• 358,603

# Posts Tagged ‘addition of angular momentum’

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## PHY452H1S Basic Statistical Mechanics. Lecture 13: Interacting spin. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 5, 2013

[Click here for a PDF of this post with nicer formatting (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.

# Interacting spin

For these notes

\begin{aligned}\boxed{\hbar = k_{\mathrm{B}} = 1}\end{aligned}

This lecture requires concepts from phy456 [1].

We’ll look at pairs of spins as a toy model of interacting spins as depicted in fig. 1.1.

Fig 1.1: Pairs of interacting spins

Example:

Simple atomic system, with the nucleus and the electron can interact with each other (hyper-fine interaction).

Consider two interacting spin $1/2$ operators $\mathbf{S}$ each with components $\hat{S}^x$, $\hat{S}^y$, $\hat{S}^z$

\begin{aligned}H = J \mathbf{S}_1 \cdot \mathbf{S}_2 - B (\hat{S}_1^z + \hat{S}_2^z)\end{aligned} \hspace{\stretch{1}}(1.2.1)

\begin{aligned}\hat{S}_1^z + \hat{S}_2^z \propto \mbox{magnetization along z}\end{aligned} \hspace{\stretch{1}}(1.2.2)

We rewrite the dot product term of the Hamiltonian in terms of just the squares of the spin operators

\begin{aligned}H = J \frac{(\mathbf{S}_1 + \mathbf{S}_2)^2 - \mathbf{S}_1^2 - \mathbf{S}_2^2}{2}- B (\hat{S}_1^z + \hat{S}_2^z)\end{aligned} \hspace{\stretch{1}}(1.2.3)

The squares $\mathbf{S}_1^2$, $\mathbf{S}_2^2$, $(\mathbf{S}_1 + \mathbf{S}_2)^2$ can be thought of as “length”s of the respective angular momentum vectors.

We write

\begin{aligned}\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2,\end{aligned} \hspace{\stretch{1}}(1.2.4)

for the total angular momentum. We recall that we have

\begin{aligned}\hat{S}^z_2 = \hat{S}^z_1 = S(S + 1),\end{aligned} \hspace{\stretch{1}}(1.2.5)

where $S = 1/2$, and $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$ implies that $S_{\mathrm{total}} \in \{0, 2\}$.

$S_{\mathrm{total}} = 0$ (singlet).
$S_{\mathrm{total}} = 1$. Triplet: $(-1, 0, +1)$.

$S_{\mathrm{total}} = 0$ state.

For $m = 0$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \downarrow - \downarrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

energies

\begin{aligned}J \frac{-3/4 -3/4}{2} = -\frac{3}{4} J\end{aligned} \hspace{\stretch{1}}(1.2.7)

For $m = 1$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.8)

energies

\begin{aligned}J \left( 1 - \frac{3}{4} \right) - B \rightarrow \frac{J}{4} - B\end{aligned} \hspace{\stretch{1}}(1.2.9)

$S_{\mathrm{total}} = 1$ state

For $m = 0$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \downarrow + \downarrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.10)

energies

\begin{aligned}\frac{J}{4} \end{aligned} \hspace{\stretch{1}}(1.2.11)

For $m = 1$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \downarrow \downarrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.12)

energies

\begin{aligned}\frac{J}{4} + B.\end{aligned} \hspace{\stretch{1}}(1.2.13)

These are illustrated schematically in fig. 1.2.

Fig 1.2: Energy levels for two interacting spins as a function of magnetic field

Our single pair partition function is

\begin{aligned}Z_1 = e^{ +\beta 3 J/4}+e^{ -\beta (J/4 - B)}e^{ -\beta 3 J/4}+e^{ -\beta (J/4 + B)}\end{aligned} \hspace{\stretch{1}}(1.2.14)

So for $N$ pairs our partition function is

\begin{aligned}Z = Z_1^N = \left( e^{ +\beta 3 J/4} +e^{ -\beta (J/4 - B)} e^{ -\beta 3 J/4} +e^{ -\beta (J/4 + B)} \right)^N.\end{aligned} \hspace{\stretch{1}}(1.2.15)

Our free energy

\begin{aligned}F = - T \ln Z = - T N \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.2.16)

\begin{aligned}-\frac{\partial {F}}{\partial {\beta}} = T N \frac{\partial {}}{\partial {\beta}} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.2.17)

