Notes

this is in gaussian units. Adapted from Shankar and Shroeder.

Postulates

1. Associated to any isolated physical system is a hilbert space which is completely described by its state vector.
2. The evolution of a closed quantum system is described by a unitary transformation. i.e. the shrodinger equation

Hamiltonian formalism

EL equations

L = T - V

\[ \dv{t}\left(\pdv{L}{\dot x}\right) = \pdv{L}{x} \]

• Generalised momentum

\[ \pdv{L}{\dot q} = p \]

• Generalised force

\[ \pdv{L}{q} = \dot{p} \]

Cyclic coordinate

When the Lagrangian does not depend on a coordinate qk:

\[ \pdv{L}{\dot q_k} = C \]

This is also generalised momentum.

q_k is thus a cyclic coordinate.

Energy conservation

\[ \pdv{E}{t} = -\pdv{L}{t} \]

\[ E = \left(\sum_{i=1}^N \pdv{L}{\dot q_i}\dot{q_i}\right) - L \]

If L has no time dependence, then E is conserved.

E is indeed total energy when the new coordinate q is not related to the old coordinate x with a time dependence \(x = x(q,t)\) or qdot dependence \(x = x(q,\dot{q})\).

Noether’s theorem

For each symmetry of the Lagrangian, there is a conserved quantity.

Let the Lagrangian be invariant to first order in the small number \(\epsilon\).

\[ q_i \to q_i + \epsilon K_i(q) \]

each K_i(q) may be a function of all the q_i and need not be constant.

The quantity:

\[ P(q,\dot{q}) = \sum_i \pdv{L}{\dot q_i}K_i(q) \]

does not change with time and has the generic name of consered momentum.

Example:

For \(L = (m/2)(5\dot{x}^2 - 2\dot{x}\dot{y} + 2\dot{y}^2) + C(2x-y)\), K_x = 1 and K_y = 2:

\[ P = \pdv{L}{\dot{x}}K_x + \pdv{L}{\dot y}K_y = m(3\dot{x} + 3\dot{y}) \]

Hamiltonian

We may write E as:

\[ E(q,\dot{q}) \equiv \pdv{L(q,\dot{q})}{\dot{q}}\dot{q} - L(q,\dot{q}) \]

and we wish to exchange all \(\dot{q}\) for \(p\) in E.

\[ p \equiv \pdv{L}{\dot{q}} \]

When written in this way, we define it as the Hamiltonian.

\[ H(q,p) \equiv p_i\dot{q_i}(q,p) - L(q,\dot{q}(q,p)) \]

Hamiltonian equations

\[ \pdv{H}{p} = \dot{q} \]

\[ \pdv{H}{q} = -\dot{p} \]

Cyclic coordinates

Same as Lagrange. if q is cyclic, p is conserved.

Let w(p,q) be some function of the state variables with no explicit dependence on t.

\[ \dot{\omega} = \{\omega,\mathcal{H}\} \]

where the Poisson brackets (PB) between two variables is:

\[ \left\{\omega,\lambda\right\} = \sum_i \left(\pdv{\omega}{q_i}\pdv{\lambda}{p_i} - \pdv{\omega}{p_i}\pdv{\lambda}{q_i}\right) \]

• Poisson bracket Identities

  Any variable whose PB with \(\mathcal{H}\) vanishes is constant in time.

  {a,b} = -{b,a} {a,b+c} = {a,b} + {a,c} {a,bc} = {a,b}c + b{a,c}

  (similar to commutators)

  \[ $\{q_i,q_j\} = \{p_i,p_j\} = 0 \]

  \[ \{q_i,p_j\}=\delta_{ij} \]

  as \(\partial{q_i}/\partial{q_j} = \delta_{ij}, \partial{q_i}/\partial{p_k}=0\)

  Hamiltonian equations may be written as:

  \begin{align*} \dot{q_i} = \{q_i,\mathcal{H}\} \\
  \dot{p_i} = \{p_i,\mathcal{H}\} \end{align*}

  In general,

  \[ [X,P] = i\hbar\{x,p\} \]

Legendre transform

Say that we wish to ‘invert’:

\[ \dv{F(x)}{x} = s(x) \]

by constructing a function G(s) which is the Legendre transform of F(x):

\[ \dv{G(s)}{s} = x(s). \]

By noting \(d(F+G) = s \dd{x} + x \dd{s} = d(sx)\), we may write:

\[ G = sx-F = F’x - F \].

