August 2020 – biology, math, and physics

Working through Sakurai

Congzhou Sha on August 26, 2020April 6, 2021

This is a combination of notes loosely based on Sakurai’s Modern Quantum Mechanics and solved problems. I’m aiming for a mathematically complete approach, so I will state theorems without proof.

The wave-particle duality

Sakurai motivates representation of states in quantum mechanics as a complete, complex inner product space using the Stern-Gerlach experiment and the analogy with classical electromagnetic waves (a vaguely historical motivation based on wave-particle duality). For example, a plane wave polarized in the $\hat{\mathbf{x}}$ direction can be written as

$$\mathbf{E}(\mathbf{x}, t)=E_0\cos(kz-\omega t)\hat{\mathbf{x}}=\mathfrak{Re}\left\{E_0e^{i(kz-\omega t)}\right\}\hat{\mathbf x}$$

This correspondence is due to the fact that $U(1)$ is isomorphic to $SO(2)$ . Applying a $\frac \pi 4$ counterclockwise rotation in the $xy$ -plane, the $\frac \pi 4$ filter gives

$$\begin{bmatrix}\hat{\mathbf x}’\\\hat{\mathbf y}’\end{bmatrix}=\begin{bmatrix}\cos\frac\pi 4&\sin\frac\pi 4\\-\sin\frac\pi 4&\cos\frac\pi 4\end{bmatrix}\begin{bmatrix}\hat{\mathbf x}\\\hat{\mathbf y}\end{bmatrix}$$

Sakurai shows that from the empirical information of Stern-Gerlach experiments, spin states in the $\hat{\mathbf z}$ direction correspond to $\hat{\mathbf{x}}$ and $\hat{\mathbf y}$ polarizations of light, the $\hat{\mathbf x}$ states correspond to the $\hat{\mathbf{x}}'$ and $\hat{\mathbf y}'$ polarizations, and $\hat{\mathbf y}$ states correspond to circularly polarized light (equivalent to multiplying the $\hat{\mathbf y}$ component by $i$ , representing a $\frac \pi 2$ phase shift).

Vector spaces

Denote the elements of a complete, complex inner product space by $|x\rangle\in V$ . To the vector space $V$ , there is a dual vector space which is the set of linear functionals $\omega: V\rightarrow \mathbb{R}$ .

Bases

With the axiom of choice (in the form of Zorn’s lemma) any (possibly infinite dimensional) vector space has a basis called a Hamel basis. When we restrict to a complete inner product space (includes all limit points), a Hilbert space, which has the additional requirement of being separable then there exists a countable (orthonormal) basis $\{e_i | i\in \mathbb{N}\}$ and the Hilbert space is isomorphic to $l^2(\mathbb{R})$ . There is a natural correspondence between a basis in the Hilbert space $V$ and a basis in the dual space $\omega_i\in V^*$ . We require that

$$\omega_i(e_j)=\delta_{ij}$$

where $\delta_{ij}$ is the Kronecker delta which is $1$ if $i=j$ and $0$ otherwise.

The Reisz representation theorem for Hilbert spaces guarantees that $V^*$ and $V$ are related such that

$$\forall \omega\in V^*, \forall v\in V, \exists w\in V\ \text{s.t. }\omega(v)=\langle w, v\rangle$$
and
$$||w||_V=||\omega||_{V^*}$$
where the norm induced by the inner product in a Hilbert space is defined by $||v||_H=\sqrt{\langle v, v\rangle}$ .

The Hilbert space inner product satisfies the following properties

$$\langle y, x\rangle = \overline{\langle x,y\rangle}$$

$$\langle ax_1+bx_2,y\rangle=a\langle x_1,y\rangle+b\langle x_2,y\rangle$$

and the inner product is positive definite.

