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# Posts Tagged ‘trace’

## 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 14: Grand canonical ensemble. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 13, 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.

This lecture had a large amount of spoken content not captured in these notes. Reference to section 4 [1] was made for additional details.

# Grand canonical ensemble

Fig 1.1: Ensemble pictures

We are now going to allow particles to move to and from the system and the reservoir. The total number of states in the system is

\begin{aligned}\Omega_tot (N, V, E) =\sum_{N_S, E_S} \Omega_S(N_S, V_S, E_S)\Omega_R(N - N_S, V_R, E - E_S),\end{aligned} \hspace{\stretch{1}}(1.2.1)

so for $N_S \ll N$, and $E_S \ll E$, we have

\begin{aligned}\Omega_R &= \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N- N_S, V_R, E - E_S) \right) \\ &\approx \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N, V_R, E) - \frac{N_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {N}}\right)_{{V, E}} - \frac{E_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {E}}\right)_{{N, V}} \right) \\ &\propto \Omega_S(N_S, V_S, E_S)e^{-\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} },\end{aligned} \hspace{\stretch{1}}(1.2.2)

where the chemical potential \index{chemical potential} and temperature \index{temperature} are defined respectively as

\begin{aligned}\frac{\mu}{T} = -\left({\partial {S_R}}/{\partial {N}}\right)_{{V,E}}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\frac{1}{T} = \left({\partial {S_R}}/{\partial {E}}\right)_{{N,V}}.\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\begin{aligned}\mathcal{P} \propto e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.4)

With $\{c\}$ as the set of all possible configuration pairs $\{N_S, E_S\}$, we define the grand partition function

\begin{aligned}Z_{\mathrm{G}} = \sum_{\{c\}}e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.5)

So that the probability of finding a given state with energy and particle numbers $\{E_S, N_S\}$ is

\begin{aligned}\mathcal{P}(E_S, N_S) = \frac{e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }}{Z_{\mathrm{G}}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

For a classical system we have

\begin{aligned}\{ c \} \rightarrow \{ x \} \{ p \},\end{aligned} \hspace{\stretch{1}}(1.0.7)

whereas in a quantum content we have

\begin{aligned}\{ c \} \rightarrow \text{eigenstate}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}Z_{\mathrm{G}}^{\mathrm{\mathrm{\mathrm{QM}}}} = {\text{Tr}}_{\{\text{energy}, N\}} \left( e^{ -\beta (\hat{H} - \mu \hat{N} } \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

We want to do this because the calculation of the number of states

\begin{aligned}\int_{\{ x \} \{ p \}} \delta\left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right),\end{aligned} \hspace{\stretch{1}}(1.0.10)

can quickly become intractable. We want to go to the canonical ensemble was because the partition function

\begin{aligned}Z_c = \int_{\{ x \} \{ p \}}e^{-\beta \left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right)},\end{aligned} \hspace{\stretch{1}}(1.0.11)

yields the same results, but can be much easier to compute. We have a similar reason to go to the grand canonical ensemble, because this computation, once we allow the number of particles to vary also becomes very hard.

We are now going to define a notion of equilibrium so that it includes

1. All forces are equal (mechanical equilibrium)
2. Temperatures are equal (no net heat flow)
3. Chemical potentials are equal (no net particle flow)

We’ll isolate a subsystem, containing a large number of particles fig. 1.2.

Fig 1.2: A subsystem to and from which particle motion is allowed

When we think about Fermions we have to respect the “Pauli exclusion” principle \index{Pauli exclusion principle}.

Suppose we have just a one dimensional Fermion system for some potential as in fig. 1.3.

Fig 1.3: Energy level filling in a quantum system

For every momentum $k$ there are two possible occupation numbers $n_k \in \{0, 1\}$

our partition function is

\begin{aligned}Z_c = \sum_{n_k,\sum_k n_k = N} e^{-\beta \sum_k \epsilon_k n_k}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

We’d find that this calculation with this $\sum_k n_k = N$ constraint becomes essentially impossible.

We’ll see that relaxing this constraint will allow this calculation to become tractable.

