Susskind's Particle Pyhsics I: Basic Concepts Note

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Lecture 1: Particles and Light

Wave Properties

\begin{align} \boxed{\frac{\lambda}{T} = c} \end{align} \begin{align} f = \frac{1}{T} \quad \boxed{\lambda f = c} \end{align} \begin{align} \omega = 2\pi f = \boxed{2\pi \frac{c}{\lambda} = \omega} \end{align}

Photon Properties

\begin{align} \boxed{E_{\mathrm{phton}} = \hbar\omega} \end{align} \begin{align} E_{\mathrm{ray}} = n \hbar \omega \end{align} \begin{align} p = \frac{E}{c} \end{align}

Lecture 2: Review of Quantum Mechanics & Harmonic Oscillator

Periodic Boundary Condition

In order to make momentum conseved, we need periodic boundary condition, so

\begin{align} \lambda = \frac{L}{N} \end{align}

\(\lambda\) is discrete.

\begin{align} p = \frac{h}{\lambda} = \frac{h}{L}N \end{align}

Harmonic Oscillator

\begin{align} E &= n \hbar\omega\\ a^{+ }|n \rangle &= \sqrt{n + 1} | n+ 1 \rangle \\ a^{- }|n \rangle &= \sqrt{n } | n - 1\rangle \\ a^{+ }a^{- } &\sim n \\ a^{- }a^{+ } &\sim n + 1 \end{align}

Lecture 3: Field Operators

Field operator

In periodic boundary conditions, a plane wave

\begin{align} e^{\mathrm{i}kL}, \quad, k = n\frac{2\pi}{L} \end{align}

One particle with momentum \(k_7\)

\begin{align} |0, 0, 0, 0, 0, 0, 1, 0, 0, \cdots \rangle \end{align}

Classical version of Fourier coefficient of field

\begin{align} \psi(x) &= \sum_k \alpha(k) e^{\mathrm{i} kx} \\ \psi^{ * }(x) = \sum_k \alpha^{ * }(k) e^{- \mathrm{i}k x} \end{align}

quantum version

\begin{align} \psi^{ -}(x) &= \sum_k \alpha^{- }(k) e^{\mathrm{i} kx} \\ \psi^{ + }(x) &= \sum_k \alpha^{ + }(k) e^{- \mathrm{i}k x} \end{align}

Let

\begin{align} |0\rangle \equiv& |0, 0, 0, 0, 0, 0\cdots \rangle\\ |k_i\rangle \equiv& |0, 0, 0, \cdots , 1, 0, 0\cdots \rangle , \quad\mathrm{one} \quad \mathrm{in}\quad i \mathrm{th} \end{align}

Create a particle at position \(x\)

\begin{align} \psi^{+ }(x) |0\rangle = \sum_k e^{-\mathrm{i} k x} a^{+ }(k)| 0 \rangle = \sum_k e^{-\mathrm{i} k x} | k \rangle \end{align}

Possible Reaction & Stimulated Emission

A particle comes in, and when it hits the target. The target happens to be localized at position \(x\) . That particle disappears. And then two particles emitted from the same spot. How would we describe this reaction?

\begin{align} \Psi^{\dagger}(x) \Psi^{\dagger}(x) \Psi(x)|\rangle \end{align}

If that a particle with momentum \(k_i\) is absorbed at position \(x\) . What is the final state? What is the various probabilities for this final state to have this or that momentum?

A photon comes in, hits an atom localized at a point and from that point two photons go off

\begin{align} \Psi^{\dagger}(x) \Psi^{\dagger}(x) \Psi(x)|k_{\mathrm{i}}\rangle =& \sum_{l,m} a^{\dagger}(l) e^{-\mathrm{i}lx} a^{\dagger}(m) e^{-\mathrm{i}mx} \sum_k a(k) e^{\mathrm{i}kx} |k_{\mathrm{i}}\rangle \\ =& \sum_{l,m} a^{\dagger}(l) e^{-\mathrm{i}lx} a^{\dagger}(m) e^{-\mathrm{i}mx} e^{\mathrm{i}k_{\mathrm{i}}x} |0\rangle \\ =& \sum_{l,m} e^{\mathrm{i}(k_{\mathrm{i}} - l - m)}|l, m\rangle \end{align}

本来想用这个说明 stimulated emission, 因为 \(l = m\) 时会有 factor \(\sqrt{2}\) , 产生相同动量的两个粒子的概率会增加. 但是翻车了, 因为 \(l \neq m\) 时, 同样也会有一个 \(2\) , 因为 \(l = 1, m = 2\) 和 \(l = 2, m = 1\) 都会有贡献.

