Second Quantization: Operators

Hello! Welcome to our next lesson.

In our last session, we established the framework of second quantization. We introduced Fock space, the arena for many-body systems, and defined the fundamental tools: creation ($c_i^{†}$) and annihilation ($c_i$) operators. We saw how their algebraic relations—commutation for bosons and anti-commutation for fermions—elegantly encode the required particle statistics, saving us from the tedious manual symmetrization of wavefunctions.

Today, we will put these tools to work. Our goal is to express single-particle and two-particle operators in the second quantization formalism. We will learn the general prescription for translating familiar operators from first quantization (like kinetic energy and potential energy) into the language of creation and annihilation operators. This will give us a complete framework for writing down and eventually solving the Hamiltonians for the complex many-body systems you are interested in, particularly in condensed matter physics.

A Quick Recap: First Quantization Operators

In the first quantization picture, an operator for a system of $N$ identical particles is typically written as a sum of terms acting on individual particles or pairs of particles.

• Single-particle operators act on one particle at a time. The total operator is a sum over all particles:
  ${\hat{O}}_1={\sum }_{i=1}^N{\hat{o}}_i$
  Examples include the total kinetic energy $\hat{T}={\sum }_i\frac{{\hat{\mathbf{p}}}_i^2}{2m}$ and the potential energy from an external field ${\hat{V}}_{\text{ext}}={\sum }_{i}V({\hat{\mathbf{r}}}_i)$.

• Two-particle operators describe pairwise interactions. The total operator is a sum over all unique pairs:
  ${\hat{O}}_2={\sum }_{i<j}{\hat{o}}_{ij}=\frac{1}{2}{\sum }_{i\ne j}{\hat{o}}_{ij}$
  The most common example is the two-body interaction potential, such as the Coulomb interaction $V({\hat{\mathbf{r}}}_i-{\hat{\mathbf{r}}}_j)$.

Our task is to find the equivalent expressions for ${\hat{O}}_1$ and ${\hat{O}}_2$ using our new $c^{†}$ and $c$ operators.

Representing Single-Particle Operators

Let's start by considering a general single-particle operator ${\hat{O}}_1$. The following video provides a clear overview of how to express the typical one-body and two-body terms of a Hamiltonian in second quantization.

As the video illustrates, the second-quantized form of a single-particle operator ${\hat{O}}_1$ is given by a universal formula:

$${\hat{O}}_1=\sum _{\alpha ,\beta }⟨\alpha |\hat{o}|\beta ⟩c_{\alpha }^{†}{c}_{\beta }$$

Here, $\{|\alpha ⟩,|\beta ⟩,\dots \}$ represents a complete basis of single-particle states (e.g., momentum eigenstates, atomic orbitals), and $\hat{o}$ is the corresponding single-particle operator.

Physical Interpretation:
This expression has a very direct physical meaning. The term $c_{\alpha }^{†}{c}_{\beta }$ corresponds to a process where a particle is annihilated from state $|\beta ⟩$ and a particle is created in state $|\alpha ⟩$. The operator "scatters" a particle from state $\beta$ to state $\alpha$. The amplitude, or "strength," of this process is given by the first-quantization matrix element $⟨\alpha |\hat{o}|\beta ⟩$. The total operator ${\hat{O}}_1$ is the sum over all possible such scattering processes.

For a more formal statement and derivation, you can consult standard textbooks. The key result is summarized as a theorem in the following notes.

The Diagonal Representation: A Crucial Simplification

The general formula simplifies dramatically if we choose a basis that diagonalizes the single-particle operator $\hat{o}$. If the basis states $|\alpha ⟩$ are the eigenstates of $\hat{o}$ with eigenvalues $o_{\alpha }$, then:
$⟨\alpha |\hat{o}|\beta ⟩=o_{\alpha }{\delta }_{\alpha \beta }$

Substituting this into the general formula, the sum over $\beta$ collapses:
${\hat{O}}_1={\sum }_{\alpha ,\beta }{o}_{\alpha }{\delta }_{\alpha \beta }{c}_{\alpha }^{†}{c}_{\beta }={\sum }_{\alpha }{o}_{\alpha }{c}_{\alpha }^{†}{c}_{\alpha }$

Recognizing that ${\hat{n}}_{\alpha }=c_{\alpha }^{†}{c}_{\alpha }$ is the number operator for state $\alpha$, we get the highly intuitive result:

$${\hat{O}}_1=\sum _{\alpha }{o}_{\alpha }{\hat{n}}_{\alpha }$$

This states that the total value of the observable is simply the sum over all states of the eigenvalue for that state, $o_{\alpha }$, multiplied by the number of particles occupying it, ${\hat{n}}_{\alpha }$.

