A is for Annihilation

Second Quantization

  • Avoid dealing explicitly with \Psi(\mathbf{r}_1,\ldots,\mathbf{r}_N)
  • Represent H and other operators using occupation numbers
  • Synonymous with Quantum Field Theory

Recap: Product States

  • Normalized state of N bosons in orthonormal states \varphi_{\alpha_n}(\mathbf{r})

\Psi^{\text{S}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}(\mathbf{r}_1,\ldots,\mathbf{r}_N)=\sqrt{\frac{\prod_{\alpha}N_{\alpha}!}{N!}}\sum_P\varphi_{\alpha_{1}}(\mathbf{r}_{P_1})\varphi_{\alpha_{2}}(\mathbf{r}_{P_2})\cdots\varphi_{\alpha_{N}}(\mathbf{r}_{P_N})

  • The normalization factor involves occupation numbers \{N_{\alpha}\} giving number of particles in state \alpha
  • Unnormalized symmetric product state using \psi_n(\mathbf{r}), not necessarily orthonormal

\Psi_\text{prod}^{\text{S}}(\mathbf{r}_1,\ldots,\mathbf{r}_N) = \frac{1}{\sqrt{N!}}\sum_P\psi_1(\mathbf{r}_{P_1})\psi_{2}(\mathbf{r}_{P_2})\cdots\psi_{N}(\mathbf{r}_{P_N})

  • Inner product with another symmetric state formed from \phi_n(\mathbf{r})

\Phi_\text{prod}^{\text{S}}(\mathbf{r}_1,\ldots,\mathbf{r}_N) = \frac{1}{\sqrt{N!}}\sum_P\phi_1(\mathbf{r}_{P_1})\phi_{2}(\mathbf{r}_{P_2})\cdots\phi_{N}(\mathbf{r}_{P_N})

\langle{\Psi_\text{prod}}\rvert\Phi_\text{prod}\rangle = \sum_P \prod_{n=1}^N\langle{\psi_n}\rvert\phi_{P_n}\rangle = \operatorname{perm} \langle{\psi_m}\rvert\phi_{n}\rangle

  • This is the permanent of the matrix \langle{\psi_m}\rvert\phi_{n}\rangle

Goal

  • For SHO

\frac{1}{\sqrt{n!}}\left(a^\dagger\right)^n\lvert{0}\rangle\longleftrightarrow \psi_n(x) representation using a^{\vphantom{\dagger}}, a^\dagger very useful; never need explicit \psi_n(x)

  • We want

??\longleftrightarrow \Psi^{\text{S}}(\mathbf{r}_1,\ldots,\mathbf{r}_N)

  • Must respect inner product

Creation & Annihilation Operators

  • Regard N-particle state as rank N tensor \Psi_{a_1 a_2 \cdots a_N} e.g. in position basis the N-body wavefunction is \Psi(\mathbf{r}_1,\ldots \mathbf{r}_N)

  • \Psi is either symmetric or antisymmetric in the labels a_i

\begin{aligned} \Psi^{\text{S}}_{a_1 a_2 \cdots a_N} &= \Psi^{\text{S}}_{(a_1 a_2 \cdots a_N)}\\ \Psi^{\text{A}}_{a_1 a_2 \cdots a_N} &= \Psi^{\text{A}}_{[a_1 a_2 \cdots a_N]}, \end{aligned}

  • Round / square brackets denote symmetrization / antisymmetrization,

T_{(a_1 \cdots a_N)} = \frac{1}{N!}\sum_P T_{a_{P_1}\cdots a_{P_N}},\qquad T_{[a_1 \cdots a_N]} = \frac{1}{N!}\sum_P (-1)^P T_{a_{P_1}\cdots a_{P_N}}

Fock space

  • Bosonic / fermionic Fock space consists of states of form

\lvert{\Psi}\rangle=(\Psi^{\text{S/A}}, \Psi^{\text{S/A}}_{a_1}, \Psi^{\text{S/A}}_{a_1 a_2}, \Psi^{\text{S/A}}_{a_1 a_2 a_3},\ldots).