Our magnetization $\mu$ is

\begin{aligned}\mu = \frac{T N}{Z_1} \left( \beta e^{-\beta(J/4 - B)} -\beta e^{-\beta(J/4 + B)} \right)\end{aligned} \hspace{\stretch{1}}(1.2.18)

The moment per particle, after $T \beta$ cancellation, is

\begin{aligned}m = \frac{\mu}{N} = \frac{1}{Z_1} \left( e^{-\beta(J/4 - B)} -e^{-\beta(J/4 + B)} \right)=2 \frac{e^{-\beta J/4}}{Z_1} \sinh\left( \frac{B}{T} \right).\end{aligned} \hspace{\stretch{1}}(1.2.19)

Low temperatures, small $B$ ($T \ll J, B \ll J$)

The $e^{3 \beta J/4}$ term will dominate.

\begin{aligned}Z_1 \approx e^{3 J \beta/4}\end{aligned} \hspace{\stretch{1}}(1.2.20)

\begin{aligned}m \approx 2 e^{-\beta J} \sinh\left( \frac{B}{T} \right).\end{aligned} \hspace{\stretch{1}}(1.2.21)

Fig 1.3: magnetic moment

The specific heat has a similar behavior

\begin{aligned}C_V \sim e^{-\beta J}.\end{aligned} \hspace{\stretch{1}}(1.2.22)

Considering a single spin $1/2$ system, we have energies as illustrated in fig. 1.4.

Fig 1.4: Single particle spin energies as a function of magnetic field

At zero temperatures we have a finite non-zero magnetization as illustrated in fig. 1.5, but as we heat the system up, the state of the system will randomly switch between the 1, and 2 states. The partition function democratically averages over all such possible states.

Fig 1.6: Single spin magnetization

Once the system heats up, the spins are democratically populated within the entire set of possible states.

We contrast this to this interacting spins problem which has a magnetization of the form fig. 1.6.

Fig 1.6: Interacting spin magnetization

For the single particle specific heat we have specific heat of the form fig. 1.7.

Fig 1.7: Single particle specific heat

We’ll see the same kind of specific heat distribution with temperature for the interacting spins problem, but the peak will be found at an energy that’s given by the difference in energies of the two states as illustrated in fig. 1.8.

\begin{aligned}\Delta E = \frac{J}{4} - \frac{-3J}{4} = J\end{aligned} \hspace{\stretch{1}}(1.2.23)

# References

[1] Peeter Joot. Quantum Mechanics II., chapter: Two spin systems, angular momentum, and Clebsch-Gordon convention. URL http://sites.google.com/site/peeterjoot2/math2011/phy456.pdf.

## PHY456H1F: Quantum Mechanics II. Lecture 18 (Taught by Prof J.E. Sipe). The Clebsch-Gordon convention for the basis elements of summed generalized angular momentum

Posted by peeterjoot on November 14, 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.

# Recap.

Recall our table

\begin{aligned}\begin{array}{| l | l | l | l | l |} \hline j = & j_1 + j_2 & j_1 + j_2 -1 & \cdots & j_1 - j_2 \\ \hline \hline \right\rangle} & & & \\ \hline \right\rangle} \right\rangle} & & \\ \hline & \right\rangle} & & \\ \hline & \vdots & & \right\rangle} \\ \hline & \vdots & & & \vdots \\ \hline & \vdots & & \right\rangle} \\ \hline & \vdots & & & \\ \hline \right\rangle} \right\rangle} & & \\ \hline \right\rangle} & & & \\ \hline \end{array}\end{aligned} \hspace{\stretch{1}}(2.1)

## First column

Let’s start with computation of the kets in the lowest position of the first column, which we will obtain by successive application of the lowering operator to the state

\begin{aligned}{\left\lvert {j_1 + j_2, j_1 + j_2} \right\rangle} = {\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.2)