Example: The Non-Interacting Hamiltonian
This is particularly powerful for the non-interacting part of a Hamiltonian, $H_0=T+V_{\text{ext}}$. Let's define the single-particle Hamiltonian $h=\frac{{\hat{\mathbf{p}}}^2}{2m}+V(\hat{\mathbf{r}})$. If we work in the basis of its eigenstates $|k⟩$ with energies ${ϵ}_k$ (i.e., $h|k⟩={ϵ}_k|k⟩$), the many-body non-interacting Hamiltonian becomes:

$${\hat{H}}_0=\sum _k{ϵ}_k{c}_k^{†}{c}_k$$

This is a cornerstone of many-body theory. The Hamiltonian is diagonal in this basis, and its eigenstates are the Fock states $|n_1,n_2,\dots ,n_k,\dots ⟩$. The total energy is simply $E={\sum }_k{ϵ}_k{n}_k$.

The following video segment provides an excellent summary of this diagonalization process.

Representing Two-Particle Operators

Now let's turn to two-particle operators, which describe interactions. The process is analogous but involves four operators instead of two.

The general second-quantized form for a two-particle operator ${\hat{O}}_2=\frac{1}{2}{\sum }_{i\ne j}{\hat{o}}_{ij}$ is:

$${\hat{O}}_2=\frac{1}{2}\sum _{\alpha ,\beta ,\gamma ,\delta }⟨\alpha ,\beta |{\hat{o}}_{12}|\delta ,\gamma ⟩c_{\alpha }^{†}{c}_{\beta }^{†}{c}_{\gamma }{c}_{\delta }$$

Physical Interpretation:
This term describes a two-particle scattering event. Particles are annihilated from states $|\gamma ⟩$ and $|\delta ⟩$, and subsequently created in states $|\alpha ⟩$ and $|\beta ⟩$. The amplitude for this process is the two-particle matrix element $⟨\alpha ,\beta |{\hat{o}}_{12}|\delta ,\gamma ⟩$.

Let's watch a detailed explanation of this structure.

An Important Warning on Operator Ordering!
The order of the annihilation operators is crucial. The convention is $c_{\alpha }^{†}{c}_{\beta }^{†}{c}_{\gamma }{c}_{\delta }$. For bosons, $c_{\gamma }{c}_{\delta }=c_{\delta }{c}_{\gamma }$, so the order doesn't matter. But for fermions, $c_{\gamma }{c}_{\delta }=-c_{\delta }{c}_{\gamma }$. Swapping them introduces a minus sign, which could incorrectly change a repulsive interaction into an attractive one! The notes from UBC (resource LINK, section on two-particle operators) explicitly warn about this. The matrix element is $⟨\alpha \beta |{\hat{o}}_{12}|\delta \gamma ⟩$ and the operators are $c_{\alpha }^{†}{c}_{\beta }^{†}{c}_{\gamma }{c}_{\delta }$. Be careful to match the indices correctly.

Test your understanding!

Consider a Hubbard-like interaction term on a two-site system: ${\hat{H}}_{\text{int}}=U{\hat{n}}_{1\uparrow }{\hat{n}}_{1\downarrow }$. This operator gives an energy penalty $U$ only if a particle with spin up AND a particle with spin down are on site 1.

Rewrite ${\hat{n}}_{1\uparrow }=c_{1\uparrow }^{†}{c}_{1\uparrow }$ and ${\hat{n}}_{1\downarrow }=c_{1\downarrow }^{†}{c}_{1\downarrow }$. Can you manipulate the resulting expression $c_{1\uparrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }^{†}{c}_{1\downarrow }$ into the standard form $\sim c^{†}{c}^{†}cc$? (Assume the particles are fermions).

