Chapter 1 Lagrangians

Functional

Functionals : \(F[f(x)]\to\) number Functional derivative : \(\displaystyle\frac{\delta F(f)}{\delta f(x)}=\lim_{ \varepsilon \to 0} \frac{F[f(x')+\varepsilon \delta(x-x') ]-F[f(x')]}{\varepsilon}\)

Example of Functional derivative

\(\displaystyle J[f]=\int [f(y)]^p\phi(y) \, \mathrm{d}y\) , \(\displaystyle \frac{\delta J}{\delta f}=p[f(y)]^{p-1}\)

\(\displaystyle H[f]=\int_{a}^{b} g(f) \, \mathrm{d}x,\, \frac{\delta H}{\delta f}=g'[f]\)

\(\displaystyle J[f]=\int g(f') \, \mathrm{d}x\) , \(\displaystyle \frac{ \delta H }{ \delta f }=-\frac{\mathrm{d}}{\mathrm{d}x}g'[f]\)

Lagrangians and least action

Action

\[ \displaystyle S=\int \, \mathrm{d}tL \quad L=T-V \quad \frac{ \delta S }{ \delta x(t)}=0\implies \frac{ \delta L }{ \delta x(t)}-\frac{ \mathrm{d}L }{ \mathrm{d}\dot{x}(t)}=0 \]

Lagrangians density

\[ \displaystyle L=\int \, \mathrm{d}^{3} \mathcal{L} \quad S=\int \mathrm{d}t \, \mathrm{d}^{3}x \mathcal{L}=\int\mathrm{d}^{4}x\mathcal{L}=\int\mathrm{d}^4x\mathcal{L}(\phi,\partial_\mu \phi) \]

\[ \implies \frac{ \delta S }{ \delta \phi}=\frac{ \partial \mathcal{L} }{ \partial \phi }-\partial_\mu\left( \frac{ \partial \mathcal{L} }{ \partial (\partial_\mu \phi)} \right)=0 \]

Example

\(\displaystyle \mathcal{L}=\frac{1}{2}(\partial_\mu \phi)^{2}-\frac{1}{2}m^{2}\phi^{2}\) \(\displaystyle \frac{ \delta \mathcal{L} }{ \delta \phi }=-m^{2}\phi\quad \partial_\mu \frac{ \partial \mathcal{L} }{ \partial (\partial_\mu \phi) }=\partial_\mu \partial^{\mu}\phi=\partial^{2}\phi\quad \frac{ \delta S }{ \delta \phi}=-m^{2}\phi-\partial^{2}\phi=0\implies(\partial^{2}+m^{2})\phi=0\)

Notes: The derivative of tensor.

\[\color{red}\begin{align}\frac{ \partial (\partial_\mu \phi)^{2} }{ \partial (\partial_\mu \phi) } = & \frac{ \partial }{ \partial (\partial_\mu \phi) } [g^{\alpha \beta}\partial _{\alpha}\phi \partial_{\beta}\phi]=g^{\alpha \beta}\frac{ \partial (\partial_{\alpha}\phi) }{ \partial (\partial_{\mu}\phi) }\partial_{\beta }\phi +g^{\alpha \beta}\partial_{\alpha}\phi\frac{ \partial (\partial_{\beta}\phi) }{ \partial (\partial_{\mu}\phi) } \\= & g^{\alpha \beta}\delta_{\alpha}^{\mu}\partial_{\beta}\phi+g^{\alpha \beta}\delta_{\beta}^{\mu}\partial_{\alpha}\phi \\= & 2g^{\alpha \beta}\delta_{\alpha}^{\mu}\partial_{\beta}\phi=2\partial^{\mu}\phi\end{align}\]

Chapter 2 Simple harmonic oscillators

First quantization: Particles behave like waves.

Second quantization: Waves behave like particles.

\(\displaystyle \hat{H}=\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}\) \(\displaystyle \hat{a}=\sqrt{ \frac{m\omega}{2\hbar} }\hat{x}+\frac{\mathrm{i}\hat{p}}{\sqrt{ 2m\omega \hbar }}\quad \hat{a}^{\dagger}=\sqrt{ \frac{m\omega}{2\hbar} }\hat{x}-\frac{\mathrm{i}\hat{p}}{\sqrt{ 2m\omega \hbar }}\) \(\displaystyle [\hat{a},\hat{a}^{\dagger}]=1\) \(\displaystyle \hat{a}\) and \(\displaystyle \hat{a}^{\dagger}\) are hermitian conjugate. \(\displaystyle \hat{a}\) and \(\displaystyle \hat{a}^{\dagger}\) operate on the state \(\displaystyle \ket{n}\) :

\[\begin{align} \hat{a}\ket{n} = & \sqrt{ n }\ket{n-1} \\ \hat{a}^{\dagger}\ket{n} = & \sqrt{ n+1 }\ket{n+1} \\ \ket{n} = & \frac{(\hat{a}^{\dagger})^{n}}{\sqrt{ n! }}\ket{0} \end{align}\]