Recall that our lowering operator was found to be (or defined as)

\begin{aligned}J_{-} {\left\lvert {j, m} \right\rangle} = \sqrt{(j+m)(j-m+1)} \hbar {\left\lvert {j, m-1} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.3)

so that application of the lowering operator gives us

\begin{aligned}{\left\lvert {j_1 + j_2, j_1 + j_2 -1} \right\rangle} &= \frac{J_{-} {\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}}{\left(2 (j_1+ j_2)\right)^{1/2} \hbar} \\ &=\frac{(J_{1-} + J_{2-}) {\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}}{\left(2 (j_1+ j_2)\right)^{1/2} \hbar} \\ &=\frac{\left( \sqrt{(j_1 + j_1)(j_1 - j_1 + 1)} \hbar {\left\lvert {j_1(j_1 - 1)} \right\rangle} \right) \otimes {\left\lvert {j_2 j_2} \right\rangle}}{\left(2 (j_1+ j_2)\right)^{1/2} \hbar} \\ &\quad+\frac{{\left\lvert {j_1 j_1} \right\rangle} \otimes \left(\sqrt{(j_2 + j_2)(j_2 - j_2 + 1)} \hbar {\left\lvert {j_2(j_2 -1)} \right\rangle}\right)}{\left(2 (j_1+ j_2)\right)^{1/2} \hbar} \\ &=\left(\frac{j_1}{j_1 + j_2}\right)^{1/2}{\left\lvert {j_1 (j_1-1)} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}+\left(\frac{j_2}{j_1 + j_2}\right)^{1/2}{\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 (j_2-1)} \right\rangle} \\ \end{aligned}

Proceeding iteratively would allow us to finish off this column.

## Second column

Moving on to the second column, the top most element in the table

\begin{aligned}{\left\lvert {j_1 + j_2 - 1, j_1 + j_2 -1} \right\rangle} ,\end{aligned} \hspace{\stretch{1}}(2.4)

can only be made up of ${\left\lvert {j_1 m_1} \right\rangle} \otimes {\left\lvert {j_2 m_2} \right\rangle}$ with $m_1 + m_2 = j_1 + j_2 -1$. There are two possibilities

\begin{aligned}\begin{array}{l l l l}m_1 &= j_1 & m_2 &= j_2 - 1 \\ m_1 &= j_1 - 1 & m_2 &= j_2\end{array}\end{aligned} \hspace{\stretch{1}}(2.5)

So for some $A$ and $B$ to be determined we must have

\begin{aligned}{\left\lvert {j_1 + j_2 - 1, j_1 + j_2 -1} \right\rangle} =A{\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 (j_2-1)} \right\rangle}+B{\left\lvert {j_1 (j_1-1)} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.6)

Observe that these are the same kets that we ended up with by application of the lowering operator on the topmost element of the first column in our table. Since ${\left\lvert {j_1 + j_2, j_1 + j_2 -1} \right\rangle}$ and ${\left\lvert {j_1 + j_2 - 1, j_1 + j_2 -1} \right\rangle}$ are orthogonal, we can construct our ket for the top of the second column by just seeking such an orthonormal superposition. Consider for example

\begin{aligned}0 &=(a {\left\langle {b} \right\rvert} + c {\left\langle {d} \right\rvert})( A {\left\lvert {b} \right\rangle} + C {\left\lvert {d} \right\rangle}) \\ &=a A + c C\end{aligned}

With $A = 1$ we find that $C = -a/c$, so we have

\begin{aligned}A {\left\lvert {b} \right\rangle} + C {\left\lvert {d} \right\rangle} &= {\left\lvert {b} \right\rangle} - \frac{a}{c} {\left\lvert {d} \right\rangle} \\ &\simc {\left\lvert {b} \right\rangle} - a {\left\lvert {d} \right\rangle} \\ \end{aligned}

So we find, for real $a$ and $c$ that

\begin{aligned}0 = (a {\left\langle {b} \right\rvert} + c {\left\langle {d} \right\rvert})( c {\left\lvert {b} \right\rangle} - a {\left\lvert {d} \right\rangle}),\end{aligned} \hspace{\stretch{1}}(2.7)

for any orthonormal pair of kets ${\left\lvert {a} \right\rangle}$ and ${\left\lvert {d} \right\rangle}$. Using this we find

\begin{aligned}{\left\lvert {j_1 + j_2 - 1, j_1 + j_2 -1} \right\rangle} =\left(\frac{j_2}{j_1 + j_2}\right)^{1/2}{\left\lvert {j_1 j_1} \right\rangle} \otimes {\left\lvert {j_2 (j_2-1)} \right\rangle}-\left(\frac{j_1}{j_1 + j_2}\right)^{1/2}{\left\lvert {j_1 (j_1-1)} \right\rangle} \otimes {\left\lvert {j_2 j_2} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.8)

This will work, although we could also multiply by any phase factor if desired. Such a choice of phase factors is essentially just a convention.

