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Posts Tagged ‘spin hamiltonian’

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.

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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.)]


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


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)


\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)


\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)


\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)


\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)


[1] Peeter Joot. Quantum Mechanics II., chapter: Two spin systems, angular momentum, and Clebsch-Gordon convention. URL

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

Posted by peeterjoot on March 3, 2013

In A compilation of notes, so far, for ‘PHY452H1S Basic Statistical Mechanics’ I posted a link this compilation of statistical mechanics course notes.

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

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PHY452H1S Basic Statistical Mechanics. Lecture 12: Helmholtz free energy. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 28, 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.)]


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

Canonical partition

We found


\begin{aligned}\frac{\sigma_{\mathrm{E}}}{E} \propto \frac{T \sqrt{C_V}}{E} k_{\mathrm{B}}^2\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}Z = \sum_{\{c\}} e^{-\beta E(c)}\end{aligned} \hspace{\stretch{1}}(1.0.1b)

\begin{aligned}C_V \sim N\end{aligned} \hspace{\stretch{1}}(1.0.1c)

\begin{aligned}E \sim N\end{aligned} \hspace{\stretch{1}}(1.0.1d)


where the partition function \index{partition function} acts as a probability distribution so that we can define an average as

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{\sum_{\{c\}} A(c) e^{-\beta E(c)}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.2)

If we suppose that the energy is typically close to the average energy as in fig. 1.1.

Fig 1.1: Peaked energy distribution

, then we can approximate the partition function as

\begin{aligned}Z \approx e^{-\beta \left\langle{{E}}\right\rangle} \sum_{\{c\}} \delta_{E, \bar{E}}= e^{-\beta \left\langle{{E}}\right\rangle} e^S/k_{\mathrm{B}},\end{aligned} \hspace{\stretch{1}}(1.0.4)

where we’ve used S = k_{\mathrm{B}} \ln \Omega to express the number of states where the energy matches the average energy \Omega = \sum \delta_{E, \bar{E}}.

This gives us

\begin{aligned}Z = e^{-\beta (\left\langle{{E}}\right\rangle - k_{\mathrm{B}} T S/k_{\mathrm{B}}) } = e^{-\beta (\left\langle{{E}}\right\rangle - T S) } \end{aligned} \hspace{\stretch{1}}(1.0.4)


\begin{aligned}\boxed{Z = e^{-\beta F},}\end{aligned} \hspace{\stretch{1}}(1.0.5)

where we define the Helmholtz free energy F as

\begin{aligned}\boxed{F = \left\langle{{E}}\right\rangle - T S.}\end{aligned} \hspace{\stretch{1}}(1.0.6)

Equivalently, the log of the partition function provides us with the partition function

\begin{aligned}F = - k_{\mathrm{B}} T \ln Z.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Recalling our expression for the average energy, we can now write that in terms of the free energy

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{\sum_{\{c\}} E(c) e^{-\beta E(c)}}{\sum_{\{c\}} e^{-\beta E(c)}}= -\frac{\partial {}}{\partial {\beta}}\ln Z=\frac{\partial {(\beta F)}}{\partial {\beta}}\end{aligned} \hspace{\stretch{1}}(1.0.8)

Quantum mechanical picture

Consider a subsystem as in fig. 1.2 where we have states of the form

Fig 1.2: subsystem in heat bath

\begin{aligned}{\left\lvert {\Psi_{\text{full}}} \right\rangle} = {\left\lvert {\chi_{\text{subsystem}}} \right\rangle} {\left\lvert {\phi_{\text{bath}}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.9)

and a total Hamiltonian operator of the form

\begin{aligned}H_{\text{full}} = H_{\text{subsystem}} + H_{\text{bath}} (+ H_{\text{coupling}})\end{aligned} \hspace{\stretch{1}}(1.0.10)

where the total energy of the state, given energy eigenvalues \mathcal{E}_n and \lambda_n for the states {\left\lvert {\chi_{\text{subsystem}}} \right\rangle} and {\left\lvert {\phi_{\text{bath}}} \right\rangle} respectively, is given by the sum

\begin{aligned}E = \mathcal{E}_m + \lambda_n.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Here \mathcal{E}_m, \lambda_n are many body energies, so that \delta E \sim \#e^{-\#N}.