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

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

Posted by peeterjoot on March 3, 2013

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

## 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.)]

# Disclaimer

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

# Canonical partition

We found

\begin{subequations}

\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)

\end{subequations}

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)

or

\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{subequations}

\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)

\end{subequations}

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)

## PHY356F: Quantum Mechanics I. Lecture 8 — Making Sense of Quantum Mechanics

Posted by peeterjoot on November 3, 2010

My notes from Lecture 8, November 2, 2010. Taught by Prof. Vatche Deyirmenjian.

## Discussion

Desai: “Quantum Theory is a linear theory …”

We can discuss SHM without using sines and cosines or complex exponentials, say, only using polynomials, but it would be HARD to do so, and much more work. We want the framework of Hilbert space, linear operators and all the rest to make our life easier.

Dirac: “Mathematics is only a tool, and one should learn the … (FIXME: LOOKUP)”

You have to be able to understand the concepts and apply the concepts as well as the mathematics.

Deyirmenjian: “Think before you compute.”

Joke: With his name included it is the 3Ds. There’s a lot of information included in the question, so read it carefully.

Q: The equation $A {\lvert {a_n} \rangle} = a_n {\lvert {a_n} \rangle}$ for operator $A$, eigenvalue $a_n$, $n = 1,2$ and eigenvector ${\lvert {a_n} \rangle}$ that is identified by the eigenvalue $a_n$ says that

\begin{itemize}
\item (a) measuring the physical quantity associated with $A$ gives result $a_n$
\item (b) $A$ acting on the state ${\lvert {a_n} \rangle}$ gives outcome $a_n$
\item (c) the possible outcomes of measuring the physical quantity associated with $A$ are the eigenvalues $a_n$
\item (d) Quantum mechanics is hard.
\end{itemize}

${\lvert {a_n} \rangle}$ is a vector in a vector space or Hilbert space identified by some quantum number $a_n, n \in 1,2, \cdots$.

The $a_n$ values could be expressions. Example, Angular momentum is describe by states ${\lvert {lm} \rangle}, l = 0,1,2,\cdots$ and $m = 0, \pm 1, \pm 2$

Recall that the problem is

\begin{aligned}\mathbf{L}^2 {\lvert {lm} \rangle} &= l(l+1) \hbar^2 {\lvert {lm} \rangle} \\ L_z {\lvert {lm} \rangle} &= m \hbar {\lvert {lm} \rangle}\end{aligned} \hspace{\stretch{1}}(4.72)

We have respectively eigenvalues $l(l+1)\hbar^2$, and $m \hbar$.

A: Answer is (c). $a_n$ isn’t a measurement itself. These represent possibilities. Contrast this to classical mechanics where time evolution is given without probabilities

\begin{aligned}\mathbf{F}_{\text{net}} &= m \mathbf{a} \\ \mathbf{x}(0), \mathbf{x}'(0) &\implies \mathbf{x}(t), \mathbf{x}'(t)\end{aligned} \hspace{\stretch{1}}(4.74)

The eigenvalues are the possible outcomes, but we only know statistically that these are the possibilities.

(a),(b) are incorrect because we do not know what the initial state is, nor what the final outcome is. We also can’t say “gives result $a_n$”. That statement is too strong!

Q: We wouldn’t say that $A$ acts on pure state ${\lvert {a_n} \rangle}$, instead. If the state of the system is ${\lvert {\psi} \rangle} = {\lvert {a_5} \rangle}$, the probability of measuring outcome $a_5$ is
\begin{itemize}
\item (a) $a_5$
\item (b) $a_5^2$
\item (c) $\left\langle{{a_5}} \vert {{\psi}}\right\rangle = \left\langle{{a_5}} \vert {{a_5}}\right\rangle = 1$.
\item (d) ${\left\lvert{\left\langle{{a_5}} \vert {{\psi}}\right\rangle}\right\rvert}^2 = {\left\lvert{\left\langle{{a_5}} \vert {{a_5}}\right\rangle}\right\rvert}^2 = {\left\lvert{1}\right\rvert}^2 = 1$.
\end{itemize}

A: (d) The eigenformula equation doesn’t say anything about any specific outcome. We want to talk about probability amplitudes. When the system is prepared in a particular pure eigenstate, then we have a guarentee that the probability of measuring that state is unity. We wouldn’t say (c) because the probability amplitudes are the absolute square of the complex number $\left\langle{{a_n}} \vert {{a_n}}\right\rangle$.

The probability of outcome $a_n$, given initial state ${\lvert {\Psi} \rangle}$ is ${\left\lvert{\left\langle{{a_n}} \vert {{\Psi}}\right\rangle}\right\rvert}^2$.