  • Each state a combination of 0-, 1-, 2-, etc, particle states

  • 1d space of “no particles” is “vacuum”, denoted \lvert{\text{VAC}}\rangle

\langle{\text{VAC}}\rvert \text{VAC} \rangle = 1

  • Inner product

\langle{\Psi}\rvert \Phi \rangle = \overline{\Psi}\Phi + \sum_{a_1} \overline{\Psi}_{a_1} \Phi_{a_1}+\sum_{a_1,a_2} \overline{\Psi}_{a_1 a_2} \Phi_{a_1 a_2} + \cdots

Symmetric tensor operations

  • Can form symmetric product of two tensors

(\Psi \odot \Phi)_{a_1 a_2 \cdots a_{M + N}} = \Psi_{\left(a_1 a_2 \cdots a_{M}\right.} \Phi_{\left. a_{M+1} a_{M+2} \cdots a_{M + N}\right).}

  • Also contract indices to form a lower rank tensor. If M\leq N:

(\bar{\Psi} \cdot \Phi)_{a_{M+1} \cdots a_{N}} = \bar{\Psi}_{a_1\cdots a_{M}}\Phi_{a_1 a_2 \cdots a_N},

  • Contraction preserves the symmetry of the tensors

Annihilation operator

  • Defined by

(a^{\vphantom{\dagger}}(\psi)\Phi)_{a_2\cdots a_{N}} = \sqrt{N}\overline{\psi}_{a_1} \Phi_{a_1\cdots a_{N}}.

  • Extended to general Fock space state by linearity

  • Think of the a^{\vphantom{\dagger}}(\psi) as “removing a particle from state \psi”

  • Inner product fixes adjoint operator a^\dagger(\psi)

(a^\dagger(\psi)\Phi)_{a_1\cdots a_{N+1}} = \sqrt{N + 1}\psi_{\left(a_{1}\right.} \Phi_{\left.a_2\cdots a_{N+1}\right)}.

Commutation relations

  • First note that a^{\vphantom{\dagger}}(\psi) and a^{\vphantom{\dagger}}(\phi) commute, as well as a^\dagger(\psi) and a^\dagger(\phi)

\begin{aligned} \left[a^\dagger(\psi),a^\dagger(\phi)\right]&=0\\ \left[a^{\vphantom{\dagger}}(\psi),a^{\vphantom{\dagger}}(\phi)\right]&=0. \end{aligned}

For commutator of a^{\vphantom{\dagger}}(\psi) and a^\dagger(\phi) compute

\begin{aligned} (a^{\vphantom{\dagger}}(\psi)a^\dagger(\phi)A)_{a_1\cdots a_N} &= (N+1)\overline{\psi}_b \phi_{(b}A_{a_1 \cdots a_N)}\\ (a^\dagger(\phi)a^{\vphantom{\dagger}}(\psi)A)_{a_1\cdots a_N} &= N\overline{\psi}_b\phi_{(a_1} A_{|b| a_2 \cdots a_N)}. \end{aligned}

  • Indices enclosed in vertical bars are excluded

  • First expression can be written as

(N+1)\overline{\psi}_b \phi_{(b}A_{a_1 \cdots a_N)} = \overline{\psi}_b \phi_b A_{a_1\cdots a_N} + N\overline{\psi}_b\phi_{(a_1} A_{|b| a_2 \cdots a_N)}

  • This gives the commutation relation

\left[a^{\vphantom{\dagger}}(\psi),a^\dagger(\phi)\right] = \overline{\psi}_b \phi_b = \langle{\psi}\rvert \phi \rangle

Check

The N particle symmetric product state \lvert{\Psi_\text{prod}^{\text{S}}}\rangle is represented in terms of creation operators as acting on the vacuum state \lvert{\text{VAC}}\rangle as

\lvert{\Psi_\text{prod}^{\text{S}}}\rangle = a^\dagger(\psi_1)\cdots a^\dagger(\psi_N)\lvert{\text{VAC}}\rangle

Show that the commutation relations reproduce the correct inner product.

