
# Second Quantization

• Avoid dealing explicitly with $\Psi(\br_1,\ldots,\br_N)$
• Represent $H$ and other operators using occupation numbers
• Synonymous with Quantum Field Theory

## Recap: Product States

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

$$\Psi^{\text{S}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}(\br_1,\ldots,\br_N)=\sqrt{\frac{\prod_{\alpha}N_{\alpha}!}{N!}}\sum_P\varphi_{\alpha_{1}}(\mathbf{r}_{P_1})\varphi_{\alpha_{2}}(\mathbf{r}_{P_2})\cdots\varphi_{\alpha_{N}}(\mathbf{r}_{P_N}), \label{A_OrthoProd}$$

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

$$\Psi^{\text{S}}(\br_1,\ldots,\br_N) = \frac{1}{\sqrt{N!}}\sum_P\psi_1(\mathbf{r}_{P_1})\psi_{2}(\mathbf{r}_{P_2})\cdots\psi_{N}(\mathbf{r}_{P_N}). \label{A_NProd}$$

• Inner product with another symmetric state formed from wavefunctions $\phi_n(\br)$

$$\Phi^{\text{S}}(\br_1,\ldots,\br_N) = \frac{1}{\sqrt{N!}}\sum_P\phi_1(\mathbf{r}_{P_1})\phi_{2}(\mathbf{r}_{P_2})\cdots\phi_{N}(\mathbf{r}_{P_N}).$$

$$\bra{\Psi}\Phi\rangle = \sum_P \prod_{n=1}^N\bra{\psi_n}\phi_{P_n}\rangle = \operatorname{perm} \bra{\psi_m}\phi_{n}\rangle, \label{A_perm}$$

• This is the permanent of the matrix $\bra{\psi_m}\phi_{n}\rangle$

## Goal

• For SHO $$\frac{1}{\sqrt{n!}}\left(\adop\right)^n\ket{0}\longleftrightarrow \psi_n(x)$$ representation using $\aop$, $\adop$ very useful; never need explicit $\psi_n(x)$

• We want

$$??\longleftrightarrow \Psi^{\text{S}}(\br_1,\ldots,\br_N)$$

• Must respect inner product

## Creation & Annihilation Operators

• We consider states with different numbers of particles

• $\ket{\text{VAC}}$ denotes state with no particles (the vacuum state)

• $\adop(\psi)$ creates particle in single particle state $\psi(\br)$

$$\psi(\br)\longleftrightarrow \adop(\psi)\ket{\text{VAC}}. \label{A_1part}$$

• Evidently, $\adop(\psi)$ must be linear in $\psi$

$$\adop\left(c_1\psi_1+c_2\psi_2\right) = c_1\adop(\psi_1)+c_2\adop(\psi_2).$$

• If $\ket{\Psi}$ has $N$ particle state, $\adop(\psi)\ket{\Psi}$ has $N+1$ particles

• Since this is orthogonal to the vacuum state $$\bra{\text{VAC}} \adop(\psi)\ket{\Psi}=0,$$ for any state $\ket{\Psi}$

• Taking the adjoint, this means $$\aop(\psi)\ket{\text{VAC}}=0.$$

• $N$ particle product state

$$\Psi^{\text{S}}(\br_1,\ldots,\br_N) \longleftrightarrow \adop(\psi_1)\cdots \adop(\psi_N)\ket{\text{VAC}}. \label{A_NPart}$$

• To be symmetric wavefunction, we must have $$\left[\adop(\psi),\adop(\phi)\right]=0 \label{A_adcommute}$$ for any states $\psi(\br)$ and $\phi(\br)$. Taking the adjoint gives

$$\left[\aop(\psi),\aop(\phi)\right]=0 \label{A_acommute}$$

• Inner product between two one particle states

$$\bra{\psi}\phi\rangle = \braket{\text{VAC}}{\aop(\psi)\adop(\phi)}{\text{VAC}}.$$