Wave function collapse: When you make a measurement of the physical quantity associated with $A$, then the state of the system will be the value ${\lvert {a_5} \rangle}$. The state is not the number (eigenvalue) $a_5$.

Example: SGZ. With a “spin-up” measurement in the z-direction, the state of the system is ${\lvert {z+} \rangle}$. The state before the measurement, by the magnet, was ${\lvert {\Psi} \rangle}$. After the measurement, the state describing the system is ${\lvert {\phi} \rangle} = {\lvert {z+} \rangle}$. The measurement outcome is $+\frac{\hbar}{2}$ for the spin angular momentum along the z-direction.

FIXME: SGZ picture here.

There is an interaction between the magnet and the silver atoms coming out of the oven. Before that interaction we have a state described by ${\lvert {\Psi} \rangle}$. After the measurement, we have a new state ${\lvert {\phi} \rangle}$. We call this the collapse of the wave function. In a future course (QM interpretations) the language used and interpretations associated with this language can be discussed.

Q: Express Hermitian operator $A$ in terms of its eigenvectors.
Q: The above question is vague because

\begin{itemize}
\item (a) The eigenvectors may form a descrete set.
\item (b) The eigenvectors may form a continuous set.
\item (c) The eigenvectors may not form a complete set.
\item (d) The eigenvectors are not given.
\end{itemize}

A: None of the above. A Hermitian operator is guarenteed to have a complete set of eigenvectors. The operator may also be both discrete and continuous (example: the complete spin wave function).

discrete:

\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} \right) \\ &= \sum_n (A {\lvert {a_n} \rangle} ){\langle {a_n} \rvert} \\ &= \sum_n (a_n {\lvert {a_n} \rangle}) {\langle {a_n} \rvert} \\ &= \sum_n a_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert}\end{aligned}

continuous:

\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \int d\alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert} \right) \\ &= \int d\alpha (A {\lvert {\alpha} \rangle} ){\langle {\alpha} \rvert} \\ &= \int d\alpha (\alpha {\lvert {\alpha} \rangle}) {\langle {\alpha} \rvert} \\ &= \int d\alpha \alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert}\end{aligned}

An example is the position eigenstate ${\lvert {x} \rangle}$, eigenstate of the Hermitian operator $X$. $\alpha$ is a label indicating the summation.

general case with both discrete and continuous:

\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} + \int d\alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert} \right) \\ &= \sum_n \left(A {\lvert {a_n} \rangle} \right){\langle {a_n} \rvert} + \int d\alpha \left(A {\lvert {\alpha} \rangle} \right){\langle {\alpha} \rvert} \\ &= \sum_n \left(a_n {\lvert {a_n} \rangle}\right) {\langle {a_n} \rvert} + \int d\alpha \left(\alpha {\lvert {\alpha} \rangle}\right) {\langle {\alpha} \rvert} \\ &= \sum_n a_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} + \int d\alpha \alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert}\end{aligned}

Problem Solving

\begin{itemize}
\item MODEL — Quantum, linear vector space
\item VISUALIZE — Operators can have discrete, continuous or both discrete and continuous eigenvectors.
\item SOLVE — Use the identity operator.
\item CHECK — Does the above expression give $A {\lvert {a_n} \rangle} = a_n {\lvert {a_n} \rangle}$.
\end{itemize}

Check

\begin{aligned}A {\lvert {a_m} \rangle}&= \sum_n a_n {\lvert {a_n} \rangle} \left\langle{{a_n}} \vert {{a_m}}\right\rangle + \int d\alpha \alpha {\lvert {\alpha} \rangle} \left\langle{{\alpha}} \vert {{a_n}}\right\rangle \\ &= \sum_n a_n {\lvert {a_n} \rangle} \delta_{nm} \\ &= a_m {\lvert {a_m} \rangle}\end{aligned}

What remains to be shown, used above, is that the continous and discrete eigenvectors are orthonormal. He has an example vector space, not yet discussed.

Q: what is ${\langle {\Psi_1} \rvert} A {\lvert {\Psi_1} \rangle}$, where $A$ is a Hermitian operator, and ${\lvert {\Psi_1} \rangle}$ is a general state.