Choosing a basis

  • Orthonormal basis by \varphi_\alpha(\mathbf{r})

a^\dagger(\varphi_\alpha)\equiv a^\dagger_\alpha,\quad a^{\vphantom{\dagger}}(\varphi_\alpha)\equiv a^{\vphantom{\dagger}}_\alpha

  • Then we have

\begin{aligned} \left[a^{\vphantom{\dagger}}_\alpha,a^{\vphantom{\dagger}}_\beta\right]=0,\quad \left[a^\dagger_\alpha,a^\dagger_\beta\right]=0,\quad \left[a^{\vphantom{\dagger}}_\alpha,a^\dagger_\beta\right] = \delta_{\alpha\beta} \end{aligned}

  • Same as ladder operators of a set of harmonic oscillators

\Psi^{\text{S}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}(\mathbf{r}_1,\ldots,\mathbf{r}_N) \longleftrightarrow\lvert{\mathbf{N}}\rangle \equiv \prod_\alpha \frac{\left(a^\dagger_\alpha\right)^{N_\alpha}}{\sqrt{N_\alpha!}}\lvert{\text{VAC}}\rangle

  • \mathsf{N}^{\vphantom{\dagger}}_{\alpha}\equiv a^\dagger_{\alpha}a^{\vphantom{\dagger}}_{\alpha} is number operator for state \alpha

\mathsf{N}^{\vphantom{\dagger}}_{\alpha}\lvert{\mathbf{N}}\rangle=N_{\alpha}\lvert{\mathbf{N}}\rangle

  • Commutation relations tell us

\begin{aligned} \left[a^{\vphantom{\dagger}}_{\alpha},\mathsf{N}^{\vphantom{\dagger}}_{\alpha}\right]=a^{\vphantom{\dagger}}_{\alpha}\qquad \left[a^\dagger_{\alpha},\mathsf{N}^{\vphantom{\dagger}}_{\alpha}\right]=-a^\dagger_{\alpha} \end{aligned}

  • I think of the first one as:

“count then destroy minus destroy then count”

Change of basis

  • Move to different basis \{\lvert{\tilde{\varphi}_{\alpha}}\rangle\}

\lvert{\tilde{\varphi}_{\alpha}}\rangle=\sum_{\beta} \langle{\varphi_{\beta}}\rvert \tilde{\varphi}_{\alpha} \rangle\lvert{\varphi_{\beta}}\rangle

  • One particle states with wavefunctions \varphi_{\alpha}(\mathbf{r}) are a^\dagger_{\alpha}\lvert{\text{VAC}}\rangle. So,

\tilde{a^{\vphantom{\dagger}}}_{\alpha}^\dagger\equiv\sum_{\beta} \langle{\varphi_{\beta}}\rvert \tilde{\varphi}_{\alpha} \rangle a^\dagger_{\beta}

  • Often we work in position eigenstates \{\lvert{\mathbf{r}}\rangle\}, so \langle{\varphi_{\beta}}\rvert \mathbf{r} \rangle=\varphi^{*}_{\beta}(\mathbf{r})

  • Denoting corresponding creation operator as \psi^\dagger(\mathbf{r})

\psi^\dagger(\mathbf{r})\equiv\sum_{\beta} \varphi^{*}_{\beta}(\mathbf{r})a^\dagger_{\beta}

  • Conjugate is

\psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\sum_{\beta} \varphi_{\beta}(\mathbf{r})a^{\vphantom{\dagger}}_{\beta}

  • These are quantum fields with commutation relations

\begin{gather*} \left[\psi^{\vphantom{\dagger}}(\mathbf{r}_1),\psi^\dagger(\mathbf{r}_2)\right]=\delta(\mathbf{r}_1-\mathbf{r}_2)\nonumber\\ \left[\psi^{\vphantom{\dagger}}(\mathbf{r}_1),\psi^{\vphantom{\dagger}}(\mathbf{r}_2)\right]=\left[\psi^\dagger(\mathbf{r}_1),\psi^\dagger(\mathbf{r}_2)\right]=0 \end{gather*}

Check

If a state \lvert{\Psi}\rangle has wavefunction \Psi(x_1,\ldots, x_N), show that the wavefunction of the state \psi^{\vphantom{\dagger}}(X)\lvert{\Psi}\rangle is the N-1 particle wavefunction

\sqrt{N}\Psi(X,x_1,\ldots, x_{N-1})

Hint: Show that this is true for a product state first.