• We impose $$\left[\aop(\psi),\adop(\phi)\right] = \inner{\psi}{\phi}. \label{A_ada}$$ together with $\aop(\psi)\ket{\text{VAC}}=0$ this gives correct inner product

Show that this also reproduces the inner product for $N$-particle product states

## Choosing a basis

• Orthonormal basis by $\varphi_\alpha(\br)$

$$\adop(\varphi_\alpha)\equiv \adop_\alpha,\quad \aop(\varphi_\alpha)\equiv \aop_\alpha. \label{A_CCRBasis}$$

• Then we have

\begin{align} \left[\aop_\alpha,\aop_\beta\right]=0,\quad \left[\adop_\alpha,\adop_\beta\right]=0,\quad \left[\aop_\alpha,\adop_\beta\right] = \delta_{\alpha\beta}. \end{align}

• Same as ladder operators of a set of harmonic oscillators

$$\Psi^{\text{S}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}(\br_1,\ldots,\br_N) \longleftrightarrow\ket{\mathbf{N}} \equiv \prod_\alpha \frac{\left(\adop_\alpha\right)^{N_\alpha}}{\sqrt{N_\alpha!}}\ket{\text{VAC}}$$

• $\Nop_{\alpha}\equiv \adop_{\alpha}\aop_{\alpha}$ is number operator for state $\alpha$

$$\label{2nd_quant_NOp} \Nop_{\alpha}\ket{\mathbf{N}}=N_{\alpha}\ket{\mathbf{N}}$$

• Commutation relations tell us

\begin{align} \left[\aop_{\alpha},\Nop_{\alpha}\right]=\aop_{\alpha}\qquad \left[\adop_{\alpha},\Nop_{\alpha}\right]=-\adop_{\alpha} \label{2nd_quant_NaComm} \end{align}

• I think of the first one as:

“count then destroy minus destroy then count”

## Change of basis

• Move to different basis ${\ket{\tilde{\varphi}_{\alpha}}}$

$$\label{2nd_quant_BasisChange} \ket{\tilde{\varphi}_{\alpha}}=\sum_{\beta} \inner{\varphi_{\beta}}{\tilde{\varphi}_{\alpha}}\ket{\varphi_{\beta}}$$

• One particle states with wavefunctions $\varphi_{\alpha}(\br)$ are $\adop_{\alpha}\ket{\text{VAC}}$. So,

$$\tilde{\aop}_{\alpha}^\dagger\equiv\sum_{\beta} \inner{\varphi_{\beta}}{\tilde{\varphi}_{\alpha}}\adop_{\beta}$$

• Often we work in position eigenstates ${\ket{\br}}$, so $\inner{\varphi_{\beta}}{\br}=\varphi^{*}_{\beta}(\br)$

• Denoting corresponding creation operator as $\pdop(\br)$

$$\pdop(\br)\equiv\sum_{\beta} \varphi^{*}_{\beta}(\br)\adop_{\beta}$$

• Conjugate is

$$\label{2nd_quant_PsiDef} \pop(\br)\equiv\sum_{\beta} \varphi_{\beta}(\br)\aop_{\beta},$$

• These are quantum fields with commutation relations

$$\begin{gather} \left[\pop(\br_1),\pdop(\br_2)\right]=\delta(\br_1-\br_2)\nonumber\\ \left[\pop(\br_1),\pop(\br_2)\right]=\left[\pdop(\br_1),\pdop(\br_2)\right]=0. \label{2nd_quant_PositionRelations} \end{gather}$$

If a state $\ket{\Psi}$ has wavefunction $\Psi(x_1,\ldots, x_N)$, show that the wavefunction of the state $\pop(X)\ket{\Psi}$ is the $N-1$ particle wavefunction

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

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

• Often use eigenstates of the free particle Hamiltonian $H=\frac{\bp^{2}}{2m}$ \begin{align} \label{2nd_quant_FreeParticleStates} \ket{\bk}=\frac{\exp(i\bk \cdot \br)}{\sqrt{V}}, \quad \bk=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{align} with $V=L_{x}L_{y}L_{z}$ (periodic b.c.)