A: ${\langle {\Psi_1} \rvert} A {\lvert {\Psi_1} \rangle} =$ average outcome for many measurements

of the physical quantity associated with $A$ such that the system is prepared in state ${\lvert {\Psi_1} \rangle}$ prior to each measurement.

Q: What if the preparation is ${\lvert {\Psi_2} \rangle}$. This isn’t neccessarily an eigenstate of $A$, it is some linear combination of eigenstates. It is a general state.
A: ${\langle {\Psi_2} \rvert} A {\lvert {\Psi_2} \rangle} =$ average of the physical quantity associated with $A$, but the preparation is ${\lvert {\Psi_2} \rangle}$, not ${\lvert {\Psi_1} \rangle}$.

Q: What if our initial state is a little bit of ${\lvert {\Psi_1} \rangle}$, and a little bit of ${\lvert {\Psi_2} \rangle}$, and a little bit of ${\lvert {\Psi_N} \rangle}$. ie: how to describe what comes out of the oven in the SG experiment. That spin is a statistical mixture. We could understand this as only a statistical mix. This is a physical relavent problem.
A: To describe that statistical situtation we have the following.

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}} = \sum_j w_j {\langle {\Psi_j} \rvert} A {\lvert {\Psi_j} \rangle}\end{aligned} \hspace{\stretch{1}}(4.76)

We sum up all the expectation values modified by statistical weighting factors. These $w_j$‘s are statistical weighting factors for a preparation associated with ${\lvert {\Psi_j} \rangle}$, real numbers (that sum to unity). Note that these states ${\lvert {\Psi_j} \rangle}$ are not neccessarily orthonormal.

With insertion of the identity operator we have

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}}&= \sum_j w_j {\langle {\Psi_j} \rvert} \mathbf{1} A {\lvert {\Psi_j} \rangle} \\ &= \sum_j w_j {\langle {\Psi_j} \rvert} \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} \right) A {\lvert {\Psi_j} \rangle} \\ &= \sum_j \sum_n w_j \left\langle{{\Psi_j}} \vert {{a_n}}\right\rangle {\langle {a_n} \rvert} A {\lvert {\Psi_j} \rangle} \\ &= \sum_j \sum_n w_j {\langle {a_n} \rvert} A {\lvert {\Psi_j} \rangle} \left\langle{{\Psi_j}} \vert {{a_n}}\right\rangle \\ &= \sum_n {\langle {a_n} \rvert} A \left( \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert} \right) {\lvert {a_n} \rangle} \\ \end{aligned}

This inner bit is called the density operator $\rho$

\begin{aligned}\rho &\equiv \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert}\end{aligned} \hspace{\stretch{1}}(4.77)

Returning to the average we have

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}} = \sum_n {\langle {a_n} \rvert} A \rho {\lvert {a_n} \rangle} \equiv \text{Tr}(A \rho)\end{aligned} \hspace{\stretch{1}}(4.78)

The trace of an operator $A$ is

\begin{aligned}\text{Tr}(A) = \sum_j {\langle {a_j} \rvert} A {\lvert {a_j} \rangle} = \sum_j A_{jj}\end{aligned} \hspace{\stretch{1}}(4.79)

## Section 5.9, Projection operator.

Returning to the last lecture. From chapter 1, we have

\begin{aligned}P_n = {\lvert {a_n} \rangle} {\langle {a_n} \rvert}\end{aligned} \hspace{\stretch{1}}(4.80)

is called the projection operator. This is physically relavent. This takes a general state and gives you the component of that state associated with that eigenvector. Observe

\begin{aligned}P_n {\lvert {\phi} \rangle} ={\lvert {a_n} \rangle} \left\langle{{a_n}} \vert {{\phi}}\right\rangle =\underbrace{\left\langle{{a_n}} \vert {{\phi}}\right\rangle}_{\text{coefficient}} {\lvert {a_n} \rangle}\end{aligned} \hspace{\stretch{1}}(4.81)

Example: Projection operator for the ${\lvert {z+} \rangle}$ state

\begin{aligned}P_{z+} = {\lvert {z+} \rangle} {\langle {z+} \rvert}\end{aligned} \hspace{\stretch{1}}(4.82)

We see that the density operator

\begin{aligned}\rho &\equiv \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert},\end{aligned} \hspace{\stretch{1}}(4.83)

can be written in terms of the Projection operators

\begin{aligned}{\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert} = \text{Projection operator for state} {\lvert {\Psi_j} \rangle}\end{aligned}

The projection operator is like a dot product, determining the quantity of a state that lines in the direction of another state.