  • Often use eigenstates of the free particle Hamiltonian H=\frac{\mathbf{p}^{2}}{2m} \begin{aligned} \lvert{\mathbf{k}}\rangle=\frac{\exp(i\mathbf{k}\cdot \mathbf{r})}{\sqrt{V}}, \quad \mathbf{k}=2\pi\left(\frac{n_{x}}{L_{x}},\frac{n_{y}}{L_{y}},\frac{n_{z}}{L_{z}}\right),\quad n_{x,y,z}\text{ integer}, \end{aligned} with V=L_{x}L_{y}L_{z} (periodic b.c.)

  • Matrix elements of transformation between to position basis \{\lvert{\mathbf{r}}\rangle\} are \langle{\mathbf{k}}\rvert\mathbf{r}\rangle=\exp(-i\mathbf{k}\cdot \mathbf{r})/\sqrt{V} \psi^\dagger(\mathbf{r})\equiv\frac{1}{\sqrt{V}}\sum_{\mathbf{k}} \exp(-i\mathbf{k}\cdot\mathbf{r})a^\dagger_{\mathbf{k}} and similarly \psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\frac{1}{\sqrt{V}}\sum_{\mathbf{k}} \exp(i\mathbf{k}\cdot\mathbf{r})a^{\vphantom{\dagger}}_{\mathbf{k}}

Check

What is the wavefunction of the two-particle state

\sum_\mathbf{k}c_\mathbf{k}a^\dagger_\mathbf{k}a^\dagger_{-\mathbf{k}}\lvert{\text{VAC}}\rangle?

Fermions

  • Trickier on account of minus signs! Seek a representation of

\Psi^{\text{A}}(\mathbf{r}_1,\ldots,\mathbf{r}_N) = \frac{1}{\sqrt{N!}}\sum_P (-1)^P\psi_1(\mathbf{r}_{P_1})\psi_{2}(\mathbf{r}_{P_2})\cdots\psi_{N}(\mathbf{r}_{P_N})

Note: overall sign fixed by labelling of states \psi_j

  • If \Psi^{\text{A}}(\mathbf{r}_1,\ldots,\mathbf{r}_N) \longleftrightarrow a^\dagger(\psi_1)\cdots a^\dagger(\psi_N)\lvert{\text{VAC}}\rangle we’ll need

\left\{a^\dagger(\psi),a^\dagger(\phi)\right\}=0, \{A,B\}\equiv AB+BA is anticommutator. Also

\left\{a^{\vphantom{\dagger}}(\psi),a^{\vphantom{\dagger}}(\phi)\right\}=0

  • \left\{a^{\vphantom{\dagger}}(\psi),a^\dagger(\phi)\right\} can be deduced from inner product

\langle{\Psi}\rvert\Phi\rangle = \sum_P (-1)^P\prod_{n=1}^N\langle{\psi_n}\rvert\phi_{P_n}\rangle = \det \langle{\psi_m}\rvert\phi_{n}\rangle,

which works out if

\left\{a^{\vphantom{\dagger}}(\psi),a^\dagger(\phi)\right\} = \langle{\psi}\rvert \phi \rangle

Check

Check this.

  • Introducing field operators in position basis as before leads to

\begin{gather*} \left\{\psi^{\vphantom{\dagger}}(\mathbf{r}_1),\psi^\dagger(\mathbf{r}_2)\right\}=\delta(\mathbf{r}_1-\mathbf{r}_2)\nonumber\\ \left\{\psi^{\vphantom{\dagger}}(\mathbf{r}_1),\psi^{\vphantom{\dagger}}(\mathbf{r}_2)\right\}=\left\{\psi^\dagger(\mathbf{r}_1),\psi^\dagger(\mathbf{r}_2)\right\}=0 \end{gather*}

How do a^{\vphantom{\dagger}}_\alpha and a^\dagger_\alpha look?

Check

Think about the form that the operators a^{\vphantom{\dagger}}_\alpha, a^\dagger_\alpha take in the basis of product states. Start with one state \varphi_\alpha. What’s the matrix form of a^\dagger_\alpha in terms of states \lvert{N_\alpha}\rangle? Now consider two states. Can you see how the commutation and anticommutation relations can be satisfied?

One Particle Operators

  • Notation: A acts on single particle states; \hat A acts on N particles as

\hat A = \sum_{j=1}^N A_j

  • Example: Hamiltonian for noninteracting particles

\hat H = \sum_{j=1}^N H_j = \sum_{j=1}^N \left[-\frac{\nabla_j^2}{2m}+V(\mathbf{r}_j)\right]

  • Operators of this type are one particle operators

  • How to represent them using creation and annihilation operators?