For \(N\) uncoupled simple harmonic oscillators. \(\displaystyle \hat{H}=\sum_{k=1}^{n}\hat{H}_{k} \quad \hat{H}_{k}=\frac{\hat{p}^{2}_{k}}{2m}+\frac{1}{2} m\omega^{2}_{k}\hat{x}^{2}_k\) Acting with an operator \(\displaystyle a_{k}\) only affects the number of quanta for the \(k\) th oscillator.

\[\begin{array}{l} \hat{a}_{k}^{\dagger}\left|n_{1}, n_{2}, \ldots, n_{rk}, \ldots\right\rangle \propto\left|n_{1}, n_{2}, \ldots, n_{k}+1, \ldots\right\rangle \\ \hat{a}_{k}\left|n_{1}, n_{2}, \ldots, n_{k}, \ldots\right\rangle \propto\left|n_{1}, n_{2}, \ldots, n_{k}-1, \ldots\right\rangle \end{array}\]

The operators will have the commutation rules:

\[\begin{aligned} {\left[\hat{a}_{k}, \hat{a}_{q}\right] } & =0 \\ {\left[\hat{a}_{k}^{\dagger}, \hat{a}_{q}^{\dagger}\right] } & =0 \\ {\left[\hat{a}_{k}, \hat{a}_{q}^{\dagger}\right] } & =\delta_{k q} \end{aligned}\]

\[ \hat{H}=\sum_{k=1}^{N} \hbar \omega_{k}\left(\hat{a}_{k}^{\dagger} \hat{a}_{k}+\frac{1}{2}\right) \]

vacuum state (ground state) \(\displaystyle \ket{0,0,\dots,0}\) (\(\displaystyle \ket{0}\)) \(\displaystyle \hat{a}_{k}\ket{0}=0\quad k=1,2,\dots,N\) A general state of the system, written as \(\mid n_{1}, n_{2}, \dots, n_{N} \rangle\) is known as the occupation number representation.

\[ \left|n_{1}, n_{2}, \cdots, n_{N}\right\rangle=\frac{1}{\sqrt{n_{1} ! n_{2} ! \cdots n_{N} !}}\left(\hat{a}_{1}^{\dagger}\right)^{n_{1}}\left(\hat{a}_{2}^{\dagger}\right)^{n_{2}} \cdots\left(\hat{a}_{N}^{\dagger}\right)^{n_{N}}|0,0, \cdots 0\rangle \]

More succinctly

\[ \left|\left\{n_{k}\right\}\right\rangle=\prod_{k} \frac{1}{\sqrt{n_{k}}}\left(\hat{a}_{k}^{\dagger}\right)^{n_{k}}|0\rangle \]

Chapter 3  Occupation number representation

Natural units , a particle in a box of size \(\displaystyle L\), then \(\displaystyle \psi(x)=\frac{1}{\sqrt{ L }}\mathrm{e}^{ \mathrm{i}px }\) boundary condition \(\displaystyle \psi (x)=\psi (x+L)\implies p_{m}=\frac{2\pi m}{L}\) Occupation number representation Describing the many-particle system by listing the number of identical particles in each quantum state. \(\displaystyle \ket{n_{1} ,n_{2},n_{3},\dots}\), \(\displaystyle n_{i}\) represent that there are \(\displaystyle n_{i}\) particles in state \(\displaystyle \ket{p_{i}}\).

\[ \hat{H}\ket{n_{1},n_{2},n_{3}\dots} =\sum_{m}n_{p_{m}}E_{p_{m}} \]

A general state builded up by using \(\displaystyle \hat{a}_{k}^{\dagger}\) acting on the vacuum state \(\displaystyle \ket{0}\)

\[ \left|n_{1} n_{2} \cdots\right\rangle=\prod_{k} \frac{1}{\left(n_{k} !\right)^{\frac{1}{2}}}\left(\hat{a}_{k}^{\dagger}\right)^{n_{k}}|0\rangle \]

Indistinguish ability and symmetry

\[ \hat{a}^{\dagger}_{p_{1}}\hat{a}^{\dagger}_{p_{2}}=\lambda \hat{a}^{\dagger}_{p_{2}}\hat{a}^{\dagger}_{p_{1}} \]

• \(\displaystyle \lambda=1\) Bosons

\[ \hat{a}^{\dagger}_{p_{1}}\hat{a}^{\dagger}_{p_{2}}= \hat{a}^{\dagger}_{p_{2}}\hat{a}^{\dagger}_{p_{1}} \]

or

\[ \hat{a}_{1}^{\dagger}\hat{a}_{2}^{\dagger}=\hat{a}_{2}^{\dagger}\hat{a}^{\dagger}_{1} \]