Or

\[ G(s) = sx(s) - F(x(s)) \]

Operators

Differential operator

\[ D\ket{f} = \ket{\dv{f}{x}} \]

implies

In the |x> basis, \[ \bra{x}D\ket{x'} = D_{x,x'}=\delta'(x-x')=\delta(x-x')\dv{x'} \]

where it is integrated over the second index (x') and pulls out the derivative of f at the first index (x).

We can turn it hermitian by turning it into the form:

\[ K = -iD \]

but in infinite dimensions, K^t = K is not sufficient to be hermitian. It is only hermitian in the space of functions that obey:

\[ -ig^*(x)f(x) | \given_a^b = 0 \]

so that <g|K|f> = <f|K|g>*.

• eigeneverything

  X-basis

  \[ K\ket{k} = k\ket{k} \]

  \[ -i\dv{x}\psi_k(x) = k\psi_k(x) \]

  where \(\psi_k(x) = \braket{x}{k}\).

  the solution is:

  \[ \psi_k(x) = Ae^{ikx} \]

  where we choose \(A = (1/2\pi)^{-1/2}\)

  and <k|k>

X basis

X|x> = x|x>

\[ \bra{x'}X\ket{x} = x\delta(x'-x) \]

Action on functions:

X|f> = |F>

F(x) = xf(x)

\[ \bra{k}X\ket{k'} = i\delta'(k-k') \]

[X,K] = iI

Statevectors

Statevector expressed in certain basis.

\[ \ket{\psi} = \ket{ + z}\braket{+ z}{\psi} + \ket{-z}\braket{-z}{\psi} = c_+\ket{+ z} + c_- \ket{-z} \]

Expectation value: \(\langle\Omega\rangle\) Square of uncertainty: \((\Delta \Omega)^2 = \langle\Omega^2\rangle - \langle\Omega\rangle^2\)

Two states that differ by an overall phase are the same state.

Rotation Operators

Rotate anticlockwise about z-axis can be defined as:

\[ \hat{R}(\dd{\phi} \bm{k}) = 1 - \frac{i}{\hbar}\hat{J_z}\dd{\phi} \] where \(\hat{J_z}\) is a hermitian operator and has |+- z> as an eigenvector with +-hbar/2 as eigenvalues.

\[ \hat{R}(\phi \bm{k}) = \lim_{N\to\infty}\left[1-\frac{i}{\hbar}\hat{J}_z\left(\frac{\phi}{N}\right)\right]^N = \exp(-i\hat{J}_z\phi/\hbar) \]

\[ \hat{R}(\phi \bm{k})\ket{\pm z} = e^{\mp i \phi/2} \ket{\pm z} \]

Notice that rotating by 360deg causes the state to pick up an overall minus sign.

Matrix representation

The matrix representation of \(\hat{A}\) in basis x is \(A_{ij} = \bra{i}\hat{A}\ket{j}\) where i j are the bases of x.

Passive and active transform

passive: rotating the axes active: rotating the vector expressed in the same axes.

1D problems

\[ i \hbar\ket{\dot{\psi}} = H\ket{\psi} \]

Steps for problem:

1. Find the classical hamiltonian. Then find the quantum hamiltonian operator analogue.
2. Expand the H in terms of its eigenbasis to solve the time-independent problem, \(H\ket{E} = E\ket{E}\) by solving the differential equation in some basis.
3. To find the state as it evolves over time, we expand \(\ket{\psi(t)}\) in terms of E. and find that the propogator is given by \(\sum_E \ketbra{E}{E}e^{-i E t / \hbar}\). The propogator is also given by \(U(t) = e^{-i H t/\hbar}\), which can be expanded.

Ehrenfest’s theorem

\[ \dv{}{t} \langle \Omega \rangle = \frac{1}{i\hbar}\langle [\Omega,H]\rangle \]

Particle in a box

n odd,positive:

\[ \psi_n(x) = \sqrt{\frac{2}{L}} \cos(\frac{n\pi x}{L}) \]

n even,positive:

\[ \psi_n(x) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}) \]

\[ E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2 \pi^2 n^2}{2mL^2} \]

Probability continuity

\[ \pdv{(\psi^* \psi)}{t} = -\div \frac{\hbar}{2mi}(\psi^* \grad\psi - \psi\grad\psi^*) \]

Theorems

1. There is no degeneracy in 1D bound states. \(\psi(x\to\infty) \to 0\)
2. The eigenfunctions of H can always be chosen pure real in the coordinate basis.