然后用另一个例子说明 stimulated emission. Supposing we have a particle which can decay and emit a photon.

\begin{align} \Psi^{\dagger}(x) |0\rangle = \sum_k e^{-\mathrm{i} kx}|k\rangle \end{align}

equal probability for all \(k\) .

If there pre-exist a particle

\begin{align} \Psi^{\dagger}(x) |l\rangle = \sum_k e^{-\mathrm{i} kx} a^+(k)|l\rangle \end{align}

if \(k\neq l\)

\begin{align} \Psi^{\dagger}(x) |l\rangle = \sum_k e^{-\mathrm{i} kx} |k, l\rangle \end{align}

equal probability. But if \(k = l\)

\begin{align} e^{-\mathrm{i} kx} \sqrt{2}|k, k\rangle \end{align}

There are no photons present then just decays with coefficient \(1\) , that's spontaneous emission. The presence of particles of a give type will increase the probability that out decay takes places with the final momentum being the same as the momentum already there. This is stimulated emission. Bosons!

Density of \(x\)

\begin{align} \Psi^{\dagger}(x) \Psi(x) \end{align} \begin{align} \frac{1}{L}\int \Psi^{\dagger}(x) \Psi(x) \cdot\mathrm{d}x = \frac{1}{L} = \mathrm{Number} \quad \mathrm{of} \quad \mathrm{particles} \end{align}

Lecture 4: Scattering Process

Simplest Quantum Field

The simplest quantum field, be a function of only one coordinate namely \(x\) .

\begin{align} \Psi (x, t) =& \sum_k a^-(k) e^{\mathrm{i}kx} e^{\mathrm{i}\omega(k)t} \quad \mathrm{(Definition)} \\ \Psi^{\dagger} (x, t) =& \sum_k a^-(k) e^{-\mathrm{i}kx} e^{\mathrm{i}\omega(k)t} \end{align}

Find the wave equation for \(\Psi\) .

\begin{align} \frac{\partial}{\partial t} \Psi &= \mathrm{i} \omega(k) \Psi \\ \frac{\partial}{\partial x} \Psi &= \mathrm{i} k \Psi \\ \frac{\partial^2}{\partial x^2} \Psi &= - k^2 \Psi \\ \end{align}

so

\begin{align} \mathrm{i}\frac{\partial}{\partial t} \Psi = \frac{\omega}{k}\frac{\partial^2}{\partial x^2} \Psi \end{align}

if \(\omega = \frac{k^2}{2m}\) , a non-relativitic particle

\begin{align} \mathrm{i}\frac{\partial}{\partial t} \Psi = \frac{1}{2m}\frac{\partial^2}{\partial x^2} \Psi \end{align}

\(\Psi\) is an operator.

A Model: Scattering

A particle absorbed by a fixed target and emitted by the target instantly. This process is energy conserved but momentum not conserved. Averaging or integrating over all possiable time (set \(x_0 = 0\))

\begin{align} \langle k_f | g \int \mathrm{d}t\cdot \Psi^{\dagger}(0, t) \Psi(0, t) | k_i \rangle = 2 \pi g \delta (\omega_f - \omega_i) \end{align}

where \(g\) is coupling constant. There is connection between the fact that conservation of eneryg and the fact there's no preference of any specified time.

Lecture 5: Fermion

Phase Velocity & Group Velocity

\begin{align} \omega = \sqrt{k^2 + m^2} \end{align}

for massless \(m = 0\)

\begin{align} v_{\mathrm{p}} &= \frac{\omega}{k} = 1 \\ v_{\mathrm{g}} &= \frac{\mathrm{d}\omega}{\mathrm{d}k} = 1 \end{align}

but if \(m \neq 0\)

\begin{align} v_{\mathrm{p}} &= \frac{\omega}{k} = \sqrt{1 + \frac{m^2}{k^2}} > 1\\ v_{\mathrm{g}} &= \frac{\mathrm{d}\omega}{\mathrm{d}k} = \sqrt{\frac{k^2}{k^2 + m^2}} < 1 \end{align}

\(1\) is the speed of light. So phase velocity carries nothing.

Fermion

\begin{align} c^{+ } |0 \rangle &= |1\rangle\\ c^{+ } |1 \rangle &= 0\\ c^{- } |0 \rangle &= 0\\ c^{0 } |1 \rangle &= |0\rangle\\ \end{align} \begin{align} \boxed{c^{+ } c^{ -} |0 \rangle = 0 } \\ \boxed{c^{- } c^{ +} |0 \rangle = |0\rangle } \end{align}

Ground state

Boson: Bose condensate.