• Matrix elements of transformation between to position basis ${\ket{\br}}$ are $\bra{\bk}\br\rangle=\exp(-i\bk \cdot \br)/\sqrt{V}$ $$\label{2nd_quant_PositionAnnihilation} \pdop(\br)\equiv\frac{1}{\sqrt{V}}\sum_{\bk} \exp(-i\bk\cdot\br)\adop_{\bk},$$ and similarly $$\label{2nd_quant_PositionCreation} \pop(\br)\equiv\frac{1}{\sqrt{V}}\sum_{\bk} \exp(i\bk\cdot\br)\aop_{\bk}.$$

What is the wavefunction of the two-particle state

$$\sum_\bk c_\bk \adop_\bk\adop_{-\bk}\ket{\text{VAC}}?$$

## Fermions

• Trickier on account of minus signs! Seek a representation of $$\Psi^{\text{A}}(\br_1,\ldots,\br_N) = \frac{1}{\sqrt{N!}}\sum_P (-1)^P\psi_1(\mathbf{r}_{P_1})\psi_{2}(\mathbf{r}_{P_2})\cdots\psi_{N}(\mathbf{r}_{P_N}). \label{A_NProdAnti}$$ Note: overall sign fixed by labelling of states $\psi_j$

• If $\Psi^{\text{A}}(\br_1,\ldots,\br_N) \longleftrightarrow \adop(\psi_1)\cdots \adop(\psi_N)\ket{\text{VAC}}$ we’ll need $$\left\{\adop(\psi),\adop(\phi)\right\}=0, \label{A_adanticommute}$$ $\{A,B\}\equiv AB+BA$ is anticommutator. Also $$\left\{\aop(\psi),\aop(\phi)\right\}=0. \label{A_aanticommute}$$

• $\left\{\aop(\psi),\adop(\phi)\right\}=0$ can be deduced from inner product $$\bra{\Psi}\Phi\rangle = \sum_P (-1)^P\prod_{n=1}^N\bra{\psi_n}\phi_{P_n}\rangle = \det \bra{\psi_m}\phi_{n}\rangle, \label{A_det}$$ which works out if

$$\left\{\aop(\psi),\adop(\phi)\right\} = \inner{\psi}{\phi}. \label{A_adaanti}$$

Check this.

• Introducing field operators in position basis as before leads to

$$\begin{gather} \left\{\pop(\br_1),\pdop(\br_2)\right\}=\delta(\br_1-\br_2)\nonumber\\ \left\{\pop(\br_1),\pop(\br_2)\right\}=\left\{\pdop(\br_1),\pdop(\br_2)\right\}=0. \label{2nd_quant_PositionRelationsAnti} \end{gather}$$

## Explicit Form of Operators

Think about the form that the operators $\aop_\alpha$, $\adop_\alpha$ take in the basis of product states. Start with one state $\varphi_\alpha$. What’s the matrix form of $\adop_\alpha$ in terms of states $\ket{N_\alpha}$? Now consider two states. Can you see how the commutation and anticommutation relations can be satisfied?

## Single 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 $$\hat H = \sum_{j=1}^N H_j = \sum_{j=1}^N \left[-\frac{\nabla_j^2}{2m}+V(\br_j)\right]. \label{A_H1}$$

• Operators of this type are single particle operators

• How to represent them using creation and annihilation operators?