Q: What is the projection operator for spin-up along the z-direction.
A:

\begin{aligned}P_{z+} = {\lvert {z+} \rangle}{\langle {z+} \rvert}\end{aligned} \hspace{\stretch{1}}(4.84)

Or in matrix form with

\begin{aligned}{\langle {z+} \rvert} &=\begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\langle {z-} \rvert} &=\begin{bmatrix}0 \\ 1\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.85)

so

\begin{aligned}P_{z+} = {\lvert {z+} \rangle}{\langle {z+} \rvert} =\begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.87)

Q: A harder problem. What is $P_\chi$, where

\begin{aligned}{\lvert {\chi} \rangle} =\begin{bmatrix}c_1 \\ c_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.88)

Note: We want normalized states, with $\left\langle{{\chi}} \vert {{\chi}}\right\rangle = {\left\lvert{c_1}\right\rvert}^2 + {\left\lvert{c_2}\right\rvert}^2 = 1$.

A:

\begin{aligned}P_{\chi} = {\lvert {\chi} \rangle}{\langle {\chi} \rvert} =\begin{bmatrix}c_1^{*} \\ c_2^{*}\end{bmatrix}\begin{bmatrix}c_1 & c_2\end{bmatrix}=\begin{bmatrix}c_1^{*} c_1 & c_1^{*} c_2 \\ c_2^{*} c_1 & c_2^{*} c_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.89)

Observe that this has the proper form of a projection operator is that the square is itself

\begin{aligned}({\lvert {\chi} \rangle}{\langle {\chi} \rvert}) ({\lvert {\chi} \rangle}{\langle {\chi} \rvert})&= {\lvert {\chi} \rangle} (\left\langle{{\chi}} \vert {{\chi}}\right\rangle ){\langle {\chi} \rvert} \\ &= {\lvert {\chi} \rangle} {\langle {\chi} \rvert}\end{aligned}

Q: Show that $P_{\chi} = a_0 \mathbf{1} + \mathbf{a} \cdot \boldsymbol{\sigma}$, where $\mathbf{a} = (a_x, a_y, a_z)$ and $\boldsymbol{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$.

A: See Section 5.9. Note the following about computing $(\boldsymbol{\sigma} \cdot \mathbf{a})^2$.

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a})^2&=(a_x \sigma_x+ a_y \sigma_y+ a_z \sigma_z)(a_x \sigma_x+ a_y \sigma_y+ a_z \sigma_z) \\ &=a_x a_x \sigma_x \sigma_x+a_x a_y \sigma_x \sigma_y+a_x a_z \sigma_x \sigma_z+a_y a_x \sigma_y \sigma_x+a_y a_y \sigma_y \sigma_y+a_y a_z \sigma_y \sigma_z+a_z a_x \sigma_z \sigma_x+a_z a_y \sigma_z \sigma_y+a_z a_z \sigma_z \sigma_z \\ &= (a_x^2 + a_y^2 + a_z^2) I+ a_x a_y ( \sigma_x \sigma_y + \sigma_y \sigma_x)+ a_y a_z ( \sigma_y \sigma_z + \sigma_z \sigma_y)+ a_z a_x ( \sigma_z \sigma_x + \sigma_x \sigma_z) \\ &= {\left\lvert{\mathbf{x}}\right\rvert}^2 I\end{aligned}

So we have

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a})^2 = (\mathbf{a} \cdot \mathbf{a}) \mathbf{1} \equiv \mathbf{a}^2\end{aligned} \hspace{\stretch{1}}(4.90)

Where the matrix representations

\begin{aligned}\sigma_x &\leftrightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ \sigma_y &\leftrightarrow \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ \sigma_z &\leftrightarrow \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.91)

would be used to show that

\begin{aligned}\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I\end{aligned} \hspace{\stretch{1}}(4.94)

and

\begin{aligned}\sigma_x \sigma_y &= -\sigma_y \sigma_x \\ \sigma_y \sigma_z &= -\sigma_z \sigma_y \\ \sigma_z \sigma_x &= -\sigma_x \sigma_z\end{aligned} \hspace{\stretch{1}}(4.95)