We can now write the total number of states as

\begin{aligned}\Omega(E) &= \underbrace{\sum_m}_{\text{subsystem}}\underbrace{\sum_n}_{\text{bath}}\delta(E - \mathcal{E}_m -\lambda_n)\\ &= \sum_m e^{\frac{1}{{k_{\mathrm{B}}}} S(E - \mathcal{E}_m)} \\ &\approx \sum_m e^{\frac{1}{{k_{\mathrm{B}}}} S(E)}e^{\beta \mathcal{E}_m}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}Z = \sum_m e^{-\beta \mathcal{E}_m} = \text{Tr} \left( e^{-\beta \hat{H}_{\text{subsystem}}} \right)\end{aligned} \hspace{\stretch{1}}(1.0.13)

We’ve ignored the coupling term in eq. 1.0.10. This is actually a problem in quantum mechanics since we require this coupling to introduce state changes.

Example: Spins

Given N spin 1/2 objects \uparrow, \downarrow, satisfying

\begin{aligned}S_z = \pm \frac{1}{{2}} \hbar\end{aligned} \hspace{\stretch{1}}(1.0.14)

Dropping \hbar we have

\begin{aligned}S_z \rightarrow \pm \frac{1}{{2}} \sigma\end{aligned} \hspace{\stretch{1}}(1.0.15)

Our system has a state {\left\lvert {\sigma_1, \sigma_2, \cdots \sigma_N} \right\rangle} where \sigma_i = \pm 1. The total number of states is 2^N.

Our Hamiltonian is

\begin{aligned}\hat{H} = - B \sum_i \hat{S}_{z_i}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

This is the associated with the Zeeman effect, where states can be split by a magnetic field, as in fig. 1.3.

Fig 1.3: Zeeman splitting

Our minimum and maximum energies are


\begin{aligned}E_{\mathrm{min}} = -\frac{B}{2} N\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}E_{\mathrm{max}} = -\frac{B}{2} N\end{aligned} \hspace{\stretch{1}}(1.0.17b)


The total energy difference is

\begin{aligned}\Delta E = B N,\end{aligned} \hspace{\stretch{1}}(1.0.23)

and the energy differences are

\begin{aligned}\delta E \sim \frac{B N}{2^N} \sim \# e^{-\# N}.\end{aligned} \hspace{\stretch{1}}(1.0.23)

This is a measure of the average energy difference between two adjacent energy levels. In a real system we cannot assume that we have non-interacting spins. Any weak interaction will split our degenerate energy levels as in fig. 1.4.

Fig 1.4: Interaction splitting

We can now express the partition function

\begin{aligned}Z &= \sum_{\{\sigma\}} e^{-\beta \left( -\frac{B}{2} \sum_i \sigma_i \right)} \\ &= \left( \sum_{\{\sigma_1\}} \exp \left( -\frac{\beta B}{2} \sigma_i \right) \right)\left( \sum_{\{\sigma_2\}} \exp \left( -\frac{\beta B}{2} \sigma_i \right) \right)\cdots \\ &= \left( \exp \left( -\frac{\beta B}{2} (1) \right) + \exp \left( -\frac{\beta B}{2} (-1) \right) \right)^N \\ &= \left( 2 \cosh\left( \frac{B}{2 k_B T} \right) \right)^N\end{aligned} \hspace{\stretch{1}}(1.0.23)

Our free energy is

\begin{aligned}F = - k_B T N \ln \left( 2 \cosh \left( \frac{B}{2 k_B T} \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

For the expected value of the spin we find

\begin{aligned}\left\langle{{S_z}}\right\rangle = \sum_i \left\langle{{S_{z_i}}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.0.23)

\begin{aligned}\left\langle{{S_{z_i}}}\right\rangle=\frac{1}{{2}} \frac{\sum_\sigma \sigma e^{\beta B \sigma/2}}{\sum_\sigma e^{\beta B \sigma/2}}= \frac{1}{{2}} \tanh \left( \frac{B}{2 k_B T} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

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PHY456H1F: Quantum Mechanics II. Lecture 15 (Taught by Prof J.E. Sipe). Rotation operator in spin space

Posted by peeterjoot on October 31, 2011

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

Rotation operator in spin space.

We can formally expand our rotation operator in Taylor series

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar}= I +\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^3+ \cdots\end{aligned} \hspace{\stretch{1}}(2.1)


\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2}&= I +\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) \sigma_0+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) + \cdots \\ &=\sigma_0 \left( 1 - \frac{1}{{2!}}\left(\frac{\theta}{2}\right)^2 + \cdots \right) +(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \left( \frac{\theta}{2} - \frac{1}{{3!}}\left(\frac{\theta}{2}\right)^3 + \cdots \right) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})\end{aligned}

where we’ve used the fact that (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2 = \sigma_0.