• Action of $A$ on states $\ket{\varphi_\alpha}$ written

$$A\ket{\varphi_\alpha} = \sum_{\beta} \ket{\varphi_\beta}\braket{\varphi_\beta}{A}{\varphi_\alpha} = \sum_\beta A_{\beta\alpha}\ket{\varphi_\beta}$$

• Action of $\hat A$ on product state $\ket{\Psi^{\text{S/A}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}}$ is

$$\hat A \ket{\Psi^{\text{S/A}}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}} = \sum_\beta \left[A_{\beta\alpha_1}\ket{\Psi^{\text{S/A}}_{\beta\alpha_{2}\cdots\alpha_{N}}} +A_{\beta\alpha_2}\ket{\Psi^{\text{S/A}}_{\alpha_1\beta\cdots\alpha_{N}}}+\cdots A_{\beta\alpha_N}\ket{\Psi^{\text{S/A}}_{\alpha_1\alpha_{2}\cdots\beta}}\right] \label{A_1OpAct}$$

• We’ll see same job is done by $$$$\hat A = \sum_{\alpha\beta}A_{\alpha\beta}\adop_\alpha\aop_\beta$$$$ acting on state $\ket{\mathbf{N}} \equiv \prod_\alpha \frac{\left(\adop_\alpha\right)^{N_\alpha}}{\sqrt{N_\alpha!}}\ket{\text{VAC}}$
• We have $$$$\left[\adop_\alpha\aop_\beta,\adop_\gamma\right]=\adop_\alpha\delta_{\beta\gamma}$$$$ for bosons and fermions

• Commute $\adop_\alpha \aop_\beta$ though each of creation operators in $\ket{\mathbf{N}}$, e.g.

\begin{align} \label{2nd_quant_} \mathop{\hat A}\adop_{\beta}\ket{\text{VAC}}&=\left(\left[\mathop{\hat A},\adop_{\beta}\right]+\adop_{\beta}\mathop{\hat A}\right)\ket{\text{VAC}}\nonumber\\ &=\sum_{\alpha} A_{\alpha \beta} \adop_{\alpha}\ket{\text{VAC}}. \end{align}

Try it for a two particle state!

$$$$\hat A = \sum_{\alpha\beta}A_{\alpha\beta}\adop_\alpha\aop_\beta$$$$

• Find matrix element $\braket{\mathbf{N}}{\hat A}{\mathbf{N’}}$ between product states made of orthonormal single particle states

• Vanishes unless $N_\beta = N'_\beta-1$ and $N_\alpha = N'_\alpha+1$ we have

$$\braket{\mathbf{N}}{\hat A}{\mathbf{N'}} = A_{\alpha\beta} \sqrt{N_\alpha N'_\beta}. \label{A_Aab}$$

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

• Same for bosons and fermions

$$$$\hat A = \sum_{\alpha\beta}A_{\alpha\beta}\adop_\alpha\aop_\beta$$$$

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

• 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

### Example 1: noninteracting Hamiltonian

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

• Second quantized form $$\label{2nd_quant_H2ndQ} \mathop{\hat H} \equiv \sum_{\alpha,\beta}\braket{\varphi_{\alpha}}{\mathop{H}}{\varphi_{\beta}} \adop_{\alpha}\aop_{\beta},$$ $H$ is single particle Hamiltonian $H=-\frac{1}{2m}\nabla_{i}^{2}+V(\mathbf{r_{i}})$

• If basis $\ket{\varphi_{\alpha}}$ is eigenbasis of $H$: $\braket{\varphi_{\alpha}}{\mathop{H}}{\varphi_{\beta}}=E_{\alpha}\delta_{\alpha \beta}$ and

\begin{align} \label{2nd_quant_Nrep} \mathop{\hat H} \equiv \sum_{\alpha} E_{\alpha} \adop_{\alpha}\aop_{\alpha}=\sum_{\alpha} E_{\alpha} \Nop_{\alpha}. \end{align}

• In the position basis \begin{align} \mathop{\hat H}&=\int d\br \left[-\frac{1}{2m}\pdop(\br)\nabla^{2}\pop(\br)+V(\br)\pdop(\br)\pop(\br)\right]\nonumber\\ &=\int d\br \left[\frac{1}{2m}\nabla\pdop(\br)\cdot\nabla\pop(\br)+V(\br)\pdop(\br)\pop(\br)\right], \label{2nd_quant_HPos} \end{align} (integration by parts)