Harmonic oscillator

\[ H = \frac{P^2}{2m} + \frac12 m\omega^2 X^2 \]

Step 1. Introduce dimensionless variables natural to the problem. Step 2. Extract the asymptotic behavior of \(\psi\). Step 3. Write iv as a product of the asymptotic form and an unknown function \(u\). The function \(u\) will usually be easier to find than \(\psi\). Step 4. Try a power series to see if it will yield a recursion relation.

Solving the time independent shrodinger’s equation using series solutions with hermite polynomials, we get:

\[ E_n = (n+\frac12)\hbar \omega \]

\[ \psi_n = \left(\frac{m\omega}{\pi\hbar 2^{2n} (n!)^2}\right)^{1/4} \exp(-\frac{m\omega x^2}{2\hbar})H_n\left[\left(\frac{m\omega}{\hbar}\right)^{1/2} x\right] \]

\[ U(x,t;x',t') = \sqrt{\frac{m\omega}{2\pi i \hbar\sin\omega T}\right} \exp\left[\frac{im\omega}{\hbar} \frac{(x^2 + x'^2)\cos\omega T -2xx'}{2\sin\omega T}\right] \]

where \(T = t-t'\)

The energies can also be found using heisenberg’s uncertainty principle, gaussian packets which saturate the relation, and trying to find the minimize energy for the minimal state.

Method of factorization

Introducing the operator,

\[ a = \sqrt{\frac{m\omega}{2\hbar}}X + i \sqrt{\frac{1}{2m\omega\hbar}}P \]

and its adjoint which satisfy \([a,a^\dagger] = 1\),

(notice \(m\omega \leftrightarrow \frac{1}{m\omega}\) as \(X \leftrightarrow P\))

Since \(a^\dagger a = \frac{H}{\hbar \omega} - \frac12\), define an operator \(\hat{H}\):

\[ \hat{H} = \frac{H}{\hbar\omega} = (a^\dagger a+1/2) \]

which has the properties,

\[ [a,\hat{H}] = a \] \[ [a^\dagger,\bar{H}] = -a^\dagger. \]

We wish to solve the eigenvalue equation for \(\hat{H}\), \(\hat{H}\ket{\epsilon} = \epsilon\ket{\epsilon}\).

It can then be shown that \(a\ket{\epsilon}\) is an eigenstate with eigenvalue \(\epsilon -1\) and so is \(a^\dagger \ket{\epsilon}\) with eigenvalue \(\epsilon +1\).

There is a lower bound to the energy, \(\epsilon_0 = \frac{1}{2}\), and thus we get back the energy levels we derived earlier.

In addition, we have,

\[ a\ket{n} = \sqrt{n}\ket{n-1} \] \[ a^\dagger\ket{n}=\sqrt{n+1}\ket{n+1} \] \[ a^\dagger a\ket{n} = n\ket{n} \]

These relations allow us to compute the matrix elements of operators in the \(\ket{n}\) basis by inverting the earlier relations for \(X\) and \(P\).

We can also express

\[ \ket{n} = \frac{(a^\dagger)^n}{\sqrt{n}}\ket{0} \]

• X-basis

  \[ \braket{x}{n} = \frac{1}{\sqrt{n!}}\left[\frac{1}{\sqrt{2}}\left(y-\dv{}{y}\right)\right]^n\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-y^2/2} \]

  where \(y = \sqrt{\frac{\hbar}{m\omega}}x\)

Path integral formulation

Finding propogators has been hard.

Stern-Gerlach

The intrinsic spin angular momentum of a particle, we write

\[ \bm{\mu} = \frac{gq}{2mc}\bm{S} \]

The force by the magnet equals:

\[ F_z = \mu_B \pdv{B}{z} \]

where u_b is the bohr magneton

Angular momentum

In general:

\[ \hat{R}(\phi \bm{n}) = e^{-i \bm{\hat{J}\cdot n} \phi / \hbar} \]

We can represent 2d rotation in the cartesian plane like this:

\begin{bmatrix} \bra{x}\hat{R}(\phi \bm{k})\ket{x} & \bra{x}\hat{R}(\phi \bm{k})\ket{y} \\
\bra{y}\hat{R}(\phi \bm{k})\ket{x} & \bra{y}\hat{R}(\phi \bm{k})\ket{y} \\
\end{bmatrix}

which just gives:

\begin{bmatrix} \cos\phi & -\sin\phi \\
\sin\phi & \cos\phi \\
\end{bmatrix}

Note that rotations and generators don’t commute as:

\[ [\hat{J}_x,\hat{J}_y] = i\hbar \hat{J}_z \]

which holds for cyclic permutations

Commuting operators

Consider two linear Hermitian operators \(\Omega, \Lambda\) which commute.