Commutation relation

\(\displaystyle [\hat{a}_{i},\hat{a}_{j}]=[\hat{a}^\dagger_{i},\hat{a}^\dagger_{j}]=0\), \(\displaystyle [\hat{a}_{i},\hat{a}^\dagger_{j}]=\delta_{ij}\)

acting on general state

\[ \hat{a}^{\dagger} _{i}\ket{n_{1},\dots,n_{i},\dots}=\sqrt{ n_i+1 }\ket{n_{1},\dots,n_{i},\dots} \quad\hat{a}_{i}\ket{n_{1},\dots,n_{i},\dots}=\sqrt{ n_{i} }\ket{n_{1},\dots,n_{i},\dots} \]

- \(\displaystyle \lambda=-1\) Fermions

\[ \hat{c}^\dagger_{i}\hat{c}^\dagger_{j}+\hat{c}^\dagger_{j}\hat{c}^\dagger_{i}=0\quad\left\{ \hat{c}^\dagger_{i},\hat{c}^\dagger_{j}\right\}=0\implies \hat{c}^\dagger_{i}\hat{c}^\dagger_{i}=0 \]

Pauli exclusion principle which means that each state can either be unoccupied or contain a single fermion. acting on general state

\[ \hat{c}^{\dagger} _{i}\ket{n_{1},\dots,n_{i},\dots}=(-1)^{\sum_{i}}\sqrt{ 1-n_{i}}\ket{n_{1},\dots,n_{i},\dots} \quad\hat{c}_{i}\ket{n_{1},\dots,n_{i},\dots}=(-1)^{\sum_{i}}\sqrt{ n_{i} }\ket{n_{1},\dots,n_{i},\dots} \]

where \(\displaystyle (-1)^{\sum_{i}}=(-1)^{n_{1}+n_{2}+\dots+n_{i-1}}\)

The continuum limit

\(\displaystyle [\hat{a}_{\boldsymbol{p}},\hat{a}_{\boldsymbol{q}}]=\delta ^{(3)}(\boldsymbol{p}-\boldsymbol{q})\), \(\displaystyle \hat{H}=\int\mathrm{d}^{3}pE_{\boldsymbol{p}}\hat{a}_{\boldsymbol{p}}\hat{a}^\dagger_{\boldsymbol{p}}\)

Chapter 4 Making second quantization work

Filed operators

\[ \hat{\psi}^{\dagger}(\boldsymbol{x})=\frac{1}{\sqrt{ \mathcal{V} }}\sum_{\boldsymbol{p} }\hat{a}^\dagger_{\boldsymbol{p}}\mathrm{e}^{ -\mathrm{i}\boldsymbol{p}\cdot \boldsymbol{x} }\quad\hat{\psi}(\boldsymbol{x})=\frac{1}{\sqrt{ \mathcal{V} }}\sum_{\boldsymbol{p} }\hat{a}_{\boldsymbol{p}}\mathrm{e}^{ \mathrm{i}\boldsymbol{p}\cdot \boldsymbol{x} } \]

creates or annihilates a particle at position \(\displaystyle \boldsymbol{x}\).

Note

Bosons: \(\displaystyle [\hat{\psi}(\boldsymbol{x}),\hat{\psi}^{\dagger}(\boldsymbol{y})]=\delta^3(\boldsymbol{x}-\boldsymbol{y})\), \(\displaystyle [\hat{\psi}(\boldsymbol{x}),\hat{\psi}(\boldsymbol{y}) ]=[\hat{\psi}^{\dagger}(\boldsymbol{x}),\hat{\psi}^{\dagger}(\boldsymbol{y})]=0\) Fermions: \(\displaystyle \left\{ \hat{\psi}(\boldsymbol{x}),\hat{\psi}^{\dagger}(\boldsymbol{y}) \right\}=\delta^{3}(\boldsymbol{x}-\boldsymbol{y})\), \(\displaystyle \left\{ \hat{\psi}^{\dagger}(\boldsymbol{x}),\hat{\psi}^{\dagger}(\boldsymbol{y}) \right\}=\left\{ \hat{\psi}(\boldsymbol{x}),\hat{\psi}(\boldsymbol{y}) \right\}=0\)

How to second quantize an operator

!!! note Single-particle operator \(\displaystyle \to\) many-particles operator

\[\hat{\mathcal{A}}=\sum_{\alpha \beta}\mathcal{A}_{\alpha \beta}\hat{a}^\dagger_{\alpha}\hat{a}_{\beta}\]

where \(\displaystyle \mathcal{A}_{\alpha \beta}=\bra{\alpha}\hat{\mathcal{A}}\ket{\beta}\)

The operator \(\displaystyle \hat{\mathcal{A}}\) is a sum over all processes in which you use \(\displaystyle \hat{a}_{\beta}\) to remove a single particle in state \(\displaystyle \ket{\beta}\), multiply it by the matrix element \(\displaystyle \mathcal{A}_{\alpha \beta}\), and then use \(\displaystyle \hat{a}^\dagger_{\alpha}\) to place that particle into a final state \(\displaystyle \ket{\alpha}\).