• Confirm equality to previous expression using

$$\pop(\br)\equiv\sum_{\beta} \varphi_{\beta}(\br)\aop_{\beta},$$

• Heisenberg equation of motion for $\pop(\br)=e^{i\hat Ht}\pop(\br) e^{-i\hat H t}$ with noninteracting Hamiltonian $$$$\label{2nd_quant_HeomFree} \begin{split} i\partial_{t}\pop(\br,t) &= -\left[\mathop{\hat H},\pop(\br,t)\right]\\ &= -\frac{1}{2m}\nabla^{2}\pop(\br,t)+V(\br)\pop(\br,t), \end{split}$$$$ just the time dependent Schrödinger equation!

### Example 2: Density

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

$$\label{2nd_quant_spDens} \rho(\mathbf{x})\equiv\delta(\mathbf{x}-\br).$$

• Expectation value in single particle state $\varphi(\br)$ is $\rho(\mathbf{x})=\abs{\varphi(\mathbf{x})}^{2}$

• Second quantized form of the operator is then

$$\label{2nd_quant_2ndQDens} \hat\rho(\mathbf{x})\equiv\pdop(\mathbf{x})\pop(\mathbf{x}).$$

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

$$\label{2nd_quant_DensIntegral} \hat N=\int d\mathbf{x}, \pdop(\mathbf{x})\pop(\mathbf{x})=\sum_{\alpha} \adop_{\alpha}\aop_{a}=\sum_{\alpha}\Nop_{\alpha},$$

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

$$\label{2nd_quant_DensityExp} \braket{N_{0},N_{1}\ldots}{ \hat\rho(\br)}{N_{0},N_{1}\ldots} = \sum_{\alpha} N_{\alpha}\left|\varphi_{\alpha}(\br)\right|^{2}.$$

Prove using $\pop(\br)\equiv\sum_{\beta} \varphi_{\beta}(\br)\aop_{\beta}$

• Interpretation: $\left|\varphi_{\alpha}(\br)\right|^{2}$ is probability to find a particle in state $\alpha$ at position $\br$. Density is weighted by occupation

### Example 3: Current operator

$$\label{2nd_quant_current} \hat{\mathbf{j}}(\br)=-i\frac{1}{2m}\left[\pdop(\br)\left(\nabla\pop(\br)\right)-\left(\nabla\pdop(\br)\right)\pop(\br)\right].$$

• Often we consider Fourier components of density or current \begin{align} \label{2nd_quant_FourierComp} \hat\rho_{\bq}\equiv\int d\br\, \hat\rho(\br)e^{-i\bq \cdot \br}=\sum_{\bk} \adop_{\bk-\bq/2}\aop_{\bk+\bq/2}\nonumber\\ \hat{\mathbf{j}}_{\bq}\equiv\int d\br\, \hat{\mathbf{j}}(\br)e^{-i\bq \cdot \br}=\sum_{\bk} \frac{\bk}{m}\adop_{\bk-\bq/2}\aop_{\bk+\bq/2}. \end{align}
• $\bq=0$ modes are total particle number and $\frac{1}{m}\times$ total momentum, respectively

### Single particle density matrix

$$g(\br,\br’) \equiv N \int d\br_{2}\cdots d\br_{N},\Psi^{*}(\br,\br_{2},\ldots,\br_{N})\Psi(\br’,\br_{2},\ldots,\br_{N}).$$

Show that this can be written as

$$\label{2nd_quant_SPDensity} g(\br,\br’)= \braket{\Psi}{\pdop(\br)\pop(\br’)}{\Psi}$$