Suppose only a single state \(\ket{\omega}\) that is an eigenstate of \(\Omega\) with eigenvalue \(\omega\).

Then from the commutativity relation: \(\Lambda\ket{\omega}\) is also an eigenstate of operator \(\Omega\). Since we presumed there is only one such state, this shows that \(\Lambda\ket{\omega} = \lambda\ket{\omega}\).

We can then label the eigenvector as \(\ket{\omega,\lambda}\).

Degeneracy

If there is more than one eigenstate of the operator \(\Omega\) with eigenvalue \(\omega\), we say there is degeneracy.

Eigenvalues and eigenstates of angular momentum

Although the generators of rotations about different axes do not commute, the following operator commutes with each of the gneerators.

\[ \bm{\hat{J}}^2 = \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2 \]

Since it commutes with each generator, these operators have simulataneous eigenstates in common.

\[ \bm{\hat{J}}^2\ket{\lambda,m} = \lambda \hbar^2 \ket{\lambda,m} \]

\[ \hat{J}_z\ket{\lambda,m} = m \hbar \ket{\lambda,m} \]

Quantum statistics

Boltzmann statistics for ideal gases

The partition function of a composite system of \(N\) noninteracting indistinguishable gas molecules is of the form,

\[ Z = \frac{1}{N!}Z_1^N, \]

where \(Z_1\) is the partition function for a single individual particle, and the factorial prefactor expresses the number of ways of interchanging N particles with each other.

For a single molecule, we may decompose the energy of a molecule in terms of its translational kinetic energy \(E_{tr}\), and its internal energy \(E_{int}\) (rotational, vibrational etc.) for a particular state. Thus \(Z_1 = Z_{tr}Z_{int}\).

Particle in a box

In a one dimensional box of length \(L\), the standing-wave patterns are limited to wavelengths of \(\lambda_n = \frac{2L}{n}\) for positive integer \(n\). The corresponding momentum is \(p_n =\frac{hn}{2L}\) and the corresponding energy is \(E_n = \frac{h^2n^2}{8mL^2}\). Computing the translational partition function

\[ Z_{1d} = \sum_n e^{-E_n/kT} \approx \sqrt{\frac{2\pi m k T}{h^2}} L \equiv L/l_Q \]

where \(l_Q\) is the quantum length. It is the de Broglie wavelength of a particle of mass m whose kinetic energy is \(kT\), aside from the \(\pi\).

In three dimensions, we have three times the momentum components so instead \(Z_{tr} = V/v_Q\), where \(v_Q = l_Q^3\). Thus

\[ Z = \frac{1}{N!}\left(\frac{VZ_{int}}{v_Q}\right)^N. \]

• Thermal properties

  From the helmholtz free energy, we get the thermal properties

  \[ \mu = -kT \ln(\frac{VZ_{int}}{Nv_Q}) \]

Gibbs factor

Bosons and fermions

However, when considering finite quantum states with the issue of particle overlap, then the earlier partition function for composite systems may not apply.

A system of \(N\) indistinguishable particles which has no restriction on the number of particles that can occupy a single quantum state is said to obey Bose-Einstein statistics. i.e. bosons. If there is a restriction of at most particle occupying a single quantum state, the system is said to obey Fermi-dirac statistics. i.e. fermions (Think Pauli exclusion principle).

For reference, Maxwell-Boltzmann statistics concerns distinguishable without restrictions.

However, when the available single-particle states is much greater than the number of particles, i.e. \(Z_1 \gg N\), then the chance of two particles occupying the same state is negligible. Thus the regualr partition function applies, i.e \(V/N \gg v_Q\). In other words, the distance between particles must be much greater than the de Broglie wavelength. Quantum gases violate this condition. Some examples of systems which also violate the condition include very dense neutron stars, very cold liquid helium or very light particles like electrons in a metal or photons in a hot oven.