Show answer

We start with ${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }^{†}{c}_{1\downarrow }$.
To get it into the standard form, we need to move the annihilation operators to the right. Let's move $c_{1\uparrow }$ past $c_{1\downarrow }^{†}$. Since the operators have different indices (one is spin-up on site 1, the other is spin-down on site 1), they anti-commute: $\{c_{1\uparrow },c_{1\downarrow }^{†}\}=0$, which means $c_{1\uparrow }{c}_{1\downarrow }^{†}=-c_{1\downarrow }^{†}{c}_{1\uparrow }$.

So, ${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}(-c_{1\downarrow }^{†}{c}_{1\uparrow })c_{1\downarrow }=-U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }$.

This is now in the standard form $c^{†}{c}^{†}cc$, but notice the minus sign that appeared! If we had commuted them as if they were simple numbers, we would have missed this crucial sign. The final form is ${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\downarrow }{c}_{1\uparrow }$. Let's check this again: $c_{1\uparrow }{c}_{1\downarrow }^{†}=-c_{1\downarrow }^{†}{c}_{1\uparrow }+\{c_{1\uparrow },c_{1\downarrow }^{†}\}$. Since the states are different, the anticommutator is 0. So $c_{1\uparrow }{c}_{1\downarrow }^{†}=-c_{1\downarrow }^{†}{c}_{1\uparrow }$. The first expression is $U{c}_{1\uparrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }^{†}{c}_{1\downarrow }=U{c}_{1\uparrow }^{†}(-c_{1\downarrow }^{†}{c}_{1\uparrow })c_{1\downarrow }=-U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }$. Wait, this can't be right. The energy should be positive.

Let's re-examine ${\hat{n}}_1{\hat{n}}_2=c_1^{†}{c}_1{c}_2^{†}{c}_2$. We commute $c_1$ and $c_2^{†}$. If the states 1 and 2 are different, $c_1{c}_2^{†}=-c_2^{†}{c}_1$. So ${\hat{n}}_1{\hat{n}}_2=-c_1^{†}{c}_2^{†}{c}_1{c}_2$. This is still strange.

Ah, the issue is how we define the two-body operator from the number operators. The standard form ${\hat{O}}_2=\frac{1}{2}\sum ⟨\dots ⟩c^{†}{c}^{†}cc$ is for operators that are inherently two-body like $V(r_1-r_2)$. A product of number operators like ${\hat{n}}_i{\hat{n}}_j$ is already in a "normal-ordered" form (all $c^{†}$ to the left of all $c$). Let's re-examine $c_1^{†}{c}_1{c}_2^{†}{c}_2$. We can commute $c_1$ and $c_2^{†}$ only if they are different states.
$c_1^{†}{c}_1{c}_2^{†}{c}_2=c_1^{†}(-c_2^{†}{c}_1)c_2=-c_1^{†}{c}_2^{†}{c}_1{c}_2$. This is still giving a minus sign. Let's try commuting the other way:
$c_1^{†}{c}_1{c}_2^{†}{c}_2=c_1^{†}{c}_1(-c_2{c}_2^{†}+{\delta }_{22})=c_1^{†}{c}_1$. No, that's not right.
The correct approach is to note that $c_{1\uparrow }$ and $c_{1\downarrow }$ operators act on different spin states, so they effectively behave as different species. Fermionic operators for distinguishable states commute. More formally, if $\alpha \ne \beta$, $\{c_{\alpha },c_{\beta }\}=0$, but also $[c_{\alpha },c_{\beta }^{†}]=0$. So $c_{1\uparrow }$ and $c_{1\downarrow }^{†}$ commute!
Therefore:
${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }^{†}{c}_{1\downarrow }=U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\uparrow }{c}_{1\downarrow }$
Finally, to match the standard form $c^{†}{c}^{†}{c}_{\gamma }{c}_{\delta }$, we just need to reorder the annihilation operators:
${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}(-c_{1\downarrow }{c}_{1\uparrow })=-U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\downarrow }{c}_{1\uparrow }$
The correct answer is simply ${\hat{H}}_{\text{int}}=U{c}_{1\uparrow }^{†}{c}_{1\downarrow }^{†}{c}_{1\downarrow }{c}_{1\uparrow }$. The matrix element $⟨1\uparrow ,1\downarrow |U|1\downarrow ,1\uparrow ⟩$ would be $-U$, and $⟨1\uparrow ,1\downarrow |U|1\uparrow ,1\downarrow ⟩$ would be $U$. The product of number operators is already in the correct form. The expression is simply what it is, and can be seen as a special case of the general two-body operator where only the diagonal terms $\alpha =\delta ,\beta =\gamma$ are non-zero.