Operators

The set of automorphisms of this inner product space are linear operators, which for finite dimensional vector spaces can be represented by invertible matrices $GL_n(\mathbb{C})$ . There is identification for operators acting on $V$ with operators acting on $V^*$ ; $A^*$ is the adjoint of an operator $A$ if

$$\forall x,y\in H: \langle Ax,y\rangle=\langle x,A^*y\rangle$$

For finite dimensional Hilbert spaces, this definition along with the first property of the inner product determines that the given the matrix representation $A_{ij}$ of an operator $A$ , its adjoint satisfies $$(A^*)_{ij}=\overline{A_{ji}}\equiv A^\dagger$$

With abuse of notation, if $A=A^*$ ( $A=A^\dagger$ ), then $A$ is self-adjoint.

For even more abstraction, see *-algebras.

modern physics, phenomenology

Congzhou Sha on August 25, 2020April 6, 2021

Bragg’s diffraction law can be derived heuristically as follows: two planes of molecules a distance $d$ apart, and each plane is evenly spaced. Using the image from Wikipedia:

For waves to interfere constructively, it must be that $|AB|+|BC|-|AC'|=n\lambda$ . $$\sin\theta=\frac{d}{|AB|}\rightarrow|AB|=\frac{d}{\sin\theta}$$ By symmetry, $$|BC|=\frac{d}{\sin\theta}$$ $$\cos\theta=\frac{|AC’|}{|AC|}$$ and $$\frac{d}{\frac{|AC|}2}=\tan\theta$$ so that
$$|AC’|=|AC|\cos\theta=\frac{2d}{\tan\theta}\cos\theta$$
$$|AC|+|BC|-|AC’|=\frac{2d}{\sin\theta}(1-\cos^2\theta)=2d\sin\theta=n\lambda$$

Single-slit diffraction can be derived as follows using a picture from this website:

The extra distance traveled from the bottom of the slit compared to the top of the slit, if the angle to the central axis of the slit is $\theta$ , is $D\sin\theta$ . Integers multiples of the wavelength interfere constructively (net), while half integer multiples of the wavelength interfere destructively.

Double slit experiment, with slits separated by distance $d$ . Say the screen is a distance $R$ away and that $d\ll R$ . The result is the same as in the single-slit experiment, but the peaks are more even in intensity (closer to $\sin \theta$ rather than $sinc\theta$ .

some fluid mechanics

Congzhou Sha on August 21, 2020August 21, 2020

The material derivative is defined considering a control volume $V$ . In this picture, the fluid element $dV$ containing a fixed amount of fluid is moving in space with time, and the quantities that describe the fluid element vary only with time. This is called the Lagrangian picture. The Eulerian picture corresponds to making the properties depend on time and space, and instead having the unit volume fixed with time.

The Reynolds transport theorem fixes the relationship between the Lagrangian and Eulerian pictures. Let $\alpha(t)$ be a fluid property in the Lagrangian picture. Then define