## The Clebsch-Gordon convention

This is the convention we will use, where we

\begin{itemize}
\item choose the coefficients to be real.
\item require the coefficient of the $m_1 = j_1$ term to be $\ge 0$
\end{itemize}

This gives us the first state in the second column, and we can proceed to iterate using the lowering operators to get all those values.

Moving on to the third column

\begin{aligned}{\left\lvert {j_1 + j_2 - 2, j_1 + j_2 -2} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.9)

can only be made up of ${\left\lvert {j_1 m_1} \right\rangle} \otimes {\left\lvert {j_2 m_2} \right\rangle}$ with $m_1 + m_2 = j_1 + j_2 -2$. There are now three possibilities

\begin{aligned}\begin{array}{l l l l}m_1 &= j_1 & m_2 &= j_2 - 2 \\ m_1 &= j_1 - 2 & m_2 &= j_2 \\ m_1 &= j_1 - 1 & m_2 &= j_2 - 1\end{array}\end{aligned} \hspace{\stretch{1}}(2.10)

and 2 orthogonality conditions, plus conventions. This is enough to determine the ket in the third column.

We can formally write

\begin{aligned}{\left\lvert {jm ; j_1 j_2} \right\rangle} = \sum_{m_1, m_2}{\left\lvert { j_1 m_1, j_2 m_2} \right\rangle}\left\langle{{ j_1 m_1, j_2 m_2}} \vert {{jm ; j_1 j_2}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.11)

where

\begin{aligned}{\left\lvert { j_1 m_1, j_2 m_2} \right\rangle} = {\left\lvert {j_1 m_1} \right\rangle} \otimes {\left\lvert {j_2 m_2} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.12)

and

\begin{aligned}\left\langle{{ j_1 m_1, j_2 m_2}} \vert {{jm ; j_1 j_2}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.13)

are the Clebsch-Gordon coefficients, sometimes written as

\begin{aligned}\left\langle{{ j_1 m_1, j_2 m_2 }} \vert {{ jm }}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.14)

Properties
\begin{enumerate}
\item $\left\langle{{ j_1 m_1, j_2 m_2 }} \vert {{ jm }}\right\rangle \ne 0$ only if $j_1 - j_2 \le j \le j_1 + j+2$

This is sometimes called the triangle inequality, depicted in figure (\ref{fig:qmTwoL18:qmTwoL18fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig1}
\caption{Angular momentum triangle inequality.}
\end{figure}

\item $\left\langle{{ j_1 m_1, j_2 m_2 }} \vert {{ jm }}\right\rangle \ne 0$ only if $m = m_1 + m_2$.

\item Real (convention).

\item $\left\langle{{ j_1 j_1, j_2 (j - j_1) }} \vert {{ j j }}\right\rangle$ positive (convention again).

\item Proved in the text. If follows that

\begin{aligned}\left\langle{{ j_1 m_1, j_2 m_2 }} \vert {{ j m }}\right\rangle=(-1)^{j_1 + j_2 - j}\left\langle{{ j_1 (-m_1), j_2 (-m_2) }} \vert {{ j (-m) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.15)

\end{enumerate}

Note that the $\left\langle{{ j_1 m_1, j_2 m_2 }} \vert {{ j m }}\right\rangle$ are all real. So, they can be assembled into an orthogonal matrix. Example

\begin{aligned}\begin{bmatrix}{\left\lvert {11} \right\rangle} \\ {\left\lvert {10} \right\rangle} \\ {\left\lvert {\overline{11}} \right\rangle} \\ {\left\lvert {00} \right\rangle}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & \frac{1}{{\sqrt{2}}} & \frac{1}{{\sqrt{2}}} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & \frac{1}{{\sqrt{2}}} & \frac{-1}{\sqrt{2}} & 0 \\ \end{bmatrix}\begin{bmatrix}{\left\lvert {++} \right\rangle} \\ {\left\lvert {++} \right\rangle} \\ {\left\lvert {-+} \right\rangle} \\ {\left\lvert {--} \right\rangle}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.16)