Distribution functions

Consider a single-particle-state (as opposed to a particle itself) system. The probabilty of a state being occupied by \(n\) particles is \(\mathcal{P}(n) = \frac{1}{\mathcal{Z}}e^{-n(\epsilon-\mu)/kT}\), where \(\epsilon\) is the energy of the state when occupied by a single particle.

For fermions, the average number of particles in the state is given by the formula

\[ \bar{n}_{FD} =\frac{1}{e^{(\epsilon-\mu)/kT}+1} \],

which is called the Fermi-Dirac distribution.

Similarly for bosons, the Bose-Einstein distribution is given by

\[ \bar{n}_{BE} = \frac{1}{e^{(\epsilon-\mu)/kT}-1} \].

For reference, the average occupancy for Boltzmann statistics is given by

\[ \bar{n}_{Boltzmann} = e^{-(\epsilon-\mu)/kT} \]

Degenerate Fermi gases

Consider a “gas” of fermions at a very low temperature such that the condition for Boltzmann statistics to apply is badly violated. At \(T=0\), all single-particle states with energy less than \(\mu(T=0)\) are occupied while all other states are unoccupied. We call \(\mu(T=0)\) the Fermi energy, denoted \(\epsilon_F\). The fermi-dirac distribution thus becomes a step function.

When the temperature is low enough that this is almost true, the gas is said to be degenerate.

Considering the energy distribution in n-space as a eighth-sphere and the relation \(\epsilon_F = \frac{h^2 n^2_{max}}{8mL^2}\) from particle in a box, we get the equation for the Fermi energy as a function of \(N\) and \(V\).

\[ \epsilon_F = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{2/3} \].

Thus the total energy integrated over n-space is

\[ U = 3/5 N\epsilon_{F}, \]

and we may verify that the Boltzmann condition is equivalent to the condition \(kT \gg \epsilon_F\). The required temperature for equality is the Fermi temperature.

The degeneracy pressure given by \(P = -(\partial U/\partial V)_{S,N}\) is given by

\[ P =\frac{2}{3}\frac{U}{V} \].

When a degenerate electron gas is compressed, the wavelength of all the wavefunctions are reduced, and hence the energies of all the wavefunctions increase. This is what keeps matter from collapsing under the huge electrostatic forces that pull electrons and protons together.

Ultraviolet catastrophe

In classical physics, we may consider EM radiation permeating a box, each standing wave behaving a harmonic oscillator with two degrees of freedom, thus with energy \(kT\). Since there are infinitely many such waves, there may seem to be an infinite amount of energy in the box.

However, in QM, a harmonic oscillator may only have energy levels of integer multiples of \(hf\) (relative to ground state). This results in the Planck distribution,

\[ \bar{n}_{Pl} =\frac{1}{e^{hf/kT}-1} \]

This freezes out the short-wavelength (high energy) modes, and allows for a finite energy.

To find the total energy of all the photons in the box, we may derive Wien’s law, by integrating over the Planck distribution with energy and a factor of two for the two independent polarisations of photons.

This is given by,

\[ \frac{U}{V} = \frac{8\pi(kT)^4}{(hc)^3}\int_0^\infty \frac{x^3}{e^x-1}\dd{x} = \frac{8\pi^5(kT)^4}{15(hc)^3} \]

under the change of variable \(x=\epsilon/kT\). The spectrum (energy density per unit photon energy) peaks at \(x=2.82\).

Other thermal properties

The heat capacity is

\[ C_V = \left(\pdv{U}{T}\right)_V = 4aT^3 \]

The entropy is given by

\[ S(T) = \frac{32\pi^5}{45}V\left(\frac{kT}{hc}\right)^3 k \]

Photons escaping through a hole (Radiation)

Volume of some chunk of photons: \(R\dd{\theta} \times R\sin\theta\dd{\phi} \times c\dd{t}\). Energy in chunk: \(U/V c\dd{t} R^2\sin\theta\dd{\theta}\dd{\phi}\) Probability of escape for photon in chunk: \(\frac{A\cos\theta}{4\pi R^2}\) Total energy: \(\frac{A}{4} \frac{U}{V}c\dd{t}\)

Thus we get intensity

\[ I = \frac{c}{4} \frac{U}{V} = \sigma T^4 \],

where the familiar Stefan-Boltzmann constant is given by

\[ \sigma =\frac{2\pi^5k^4}{15h^3c^2} = 5.67\times 10^{-8} Wm^{-2}K^{-4} \]