  • We show that the one-particle operator \hat A has representation

\hat A = \int \psi^\dagger(x_1) [A_{x_1} \psi^{\vphantom{\dagger}}({x_1})]\, d{x_1}

  • Evaluate matrix element \lvert{\Psi}\rangle and \lvert{\Phi}\rangle

\begin{aligned} \langle{\Psi}\rvert \hat A \lvert \Phi \rangle &= \int \langle{\Psi}\rvert \psi^\dagger(x_1) [A_{x_1} \psi^{\vphantom{\dagger}}(x_1)] \lvert \Phi \rangle\, dx_1 \\ &= N \int \Psi^*(x_1,x_2,\ldots x_{N})\left[A_{x_1} \Phi(x_1,x_2,\dots x_{N})\right]dx_1 dx_2\ldots dx_{N}\\ &=\sum_{j=1}^N \int \Psi^*(x_1,x_2,\ldots x_{N})\left[A_{x_j} \Phi(x_1,x_2,\dots x_{N})\right]dx_1 dx_2\ldots dx_{N}. \end{aligned}

  • Any sign in the fermionic case is squared and therefore has no effect.

\hat A = \int \psi^\dagger(x_1) [A_{x_1} \psi^{\vphantom{\dagger}}({x_1})]\, d{x_1}

  • Substitute \psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\frac{1}{\sqrt{V}}\sum_{\mathbf{k}} \exp(i\mathbf{k}\cdot\mathbf{r})a^{\vphantom{\dagger}}_{\mathbf{k}} \hat A = \sum_{\alpha\beta}A_{\alpha\beta}a^\dagger_\alpha a^{\vphantom{\dagger}}_\beta where A_{\alpha\beta}=\langle{\varphi_\alpha}\rvert A \lvert \varphi_\beta \rangle are matrix elements

  • Like expectation value of \mathop{A} in single particle state \sum_{\alpha}a_{\alpha}\lvert{\varphi_{\alpha}}\rangle

  • But a_\alpha are operators not numbers. This is the origin of the name second quantization

  • Resemblance makes it easy to write down second quantized form

\hat A = \sum_{\alpha\beta}A_{\alpha\beta}a^\dagger_\alpha a^{\vphantom{\dagger}}_\beta

  • Matrix element \langle{\mathbf{N}}\rvert \hat A \lvert \mathbf{N'} \rangle vanishes unless \mathbf{N} and \mathbf{N'} differ by the movement of of one particle

  • If N_\beta = N'_\beta-1 and N_\alpha = N'_\alpha+1 we have

\langle{\mathbf{N}}\rvert \hat A \lvert \mathbf{N'} \rangle = A_{\alpha\beta} \sqrt{N_\alpha N'_\beta}.

Check

This formula is not so easy to work out in the first quantized representation. Try it!

Example 1: noninteracting Hamiltonian

\hat H = \sum_{j=1}^N H_j = \sum_{j=1}^N \left[-\frac{\nabla_j^2}{2m}+V(\mathbf{r}_j)\right]

  • Second quantized form \mathop{\hat H} \equiv \sum_{\alpha,\beta}\langle{\varphi_{\alpha}}\rvert \mathop{H} \lvert \varphi_{\beta} \rangle a^\dagger_{\alpha}a^{\vphantom{\dagger}}_{\beta}, H is single particle Hamiltonian H=-\frac{1}{2m}\nabla_{i}^{2}+V(\mathbf{r_{i}})

  • If basis \lvert{\varphi_{\alpha}}\rangle is eigenbasis of H: \langle{\varphi_{\alpha}}\rvert \mathop{H} \lvert \varphi_{\beta} \rangle=E_{\alpha}\delta_{\alpha \beta} and

\begin{aligned} \mathop{\hat H} \equiv \sum_{\alpha} E_{\alpha} a^\dagger_{\alpha}a^{\vphantom{\dagger}}_{\alpha}=\sum_{\alpha} E_{\alpha} \mathsf{N}^{\vphantom{\dagger}}_{\alpha} \end{aligned}