The Full Hamiltonian in Second Quantization

We can now write down a complete many-body Hamiltonian. It is the sum of the one-body and two-body parts:

$$\hat{H}=\sum _{\alpha ,\beta }⟨\alpha |\hat{h}|\beta ⟩c_{\alpha }^{†}{c}_{\beta }+\frac{1}{2}\sum _{\alpha ,\beta ,\gamma ,\delta }⟨\alpha ,\beta |\hat{v}|\delta ,\gamma ⟩c_{\alpha }^{†}{c}_{\beta }^{†}{c}_{\gamma }{c}_{\delta }$$

where $\hat{h}=\frac{{\hat{\mathbf{p}}}^2}{2m}+V_{\text{ext}}(\hat{\mathbf{r}})$ and $\hat{v}$ is the pairwise interaction.

This form is completely general. To make it concrete, one must choose a basis. For your interest in condensed matter, the plane wave (momentum) basis is extremely common. The UBC notes (resource LINK) walk through this for the kinetic energy and Coulomb interaction in sections (a) and (c). The result for the full electronic Hamiltonian in a solid (the "jellium" model) is:

$$\hat{H}=\sum _{\mathbf{k},\sigma }\frac{{\hslash }^2{k}^2}{2m}{c}_{\mathbf{k}\sigma }^{†}{c}_{\mathbf{k}\sigma }+\frac{1}{2V}\sum _{\begin{array}{c}\mathbf{k},{\mathbf{k}}^{\prime },\mathbf{q}\\ \sigma ,{\sigma }^{\prime }\end{array}}v(\mathbf{q})c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{{\mathbf{k}}^{\prime }-\mathbf{q},{\sigma }^{\prime }}^{†}{c}_{{\mathbf{k}}^{\prime },{\sigma }^{\prime }}{c}_{\mathbf{k},\sigma }$$

Here, $c_{\mathbf{k}\sigma }^{†}$ creates an electron with momentum $\mathbf{k}$ and spin $\sigma$, $V$ is the volume, and $v(\mathbf{q})$ is the Fourier transform of the Coulomb potential. Notice how the interaction term naturally enforces momentum conservation: two particles with momenta $\mathbf{k}$ and ${\mathbf{k}}^{\prime }$ scatter, exchanging momentum $\mathbf{q}$, resulting in final momenta $\mathbf{k}+\mathbf{q}$ and ${\mathbf{k}}^{\prime }-\mathbf{q}$.

Conclusion

Today we have translated the language of physical observables into the second quantization formalism. This is a pivotal step that completes our new toolkit for many-body quantum mechanics.

Key Takeaways:

• A single-particle operator ${\hat{O}}_1$ takes the form ${\sum }_{\alpha ,\beta }⟨\alpha |\hat{o}|\beta ⟩c_{\alpha }^{†}{c}_{\beta }$. In its own eigenbasis, this simplifies to the intuitive form ${\sum }_{\alpha }{o}_{\alpha }{\hat{n}}_{\alpha }$.
• A two-particle operator ${\hat{O}}_2$ takes the form $\frac{1}{2}{\sum }_{\alpha ,\beta ,\gamma ,\delta }⟨\alpha ,\beta |\hat{o}|\delta ,\gamma ⟩c_{\alpha }^{†}{c}_{\beta }^{†}{c}_{\gamma }{c}_{\delta }$. This describes a two-particle scattering process.
• The complete many-body Hamiltonian can be written using these forms, providing a compact and powerful representation that is independent of the number of particles in the system.

Preview of the Next Lesson:

We now have the formal expressions for operators, but how do we calculate expectation values like $⟨\Psi |\hat{O}|\Psi ⟩$? This often involves evaluating long strings of creation and annihilation operators. In our next lesson, we will learn Wick's theorem, a powerful combinatorial tool that provides a systematic recipe for evaluating the expectation values of any product of creation and annihilation operators in a non-interacting ground state. This theorem is the engine behind diagrammatic perturbation theory, a key technique in advanced condensed matter and quantum field theory.