So our representation of the spin operator is

\begin{aligned}\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} &\rightarrow \cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) \left(n_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + n_y \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + n_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \right) \\ &=\begin{bmatrix}\cos(\theta/2) -i n_z \sin(\theta/2) & -i (n_x -i n_y) \sin(\theta/2) \\ -i (n_x + i n_y) \sin(\theta/2) & \cos(\theta/2) +i n_z \sin(\theta/2) \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.2)

Note that, in particular,

\begin{aligned}e^{-2 \pi i \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} \rightarrow \cos\pi \sigma_0 = -\sigma_0\end{aligned} \hspace{\stretch{1}}(2.3)

This “rotates” the ket, but introduces a phase factor.

Can do this in general for other degrees of spin, for s = 1/2, 3/2, 5/2, \cdots.

Unfortunate interjection by me

I mentioned the half angle rotation operator that requires a half angle operator sandwich. Prof. Sipe thought I might be talking about a Heisenberg picture representation, where we have something like this in expectation values

\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}}(2.4)

so that

\begin{aligned}{\left\langle {\psi'} \right\rvert}\mathcal{O}{\left\lvert {\psi'} \right\rangle} = {\left\langle {\psi} \right\rvert} e^{i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} \mathcal{O}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.5)

However, what I was referring to, was that a general rotation of a vector in a Pauli matrix basis

\begin{aligned}R(\sum a_k \sigma_k) = R( \mathbf{a} \cdot \boldsymbol{\sigma})\end{aligned} \hspace{\stretch{1}}(2.6)

can be expressed by sandwiching the Pauli vector representation by two half angle rotation operators like our spin 1/2 operators from class today

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2} \mathbf{a} \cdot \boldsymbol{\sigma} e^{\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2}\end{aligned} \hspace{\stretch{1}}(2.7)

where \hat{\mathbf{u}} and \hat{\mathbf{v}} are two non-colinear orthogonal unit vectors that define the oriented plane that we are rotating in.

For example, rotating in the x-y plane, with \hat{\mathbf{u}} = \hat{\mathbf{x}} and \hat{\mathbf{v}} = \hat{\mathbf{y}}, we have

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \sigma_1 \sigma_2/2} (a_1 \sigma_1 + a_2 \sigma_2 + a_3 \sigma_3) e^{\theta \sigma_1 \sigma_2/2} \end{aligned} \hspace{\stretch{1}}(2.8)

Observe that these exponentials commute with \sigma_3, leaving

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) &= (a_1 \sigma_1 + a_2 \sigma_2) e^{\theta \sigma_1 \sigma_2} +  a_3 \sigma_3 \\ &= (a_1 \sigma_1 + a_2 \sigma_2) (\cos\theta + \sigma_1 \sigma_2 \sin\theta)+a_3 \sigma_3 \\ &= \sigma_1 (a_1 \cos\theta - a_2 \sin\theta)+ \sigma_2 (a_2 \cos\theta + a_1 \sin\theta)+ \sigma_3 (a_3)\end{aligned}

yielding our usual coordinate rotation matrix. Expressed in terms of a unit normal to that plane, we form the normal by multiplication with the unit spatial volume element I = \sigma_1 \sigma_2 \sigma_3. For example:

\begin{aligned}\sigma_1 \sigma_2 \sigma_3( \sigma_3 )=\sigma_1 \sigma_2 \end{aligned} \hspace{\stretch{1}}(2.9)

and can in general write a spatial rotation in a Pauli basis representation as a sandwich of half angle rotation matrix exponentials

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} (\mathbf{a} \cdot \boldsymbol{\sigma})e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} \end{aligned} \hspace{\stretch{1}}(2.10)

when \hat{\mathbf{n}} \cdot \mathbf{a} = 0 we get the complex-number like single sided exponential rotation exponentials (since \mathbf{a} \cdot \boldsymbol{\sigma} commutes with \mathbf{n} \cdot \boldsymbol{\sigma} in that case)

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = (\mathbf{a} \cdot \boldsymbol{\sigma} )e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})} \end{aligned} \hspace{\stretch{1}}(2.11)

I believe it was pointed out in one of [1] or [2] that rotations expressed in terms of half angle Pauli matrices has caused some confusion to students of quantum mechanics, because this 2 \pi “rotation” only generates half of the full spatial rotation. It was argued that this sort of confusion can be avoided if one observes that these half angle rotations exponentials are exactly what we require for general spatial rotations, and that a pair of half angle operators are required to produce a full spatial rotation.