  • In the position basis

\begin{aligned} \mathop{\hat H}&=\int d\mathbf{r}\left[-\frac{1}{2m}\psi^\dagger(\mathbf{r})\nabla^{2}\psi^{\vphantom{\dagger}}(\mathbf{r})+V(\mathbf{r})\psi^\dagger(\mathbf{r})\psi^{\vphantom{\dagger}}(\mathbf{r})\right]\nonumber\\ &=\int d\mathbf{r}\left[\frac{1}{2m}\nabla\psi^\dagger(\mathbf{r})\cdot\nabla\psi^{\vphantom{\dagger}}(\mathbf{r})+V(\mathbf{r})\psi^\dagger(\mathbf{r})\psi^{\vphantom{\dagger}}(\mathbf{r})\right], \end{aligned} (integration by parts)

  • Confirm equality to previous expression using

\psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\sum_{\beta} \varphi_{\beta}(\mathbf{r})a^{\vphantom{\dagger}}_{\beta}

  • Heisenberg equation of motion for \psi^{\vphantom{\dagger}}(\mathbf{r},t)=e^{i\hat Ht}\psi^{\vphantom{\dagger}}(\mathbf{r}) e^{-i\hat H t} with noninteracting Hamiltonian

\begin{split} i\partial_{t}\psi^{\vphantom{\dagger}}(\mathbf{r},t) &= -\left[\mathop{\hat H},\psi^{\vphantom{\dagger}}(\mathbf{r},t)\right]\\ &= -\frac{1}{2m}\nabla^{2}\psi^{\vphantom{\dagger}}(\mathbf{r},t)+V(\mathbf{r})\psi^{\vphantom{\dagger}}(\mathbf{r},t) \end{split} just the time dependent Schrödinger equation!

Example 2: Density

  • One particle operator for density at \mathbf{x} is

\rho(\mathbf{x})\equiv\delta(\mathbf{x}-\mathbf{r})

  • Expectation value in single particle state \varphi(\mathbf{r}) is \rho(\mathbf{x})=\lvert{\varphi(\mathbf{x})}\rvert^{2}

  • Second quantized form of the operator is then

\hat\rho(\mathbf{x})\equiv\psi^\dagger(\mathbf{x})\psi^{\vphantom{\dagger}}(\mathbf{x})

  • Check: integrating over position should give total number of particles

\hat N=\int d\mathbf{x}\, \psi^\dagger(\mathbf{x})\psi^{\vphantom{\dagger}}(\mathbf{x})=\sum_{\alpha} a^\dagger_{\alpha}a^{\vphantom{\dagger}}_{a}=\sum_{\alpha}\mathsf{N}^{\vphantom{\dagger}}_{\alpha}

  • Expectation value of density in product state \lvert{N_{0},N_{1}\ldots}\rangle

\langle{N_{0},N_{1}\ldots}\rvert \hat\rho(\mathbf{r}) \lvert N_{0},N_{1}\ldots \rangle = \sum_{\alpha} N_{\alpha}\left|\varphi_{\alpha}(\mathbf{r})\right|^{2}

Check

Prove using \psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\sum_{\beta} \varphi_{\beta}(\mathbf{r})a^{\vphantom{\dagger}}_{\beta}

  • Interpretation: \left|\varphi_{\alpha}(\mathbf{r})\right|^{2} is probability to find a particle in state \alpha at position \mathbf{r}

  • Density is weighted by occupation

Example 3: Current operator

\hat{\mathbf{j}}(\mathbf{r})=-i\frac{1}{2m}\left[\psi^\dagger(\mathbf{r})\left(\nabla\psi^{\vphantom{\dagger}}(\mathbf{r})\right)-\left(\nabla\psi^\dagger(\mathbf{r})\right)\psi^{\vphantom{\dagger}}(\mathbf{r})\right].