Fermion: Fermi sphere

Dirac Equation

move speed of light( \(\omega = k\) , \(c = 1\) ), carry electric charge, only move to one direction filed \(\Psi\)

\begin{align} \Psi = e^{\mathrm{i}(k x - \omega t)} = e^{\mathrm{i} k (x - t)} \end{align}

so

\begin{align} \frac{\partial \Psi}{\partial t} = - \frac{\partial \Psi}{\partial x}, \quad (\mathrm{i}\omega = \mathrm{i}k) \end{align}

describe both positive and negative energy. Negative is filled.

Lecture 6: Dirac Equation & Higgs Boson

Move right field

\begin{align} \frac{\partial \Psi}{\partial t} = - \frac{\partial \Psi}{\partial x} \end{align}

Move left field

\begin{align} \frac{\partial \Psi}{\partial t} = \frac{\partial \Psi}{\partial x} ,\quad (\omega = -k) \end{align} \begin{align} \dot{\Psi} \equiv \begin{pmatrix} \dot{\Psi}_R \\ \dot{\Psi}_L \end{pmatrix} = - \begin{pmatrix} 1 &0 \\0 & -1 \end{pmatrix} \begin{pmatrix} \frac{\partial \Psi_R}{\partial x} \\\frac{\partial \Psi_L}{\partial x} \end{pmatrix} \equiv - \alpha \frac{\partial}{\partial x} \Psi \end{align} \begin{align} \omega = \alpha k \end{align}

we want (not massless) \(\omega = \sqrt{k^2 + m^2}\) , so let \(\omega = \alpha k + \beta m\)

\begin{align} \omega^2 = k^2 + m^2 = (\alpha k + \beta m)^2 \Rightarrow \alpha^2 = 1, \beta^2 = 1, \alpha\beta + \beta\alpha = 0 \end{align}

so \(\beta\) can be

\begin{align} \beta = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{align}

so

\begin{align} \mathrm{i} \begin{pmatrix} \dot{\Psi}_R \\ \dot{\Psi}_L \end{pmatrix} = - \mathrm{i} \alpha \frac{\partial}{\partial x}\Psi + \beta m \Psi \end{align} \begin{align} \mathrm{i}\dot{\Psi}_R =& -\mathrm{i}\partial_x \Psi_R + m\Psi_L \\ \mathrm{i}\dot{\Psi}_L =& \mathrm{i}\partial_x \Psi_L + m\Psi_R \end{align}

For a particle at rest ( \(k = 0\) )

\begin{align} \mathrm{i} \frac{\partial}{\partial t} \Psi \equiv \mathrm{i} \begin{pmatrix} \dot{\Psi}_R \\\dot{\Psi}_L \end{pmatrix}= \beta m \Psi \equiv m \begin{pmatrix} \Psi_L \\ \Psi_R \end{pmatrix} \end{align}

\(\Psi_L\) and \(\Psi_R\) are coupled. We can decouple them by \(\Psi_+ = \Psi_L + \Psi_R\) , \(\Psi_- = \Psi_L - \Psi_R\)

\begin{align} \mathrm{i}\dot{\Psi}_+ &= m\Psi_+ \\ \mathrm{i}\dot{\Psi}_- &= -m\Psi_- \end{align}

Dirac Equation in 3 Dimensions

Let

\begin{align} \omega =& \sqrt{k_1^2 + k_2^2 + k_3^2 + m^2} \\ \omega =& \alpha k + \beta m = \alpha_1 k_1 + \alpha_2k_2 + \alpha_3 k_3 + \beta m \end{align}

so

\begin{align} \beta = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}, \quad \alpha_i = \begin{pmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{pmatrix} \end{align}

where \(I\) is \(2\times 2\) identity matrix, \(\sigma_i\) is Pauli matrix.

\begin{align} \mathrm{i}\frac{\partial }{\partial t} \Psi_p = - i (\alpha_i)_{pq} \frac{\partial}{\partial x^i}\Psi_q + \beta_q m \Psi_q \end{align}

Lecture 7: Angular Momentum

Angular momentum

  • Orbital
  • Spin
\begin{align} [x_i , x_j] &= 0 \\ [p_i , p_j] &= 0 \\ [x, p_x] &= \mathrm{i}\hbar \\ [x_i, p_j] &= \mathrm{i}\hbar\delta_{ij} \\ [L_x, L_y] &= \mathrm{i}\hbar L_z \end{align} \begin{align} L_+ &= L_x + \mathrm{i} L_y \\ L_- &= L_x - \mathrm{i} L_y \\ \end{align} \begin{align} [L_{+} , L_z] &= - L_{ +} \\ [L_{- }, L_z] &= + L_- \end{align}

Lecture 8: Spin

Review spin.

Lecture 9: Interaction

Diagrams but abstract.

Lecture 10: Path Integral

Abstract.

Reference