$$\int_V \frac{D}{Dt}\alpha(t) dV= \lim_{\Delta t\rightarrow 0}\frac 1 {\Delta t}\left(\int_{V+\Delta V} \alpha(t+\Delta t) dV-\int_{V}\alpha(t)dV\right)$$
where we have made use of the fact that the volume $V$ of the fluid element changes with time. The trick is to add and subtract $\int_{V}\alpha(t+dt)dV$ to consider the changes in $t$ and $V$ separately.
$$\int_V \frac{D}{Dt}\alpha(t) dV= \lim_{\Delta t\rightarrow 0}\frac 1 {\Delta t}\left(\int_{V+\Delta V} \alpha(t+\Delta t) dV-\int_{V}\alpha(t+\Delta t)dV+\int_{V}\alpha(t+\Delta t)dV-\int_{V}\alpha(t)dV\right)$$
We can now simplify the last two terms, since
$$\int_{V}\alpha(t+\Delta t)dV-\int_{V}\alpha(t)dV=\int_V\alpha(t+\Delta t)-\alpha(t)dV$$
so that with the limit and factor of $\frac 1 {\Delta t}$ ,
$$\lim_{\Delta t\rightarrow 0} \frac{\alpha(t+\Delta t)-\alpha(t)}{\Delta t}=\frac{\partial \alpha}{\partial t}$$
To deal with the first two terms, given the velocity of the fluid element $\mathbf{u}$ ,
$$\int_{V+\Delta V} \alpha(t+\Delta t) dV-\int_{V}\alpha(t+\Delta t)dV=\int_{\Delta V}\alpha(t+\Delta t)dV$$
the element $dV=\mathbf{u}\cdot\mathbf{n}\Delta t dS$ , thus instead of integrating over $\Delta V$ , integrate over the surface $S(t)$ .
$$\lim_{\Delta t \rightarrow 0} \frac{\int_{S(t)}\alpha(t+\Delta t)\mathbf{u}\cdot\mathbf{n}\Delta t dS}{\Delta t}=\int_{S(t)}\alpha(t)\mathbf{u}\cdot\mathbf{n}dS$$
Using the divergence theorem,
$$\int_{S(t)}\alpha(t)\mathbf{u}\cdot\mathbf{n}dS=\int_V \nabla\cdot(\alpha\mathbf{u})dV$$
Since the preceding argument is true for any control volume $V$ , the integrands must be equal.
$$\frac{D}{Dt}\alpha(t)=\frac \partial{\partial t}\alpha(t)+\nabla\cdot(\alpha(t)\mathbf{u})$$
For conservation of mass, we require the material derivative of mass be equal to $0$ , since no mass flows in or out of a control volume. $\alpha(t)$ is a property per unit volume, so the quantity to examine is $\rho=\frac{m}{V}$ .
$$\frac{D}{Dt}\rho=\frac{\partial}{\partial t}\rho+\nabla\cdot(\rho\mathbf{u})=0$$
This is the continuity equation.

some statistical mechanics

Congzhou Sha on August 16, 2020August 24, 2020

Based on Reif, chapters 3 and 5.

Let $\Omega$ be the number of microstates. Let $\mathchar'26\mkern-12mu d Q$ be the infinitesimal heat (change in occupation numbers of energy levels, without change in the energy levels). For the microcanonical ensemble, the change in internal energy is $dE=\mathchar'26\mkern-12mu dQ-PdV$ , which is clear from conservation of energy. $\beta=\frac{d\ln\Omega}{dE}$ for convenience. This has units of inverse energy, so let $\beta=\frac{1}{kT}$ , where $k$ is a constant with units of energy and $T$ is a unitless parameter. For more convenience, define $S=k\ln \Omega$ .

Consider the change in $\ln\Omega$ when we increase the energy by $\mathchar'26\mkern-12mu d Q$ , which is

$\ln\Omega(E+\ \mathchar'26\mkern-12mu d Q)-\ln\Omega(E)=\frac{1}{k}dS$

Taylor expansion gives

$\ln\Omega(E+\ \mathchar'26\mkern-12mu d Q)-\ln\Omega(E)=\frac{d\ln \Omega}{dE}\ \mathchar'26\mkern-12mu d Q+O(\ \mathchar'26\mkern-12mu d Q^2)$
But substituting our definition for $\beta$ ,
$\frac{1}{k}dS=\beta\ \mathchar'26\mkern-12mu d Q$
$dS=\frac{\mathchar'26\mkern-12mu d Q}{T}$
This connects the statistical definition of entropy $S$ with the thermodynamic definition of heat $\mathchar'26\mkern-12mu d Q$ .

Now, consider $N$ particles in a volume $V$ which is composed of cells of side length $l$ . There are thus $\frac{V}{l^3}$ possible cells. We can compute $\Omega$ directly, assuming indistinguishable particles, as

$\Omega=\frac{1}{N!}\left[\frac{V}{l^3}\right]^N$

Assuming that this is an isolated system of an ideal gas, use the first law of thermodynamics for the microcanonical ensemble to solve for $P$ .