  • Often we consider Fourier components of density or current

\begin{aligned} \hat\rho_{\mathbf{q}}\equiv\int d\mathbf{r}\, \hat\rho(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}=\sum_{\mathbf{k}} a^\dagger_{\mathbf{k}-\mathbf{q}}a^{\vphantom{\dagger}}_{\mathbf{k}}\nonumber\\ \hat{\mathbf{j}}_{\mathbf{q}}\equiv\int d\mathbf{r}\, \hat{\mathbf{j}}(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}=\frac{1}{m}\sum_{\mathbf{k}} (\mathbf{k}-\mathbf{q}/2)a^\dagger_{\mathbf{k}-\mathbf{q}}a^{\vphantom{\dagger}}_{\mathbf{k}} \end{aligned}

  • \mathbf{q}=0 modes are total particle number and \frac{1}{m}\times total momentum, respectively

Single particle density matrix

g(\mathbf{r},\mathbf{r}') \equiv N \int d\mathbf{r}_{2}\cdots d\mathbf{r}_{N}\,\Psi^{*}(\mathbf{r},\mathbf{r}_{2},\ldots,\mathbf{r}_{N})\Psi(\mathbf{r}',\mathbf{r}_{2},\ldots,\mathbf{r}_{N}).

Check

Show that this can be written as

g(\mathbf{r},\mathbf{r}')= \langle{\Psi}\rvert \psi^\dagger(\mathbf{r})\psi^{\vphantom{\dagger}}(\mathbf{r}') \lvert \Psi \rangle

  • Similar calculation as for density gives

g(\mathbf{r},\mathbf{r}') = \sum_{\alpha} N_{\alpha}\varphi_{\alpha}^{*}(\mathbf{r})\varphi^{}_{\alpha}(\mathbf{r}')

  • For 3D N_{\mathbf{k}}=1 for \lvert{\mathbf{k}}\rvert<k_{F}, and 0 otherwise

\lvert{\text{Fermi sea}}\rangle = \prod_{|\mathbf{k}|<k_F} a^\dagger_\mathbf{k}\lvert{\text{VAC}}\rangle

Check

\begin{aligned} g(\mathbf{r},\mathbf{r}')&=\frac{1}{V}\sum_{|\mathbf{k}|<k_{F}} e^{i\mathbf{k}\cdot(\mathbf{r}'-\mathbf{r})}=\int_{|\mathbf{k}|<k_{F}} \frac{d\mathbf{k}}{(2\pi)^{3}}\,e^{i\mathbf{k}\cdot(\mathbf{r}'-\mathbf{r})}\nonumber\\ &=\frac{k_{F}^{3}}{2\pi^{2}}\left[\frac{\sin\left(k_{F}|\mathbf{r}'-\mathbf{r}|\right)}{(k_{F}|\mathbf{r}'-\mathbf{r}|)^{3}}-\frac{\cos\left(k_{F}|\mathbf{r}'-\mathbf{r}|\right)}{(k_{F}|\mathbf{r}'-\mathbf{r}|)^{2}}\right]. \end{aligned}

Note that g(\mathbf{r},\mathbf{r})=\frac{k_{F}^{3}}{6\pi^{2}}=n

  • c.f. earlier calculation in 1D using Slater determinant

Two Particle Operators

  • Acts pairwise on the particles (B_{jk}=B_{kj} for indistinguishable particles.)

\hat B = \sum_{j<k} B_{jk}.

  • Has representation

\hat B =\frac{1}{2} \int \psi^\dagger(x_1)\psi^\dagger(x_2) [B_{x_1,x_2} \psi^{\vphantom{\dagger}}(x_2)\psi^{\vphantom{\dagger}}(x_1)]\, dx_1 dx_2, where B_{x_1,x_2} is an operator that acts on a pair of wavefunctions

  • Taking matrix elements yields

\begin{aligned} \langle{\Psi}\rvert \hat B \lvert \Phi \rangle &= \frac{1}{2} \int \langle{\Psi}\rvert \psi^\dagger(x_1)\psi^\dagger(x_2) [B_{x_1,x_2} \psi^{\vphantom{\dagger}}(x_2)\psi^{\vphantom{\dagger}}(x_1)] \lvert \Phi \rangle\, dx_1 dx_2\\ &= \frac{1}{2}N(N-1) \int \Psi^*(x_1,x_2,\ldots x_{N})\left[B_{x_1,x_2} \Phi(x_1,x_2,\dots x_{N})\right]dx_1 dx_2\ldots dx_{N}\\ &=\sum_{j<k} \int \Psi^*(x_1,x_2,\ldots x_{N})\left[B_{x_j,x_k} \Phi(x_1,x_2,\dots x_{N})\right]dx_1 dx_2\ldots dx_{N}. \end{aligned}