• Similar calculation as for density gives

$$\label{2nd_quant_SPFock} g(\br,\br’) = \sum_{\alpha} N_{\alpha}\varphi_{\alpha}^{*}(\br)\varphi^{}_{\alpha}(\br’).$$

• For 3D $N_{\bk}=1$ for $\abs{\bk}<k_{F}$, and $0$ otherwise

$$\ket{\text{Fermi sea}} = \prod_{|\bk|<k_F} \adop_\bk\ket{\text{VAC}}$$

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

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

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

## Two Particle Operators

• Acts pairwise on the particles ($B_{jk}=B_{kj}$ for indistinguishable particles.) $$\hat B = \sum_{j<k} B_{jk}.$$

• Action of $\hat B$ on product state $\ket{\varphi_{\alpha}}_1\ket{\varphi_{\beta}}_2$ in terms of matrix elements \begin{align} B_{\alpha\beta,\gamma\delta} &= \bra{\varphi_\alpha}_1\bra{\varphi_\beta}_2 B_{12} \ket{\varphi_\gamma}_1\ket{\varphi_\delta}_2\\ \hat B &= \frac{1}{2}\sum_{\alpha\beta\gamma\delta} B_{\alpha\beta,\gamma\delta}\adop_\alpha\adop_\beta\aop_\delta\aop_\gamma,\qquad B_{\alpha\beta,\gamma\delta} = B_{\beta\alpha,\delta\gamma} \end{align} (Note order, which is important for fermions!).

\begin{align} B_{\alpha\beta,\gamma\delta} &= \bra{\varphi_\alpha}_1\bra{\varphi_\beta}_2 B_{12} \ket{\varphi_\gamma}_1\ket{\varphi_\delta}_2\\ \hat B &= \frac{1}{2}\sum_{\alpha\beta\gamma\delta} B_{\alpha\beta,\gamma\delta}\adop_\alpha\adop_\beta\aop_\delta\aop_\gamma. \end{align}

Check this works on product states ($N=2$ first)

• Remember that for one particle operators

$$\braket{\mathbf{N}}{\hat A}{\mathbf{N’}} = A_{\alpha\beta} \sqrt{N_\alpha N’_\beta}$$

• For two particle operators $$\braket{\mathbf{N}}{\hat B}{\mathbf{N'}} = \left[B_{\alpha\beta,\gamma\delta}\pm B_{\alpha\beta,\delta\gamma}\right] \sqrt{N_\alpha N_\beta N'_\gamma N'_\delta}. \label{A_Babcd}$$ with $N_{\gamma,\delta} = N'_{\gamma,\delta}-1$ and $N_{\alpha,\beta} = N'_{\alpha,\beta}+1$

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

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

• 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

$$\hat H_\text{int.} = \sum_{j<k} U(\br_j-\br_k).$$

• Expressing in position basis

$$\hat H_\text{int.} = \frac{1}{2}\int d\br_1 d\br_2, U(\br_1-\br_2)\pdop(\br_1)\pdop(\br_2)\pop(\br_2)\pop(\br_1)$$

• Remembering that $\rho(\br) = \pdop(\br)\pop(\br)$, this is almost $$\hat H_\text{int.} = \frac{1}{2}\int d\br_1 d\br_2\, U(\br_1-\br_2)\rho(\br_1)\rho(\br_2) \label{A_VNotNormal}$$ Operator order prevents a particle from interacting with itself!

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

• Hamiltonian of interacting bosons from Lecture 1 $$H = -\frac{1}{2m}\sum_j \frac{\partial^2}{\partial x_j^2} + \overbrace{c\sum_{j<k}\delta(x_j-x_k)}^{\equiv H_\text{int}}. \label{many_LL}$$ has second quantized form $$H = \int dx \left[\frac{1}{2}\partial_x\pdop(x)\partial_x\pop(x) + \frac{c}{2}\pdop(x)\pdop(x)\pop(x)\pop(x)\right].$$ $\pop(x)$, $\pdop(x)$ 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!