$P=\frac{\partial Q}{\partial V}-\frac{\partial E}{\partial V}$
Assume that $E$ does not depend on $V$ , which is clear from assumptions that there is no interaction energy between individual particles in an ideal gas. Also, apply the relation between entropy and heat from above:
$P=T\frac{\partial S}{\partial V}=kT\frac{\partial \ln \Omega}{\partial V}=kT\frac{\partial }{\partial V}\left[N(\ln V-\ln l^3)-\ln N!\right]$
$P=\frac{NkT}{V}$
$PV=NkT$

Now we prove that $c_P=c_V+R$ . We know the equation of state for one mole of an ideal gas, $PV=RT$ . Start with the equation

$dQ=TdS$
We want to find
$c_V=(\frac{\partial Q}{\partial T})_V, c_P=(\frac{\partial Q}{\partial T})_P$
In the end, we need to get a bunch of differentials $dT$ so that we can relate $c_P$ and $c_V$ , as they are derivatives of $Q$ with respect to $T$ . Additionally, we want to use $dV=0$ or $dP=0$ . Thus expand in terms of $T$ and $P$ first.
$dQ=TdS=T\left[(\frac{\partial S}{\partial T})_PdT+(\frac{\partial S}{\partial P})_TdP\right]$
$T(\frac{\partial S}{\partial T})_PdT=(\frac{\partial Q}{\partial T})_PdT=c_PdT$

So now we have $c_P$ in the equation. Next, expand $dP$ in terms of $T,V$ .
$dP=(\frac{\partial P}{\partial T})_VdT+(\frac{\partial P}{\partial V})_TdV$
Putting this all together, we have
$dQ=c_PdT+T(\frac{\partial S}{\partial P})_T\left[(\frac{\partial P}{\partial T})_VdT+(\frac{\partial P}{\partial V})_TdV\right]$
If we require that $dV=0$ , this equation implies
$c_V=(\frac{\partial Q}{\partial T})_V=c_P+T(\frac{\partial S}{\partial P})_T\left[(\frac{\partial P}{\partial T})_V\right]$
We can easily calculate $(\frac{\partial P}{\partial T})_V=\frac{R}{V}$ from the equation of state. All that remains is $(\frac{\partial S}{\partial P})_T$ , which we can find using a Maxwell relation:
$G=U-TS+PV\implies dG=-SdT+VdP$
$(\frac{\partial G}{\partial T})_P=-S, (\frac{\partial G}{\partial P})_T=V$
$-(\frac{\partial S}{\partial P})_T=(\frac{\partial}{\partial P})_T(\frac{\partial G}{\partial T})_P=\frac{\partial^2 G}{\partial P\partial T}=\frac{\partial^2 G}{\partial T\partial P}=(\frac{\partial}{\partial T})_P(\frac{\partial G}{\partial P})_T=(\frac{\partial V}{\partial T})_P$

Using the equation of state again, we find that $(\frac{\partial S}{\partial P})_T=-(\frac{\partial V}{\partial T})_P=-\frac{R}{P}$ . Thus, using the equation of state again,
$c_V=c_P-T\frac{R}{P}\frac{R}{V}=c_P-\frac{R^2T}{PV}=c_P-\frac{R(RT)}{RT}=c_P-R$
$c_P=c_V+R$

Most derivations I’ve seen handwave and say something about the internal energy of an ideal gas only depending on the temperature, but it’s nice to see that this fact is implicitly buried in the equation of state.

Maxwell relations are easy to remember: just remember we want $dV, dP, dS, dT$ . There are four possible pairs, so define energies

$dE=TdS-PdV,\\H=E+PV\implies dH=TdS+VdP\\F=E-TS\implies dF=-SdT-PdV,\\G=E-TS+PV\implies dG=-SdT+VdP$

From there, the second derivatives with respect to each pair work out so that $dU=WdX+YdZ\implies (\frac{\partial W}{\partial Z})_X=(\frac{\partial Y}{\partial X})_Z$ .

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