  • All terms in sum over pairs are identical by (anti-)symmetry of the wavefunctions
  • Remember that for one particle operators

\langle{\mathbf{N}}\rvert \hat A \lvert \mathbf{N'} \rangle = A_{\alpha\beta} \sqrt{N_\alpha N'_\beta}

  • For two particle operators \langle{\mathbf{N}}\rvert \hat B \lvert \mathbf{N'} \rangle = \left[B_{\alpha\beta,\gamma\delta}\pm B_{\alpha\beta,\delta\gamma}\right] \sqrt{N_\alpha N_\beta N'_\gamma N'_\delta} with N_{\gamma,\delta} = N'_{\gamma,\delta}-1 and N_{\alpha,\beta} = N'_{\alpha,\beta}+1 \begin{aligned} B_{\alpha\beta,\gamma\delta} &= \langle{\varphi_\alpha}\rvert\langle{\varphi_\beta}\rvert B_{12} \lvert{\varphi_\gamma}\rangle\lvert{\varphi_\delta}\rangle\nonumber\\ &=\int d\mathbf{r}d\mathbf{r}' \varphi_\alpha^*(\mathbf{r})\varphi_\beta^*(\mathbf{r}')B_{12}\varphi_\gamma(\mathbf{r})\varphi_\delta(\mathbf{r}') \end{aligned}

  • For fermions the overall sign depends on convention: best to write states explicitly rather than matrix elements

\langle{\mathbf{N}}\rvert \hat B \lvert \mathbf{N'} \rangle = \left[B_{\alpha\beta,\gamma\delta}\pm B_{\alpha\beta,\delta\gamma}\right] \sqrt{N_\alpha N_\beta N'_\gamma N'_\delta}.

  • Strictly we have

\begin{gather*} N'_\gamma N'_\delta \to N'_\gamma (N'_\gamma-1) && \gamma=\delta\\ N_\alpha N_\beta \to N_\alpha (N_\alpha-1) && \alpha=\beta \end{gather*} In thermodynamic limit these terms usually make a vanishing contribution when sums replaced with integrals

  • Exceptions: when a finite fraction of particles are in one state (which occurs for Bose—Einstein condensates). In those cases we end up neglecting N_\alpha-1 relative to N_\alpha, however!

Example: pairwise interaction

\hat H_\text{int.} = \sum_{j<k} U(\mathbf{r}_j-\mathbf{r}_k).

  • Expressing in position basis

\hat H_\text{int.} = \frac{1}{2}\int d\mathbf{r}_1 d\mathbf{r}_2\, U(\mathbf{r}_1-\mathbf{r}_2)\psi^\dagger(\mathbf{r}_1)\psi^\dagger(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_1)

  • Remembering that \rho(\mathbf{r}) = \psi^\dagger(\mathbf{r})\psi^{\vphantom{\dagger}}(\mathbf{r}), this is almost

\frac{1}{2}\int d\mathbf{r}_1 d\mathbf{r}_2\, U(\mathbf{r}_1-\mathbf{r}_2)\rho(\mathbf{r}_1)\rho(\mathbf{r}_2) Operator order prevents a particle from interacting with itself!

\hat H_\text{int.} = \frac{1}{2}\int d\mathbf{r}_1 d\mathbf{r}_2\, U(\mathbf{r}_1-\mathbf{r}_2)\psi^\dagger(\mathbf{r}_1)\psi^\dagger(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_1)

Check

\hat H_\text{int} has zero expectation for 1 particle

  • Hamiltonian of interacting bosons from Lecture 1

H = -\frac{1}{2m}\sum_j \frac{\partial^2}{\partial x_j^2} + \overbrace{c\sum_{j<k}\delta(x_j-x_k)}^{\equiv H_\text{int}}. has second quantized form H = \int dx \left[\frac{1}{2}\partial_x\psi^\dagger(x)\partial_x\psi^{\vphantom{\dagger}}(x) + \frac{c}{2}\psi^\dagger(x)\psi^\dagger(x)\psi^{\vphantom{\dagger}}(x)\psi^{\vphantom{\dagger}}(x)\right] \psi^{\vphantom{\dagger}}(x), \psi^\dagger(x) satisfy the canonical bosonic commutation relations

QFT is a language, but writing something in a new way doesn’t (necessarily) make it easier to solve!