The book [1] takes this a lot further, and produces a formulation of spin operators that is devoid of the normal scalar imaginary i (using the Clifford algebra spatial unit volume element instead), and also does not assume a specific matrix representation of the spin operators. They argue that this leads to some subtleties associated with interpretation, but at the time I was attempting to read that text I did know enough QM to appreciate what they were doing, and haven’t had time to attempt a new study of that content.

Spin dynamics

At least classically, the angular momentum of charged objects is associated with a magnetic moment as illustrated in figure (\ref{fig:qmTwoL15:qmTwoL15fig1})

\caption{Magnetic moment due to steady state current}

\begin{aligned}\boldsymbol{\mu} = I A \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.12)

In our scheme, following the (cgs?) text conventions of [3], where the \mathbf{E} and \mathbf{B} have the same units, we write

\begin{aligned}\boldsymbol{\mu} = \frac{I A}{c} \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.13)

For a charge moving in a circle as in figure (\ref{fig:qmTwoL15:qmTwoL15fig2})
\caption{Charge moving in circle.}

\begin{aligned}\begin{aligned}I &= \frac{\text{charge}}{\text{time}} \\ &= \frac{\text{distance}}{\text{time}} \frac{\text{charge}}{\text{distance}} \\ &= \frac{q v}{ 2 \pi r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.14)

so the magnetic moment is

\begin{aligned}\begin{aligned}\mu &= \frac{q v}{ 2 \pi r} \frac{\pi r^2}{c}  \\ &= \frac{q }{ 2 m c } (m v r) \\ &= \gamma L\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.15)

Here \gamma is the gyromagnetic ratio

Recall that we have a torque, as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig3})
\caption{Induced torque in the presence of a magnetic field.}

\begin{aligned}\mathbf{T} = \boldsymbol{\mu} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.16)

tending to line up \boldsymbol{\mu} with \mathbf{B}. The energy is then

\begin{aligned}-\boldsymbol{\mu} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.17)

Also recall that this torque leads to precession as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig4})

\begin{aligned}\frac{d{\mathbf{L}}}{dt} = \mathbf{T} = \gamma \mathbf{L} \times \mathbf{B},\end{aligned} \hspace{\stretch{1}}(3.18)

\caption{Precession due to torque.}

with precession frequency

\begin{aligned}\boldsymbol{\omega} = - \gamma \mathbf{B}.\end{aligned} \hspace{\stretch{1}}(3.19)

For a current due to a moving electron

\begin{aligned}\gamma = -\frac{e}{2 m c} < 0\end{aligned} \hspace{\stretch{1}}(3.20)

where we are, here, writing for charge on the electron -e.

Question: steady state currents only?. Yes, this is only true for steady state currents.

For the translational motion of an electron, even if it is not moving in a steady way, regardless of it’s dynamics

\begin{aligned}\boldsymbol{\mu}_0 = - \frac{e}{2 m c} \mathbf{L}\end{aligned} \hspace{\stretch{1}}(3.21)

Now, back to quantum mechanics, we turn \boldsymbol{\mu}_0 into a dipole moment operator and \mathbf{L} is “promoted” to an angular momentum operator.

\begin{aligned}H_{\text{int}} = - \boldsymbol{\mu}_0 \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.22)

What about the “spin”?


\begin{aligned}\boldsymbol{\mu}_s = \gamma_s \mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.23)

we write this as

\begin{aligned}\boldsymbol{\mu}_s = g \left( -\frac{e}{ 2 m c} \right)\mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.24)

so that

\begin{aligned}\gamma_s = - \frac{g e}{ 2 m c} \end{aligned} \hspace{\stretch{1}}(3.25)

Experimentally, one finds to very good approximation

\begin{aligned}g = 2\end{aligned} \hspace{\stretch{1}}(3.26)

There was a lot of trouble with this in early quantum mechanics where people got things wrong, and canceled the wrong factors of 2.

In fact, Dirac’s relativistic theory for the electron predicts g=2.

When this is measured experimentally, one does not get exactly g=2, and a theory that also incorporates photon creation and destruction and the interaction with the electron with such (virtual) photons. We get

\begin{aligned}\begin{aligned}g_{\text{theory}} &= 2 \left(1.001159652140 (\pm 28)\right) \\ g_{\text{experimental}} &= 2 \left(1.0011596521884 (\pm 43)\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)

Richard Feynman compared the precision of quantum mechanics, referring to this measurement, “to predicting a distance as great as the width of North America to an accuracy of one human hair’s breadth”.


[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

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

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