf{S}$?
\end{itemize}

FIXME: F1: standard spherical projection picture, with $\hat{\mathbf{n}}$ projected down onto the $x,y$ plane at angle $\phi$ and at an angle $\theta$ from the $z$ axis.

The eigenvalues will still be $\pm \hbar/2$ since there is nothing special about the $z$ direction.

\begin{aligned}\begin{aligned}\hat{\mathbf{n}} \cdot \mathbf{S} &= n_x S_x+n_y S_y+n_z S_z \\ &\rightarrow\frac{\hbar}{2} \begin{bmatrix}n_z & n_x - i n_y \\ n_x + i n_y & -n_z\end{bmatrix} \\ &=\frac{\hbar}{2} \begin{bmatrix}\cos\theta & \sin\theta e^{-i\phi}\sin\theta e^{i\phi} & -\cos\theta\end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.33)

To find the eigenkets we diagonalize this, and we find representations of the eigenkets are

\begin{aligned}{\left\lvert {\hat{\mathbf{n}}+} \right\rangle} &\rightarrow \begin{bmatrix}\cos\left(\frac{\theta}{2}\right) e^{-i\phi/2} \\ \sin\left(\frac{\theta}{2}\right) e^{i\phi/2} \end{bmatrix} \\ {\left\lvert {\hat{\mathbf{n}}-} \right\rangle} &\rightarrow \begin{bmatrix}-\sin\left(\frac{\theta}{2}\right) e^{-i\phi/2} \\ \cos\left(\frac{\theta}{2}\right) e^{i\phi/2} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.34)

with eigenvalues $\hbar/2$ and $-\hbar/2$ respectively.

So in the abstract notation, tossing the specific representation, we have

\begin{aligned}{\left\lvert {\hat{\mathbf{n}}+} \right\rangle} &\rightarrow \cos\left(\frac{\theta}{2}\right) e^{-i\phi/2} {\left\lvert {+} \right\rangle}\sin\left(\frac{\theta}{2}\right) e^{i\phi/2} {\left\lvert {-} \right\rangle} \\ {\left\lvert {\hat{\mathbf{n}}-} \right\rangle} &\rightarrow -\sin\left(\frac{\theta}{2}\right) e^{-i\phi/2} {\left\lvert {+} \right\rangle}\cos\left(\frac{\theta}{2}\right) e^{i\phi/2} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.36)

# Representation of two state kets

Every ket

\begin{aligned}{\left\lvert {\chi} \right\rangle} \rightarrow \begin{bmatrix}\alpha \\ \beta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.38)

for which

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 + {\left\lvert{\beta}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(3.39)

can be written in the form 2.34 for some $\theta$ and $\phi$, neglecting an overall phase factor.

For any ket in $H_s$, that ket is “spin up” in some direction.

FIXME: show this.

# Pauli spin matrices.

It is useful to write

\begin{aligned}S_x &= \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_x \\ S_y &= \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_y \\ &= \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_z \end{aligned} \hspace{\stretch{1}}(4.40)

where

\begin{aligned}\sigma_x &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ \sigma_y &= \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ \sigma_z &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.43)

These are the Pauli spin matrices.

## Interesting properties.

\begin{itemize}
\item

\begin{aligned}\left[{\sigma_i},{\sigma_j}\right] = \sigma_i \sigma_j + \sigma_j \sigma_i = 0, \qquad \mbox{ iflatex i < j}\end{aligned} \hspace{\stretch{1}}(4.46)

\item

\begin{aligned}\sigma_x \sigma_y = i \sigma_z\end{aligned} \hspace{\stretch{1}}(4.47)

(and cyclic permutations)

\item

\begin{aligned}\text{Tr}(\sigma_i) = 0\end{aligned} \hspace{\stretch{1}}(4.48)

\item

\begin{aligned}(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2 = \sigma_0\end{aligned} \hspace{\stretch{1}}(4.49)

where

\begin{aligned}\hat{\mathbf{n}} \cdot \boldsymbol{\sigma} \equiv n_x \sigma_x + n_y \sigma_y + n_z \sigma_z,\end{aligned} \hspace{\stretch{1}}(4.50)

and

\begin{aligned}\sigma_0 = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.51)

(note $\text{Tr}(\sigma_0) \ne 0$)

\item

\begin{aligned}\left[{\sigma_i},{\sigma_j}\right] &= 2 \delta_{ij} \sigma_0 \\ \left[{\sigma_x},{\sigma_y}\right] &= 2 i \sigma_z\end{aligned} \hspace{\stretch{1}}(4.52)

(and cyclic permutations of the latter).

Can combine these to show that

\begin{aligned}(\mathbf{A} \cdot \boldsymbol{\sigma})(\mathbf{B} \cdot \boldsymbol{\sigma})=(\mathbf{A} \cdot \mathbf{B}) \sigma_0 + i (\mathbf{A} \times \mathbf{B}) \cdot \boldsymbol{\sigma}\end{aligned} \hspace{\stretch{1}}(4.54)

where $\mathbf{A}$ and $\mathbf{B}$ are vectors (or more generally operators that commute with the $\boldsymbol{\sigma}$ matrices).

\item

\begin{aligned}\text{Tr}(\sigma_i \sigma_j) = 2 \delta_{ij}\end{aligned} \hspace{\stretch{1}}(4.55)

\item

\begin{aligned}\text{Tr}(\sigma_\alpha \sigma_\beta) = 2 \delta_{\alpha \beta},\end{aligned} \hspace{\stretch{1}}(4.56)

where $\alpha, \beta = 0, x, y, z$
\end{itemize}

Note that any complext matrix $M$ can be written as

\begin{aligned}\begin{aligned}M &= \sum_\alpha m_a \sigma_\alpha \\ &=\begin{bmatrix}m_0 + m_z & m_x - i m_y \\ m_x + i m_y & m_0 - m_z\end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.57)

for any four complex numbers $m_0, m_x, m_y, m_z$

where

\begin{aligned}m_\beta = \frac{1}{{2}} \text{Tr}(M \sigma_\beta).\end{aligned} \hspace{\stretch{1}}(4.58)

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

Advertisements