## Example. Electrons

Consider the special case of an electron, a spin $1/2$ particle with $s = 1/2$ and $m_s = \pm 1/2$ where we have

\begin{aligned}\mathbf{J} = \mathbf{L} + \mathbf{S}\end{aligned} \hspace{\stretch{1}}(2.17)

\begin{aligned}{\left\lvert {lm} \right\rangle} \otimes {\left\lvert {\frac{1}{2} m_s} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.18)

possible values of $j$ are $l \pm 1/2$

\begin{aligned}l \otimes \frac{1}{2} = \left(l + \frac{1}{2}\right)\oplus\left(l - \frac{1}{2}\right)\end{aligned} \hspace{\stretch{1}}(2.19)

Our table representation is then

\begin{aligned}\begin{array}{| l | l | l |} \hline j = & l + \frac{1}{2} & l - \frac{1}{2} \\ \hline \hline {2}, l + \frac{1}{2}} \right\rangle} & \\ \hline {2}, l + \frac{1}{2} - 1} \right\rangle} {2}, l - \frac{1}{2}} \right\rangle} \\ \hline & {2}, -(l - \frac{1}{2}} \right\rangle} \\ \hline {2}, -(l + \frac{1}{2})} \right\rangle} & \\ \hline \end{array}\end{aligned} \hspace{\stretch{1}}(2.20)

Here ${\left\lvert {l + \frac{1}{2}, m} \right\rangle}$

can only have contributions from

\begin{aligned}{\left\lvert {l, m-\frac{1}{2}} \right\rangle} &\otimes {\left\lvert {\frac{1}{2}\frac{1}{2}} \right\rangle} \\ {\left\lvert {l, m+\frac{1}{2}} \right\rangle} &\otimes {\left\lvert {\frac{1}{2}\overline{\frac{1}{2}}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.21)

${\left\lvert {l - \frac{1}{2}, m} \right\rangle}$ from the same two. So using this and conventions we can work out (in section 28 page 524, of our text [1]).

\begin{aligned}{\left\lvert {l\pm \frac{1}{2}, m} \right\rangle} =\frac{1}{{\sqrt{2 l + 1}}}\left(\pm (l + \frac{1}{2} \pm m)^{1/2}{\left\lvert {l, m - \frac{1}{2}} \right\rangle} \times {\left\lvert {\frac{1}{2}\frac{1}{2}} \right\rangle}\pm (l + \frac{1}{2} \mp m)^{1/2}{\left\lvert {l, m + \frac{1}{2}} \right\rangle} \times {\left\lvert {\frac{1}{2} \overline{\frac{1}{2}}} \right\rangle}\right)\end{aligned} \hspace{\stretch{1}}(2.23)

# Tensor operators

section 29 of the text.

Recall how we characterized a rotation

\begin{aligned}\mathbf{r} \rightarrow \mathcal{R}(\mathbf{r}).\end{aligned} \hspace{\stretch{1}}(3.24)

Here we are using an active rotation as depicted in figure (\ref{fig:qmTwoL18:qmTwoL18fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig2}
\caption{active rotation}
\end{figure}

Suppose that

\begin{aligned}{\begin{bmatrix}\mathcal{R}(\mathbf{r})\end{bmatrix}}_i= \sum_j M_{ij} r_j\end{aligned} \hspace{\stretch{1}}(3.25)

so that

\begin{aligned}U = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar}\end{aligned} \hspace{\stretch{1}}(3.26)

rotates in the same way. Rotating a ket as in figure (\ref{fig:qmTwoL18:qmTwoL18fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig3}
\caption{Rotating a wavefunction.}
\end{figure}

Rotating a ket

\begin{aligned}{\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.27)

using the prescription

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.28)

and write

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = U[M] {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.29)

Now look at

\begin{aligned}{\left\langle {\psi} \right\rvert} \mathcal{O} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.30)

and compare with

\begin{aligned}{\left\langle {\psi'} \right\rvert} \mathcal{O} {\left\lvert {\psi'} \right\rangle}={\left\langle {\psi} \right\rvert} \underbrace{U^\dagger[M] \mathcal{O} U[M]}_{{*}} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.31)

We’ll be looking in more detail at ${*}$.

# References

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