2.1.3 Second Quantization#

Prompts

What is the key difference between first-quantized and second-quantized descriptions of a many-body system? Why is the occupation-number representation more natural for identical particles?
Define the creation operator \(\hat{a}^\dagger\) and annihilation operator \(\hat{a}\). How do they act on Fock states for bosons versus fermions?
Why do boson operators satisfy commutation relations while fermion operators satisfy anticommutation relations? How does each algebra enforce the correct quantum statistics?
What is boson enhancement? If you apply \(\hat{b}^\dagger\) to a state with \(n\) bosons, why is the amplitude \(\sqrt{n+1}\) and what physical phenomenon does this explain?
Show that \((\hat{c}^\dagger_\alpha)^2 = 0\) for fermions and explain why this is the Pauli exclusion principle in operator language.

Lecture Notes#

Overview#

Second quantization reformulates quantum mechanics so that identical-particle statistics are automatic. Instead of labeling individual particles and then symmetrizing or antisymmetrizing by hand, we simply count how many particles occupy each single-particle mode. This shift from “particle labels” to “occupation numbers” handles variable particle number naturally and leads directly to the operator algebra of creation and annihilation.

First vs Second Quantization#

First quantization asks: “Which particle is in which state?” An \(N\)-particle state is a tensor product

\[\vert\Psi\rangle = \vert\psi_1\rangle \otimes \vert\psi_2\rangle \otimes \cdots \otimes \vert\psi_N\rangle\]

that must be manually symmetrized (bosons) or antisymmetrized (fermions).

Second quantization asks: “How many particles occupy each state?” A state is specified by occupation numbers

\[\vert n_1, n_2, \ldots, n_D\rangle\]

where \(n_\alpha\) counts particles in single-particle mode \(\alpha\). Symmetry is built into the operator algebra—no manual (anti)symmetrization needed.

Fock Space#

Fock Space

The Fock space is the direct sum of all \(N\)-particle Hilbert spaces:

\[\mathcal{F} = \bigoplus_{N=0}^\infty \mathcal{H}_N\]

Its basis states are Fock states \(\vert n_1, n_2, \ldots, n_D\rangle\), labeled by occupation numbers \(\{n_\alpha\}\). These form a complete orthonormal basis:

\[\langle n_1', n_2', \ldots \vert n_1, n_2, \ldots \rangle = \prod_\alpha \delta_{n_\alpha' n_\alpha}\]

The vacuum \(\vert 0\rangle \equiv \vert 0, 0, \ldots, 0\rangle\) has no particles.

Creation and Annihilation Operators#

The creation operator \(\hat{a}^\dagger_\alpha\) adds a particle to mode \(\alpha\); the annihilation operator \(\hat{a}_\alpha\) removes one. Their action on Fock states differs for bosons and fermions.

Bosonic Operators (\(\hat{b}_\alpha\), \(\hat{b}^\dagger_\alpha\))#

For bosons, creation and annihilation operators are defined from the symmetric insertion/deletion operators \(\rhd_+\) and \(\lhd_+\). On an \(N\)-particle first-quantized state \(\vert\Psi_N\rangle\),

(34)#\[\begin{split} \begin{split} \hat{b}_\alpha^\dagger\vert\Psi_N\rangle&=\frac{1}{\sqrt{N+1}}\left(\vert\alpha\rangle\rhd_+\vert\Psi_N\rangle\right),\\ \hat{b}_\alpha\vert\Psi_N\rangle&=\frac{1}{\sqrt{N}}\left(\vert\alpha\rangle\lhd_+\vert\Psi_N\rangle\right). \end{split} \end{split}\]

Therefore in occupation-number basis,

\[ \hat{b}^\dagger_\alpha \vert \ldots, n_\alpha, \ldots\rangle = \sqrt{n_\alpha + 1}\;\vert \ldots, n_\alpha + 1, \ldots\rangle \]

\[ \hat{b}_\alpha \vert \ldots, n_\alpha, \ldots\rangle = \sqrt{n_\alpha}\;\vert \ldots, n_\alpha - 1, \ldots\rangle \]

Derivation: Bosonic action on Fock states

Using the bosonic operator definition in (34), we now compute the action explicitly from insertion/deletion rules.

For a single mode, denote \(\vert n_\alpha\rangle\equiv\vert\alpha\rangle^{\otimes n_\alpha}\).

Claim.

\[\begin{split} \begin{split} \hat{b}_\alpha^\dagger\vert n_\alpha\rangle&=\sqrt{n_\alpha+1}\;\vert n_\alpha+1\rangle,\\ \hat{b}_\alpha\vert n_\alpha\rangle&=\sqrt{n_\alpha}\;\vert n_\alpha-1\rangle. \end{split} \end{split}\]

Proof. Using (34),

\[ \hat{b}_\alpha^\dagger\vert n_\alpha\rangle =\frac{1}{\sqrt{n_\alpha+1}}\left(\vert\alpha\rangle\rhd_+\vert\alpha\rangle^{\otimes n_\alpha}\right). \]

The recursive insertion rule generates one term for each insertion position. For \(\vert\alpha\rangle^{\otimes n_\alpha}\) there are exactly \(n_\alpha+1\) insertion slots (before the 1st factor, between adjacent factors, and after the last factor), and all produced tensors are identical in the bosonic single-mode case:

\[ \vert\alpha\rangle\rhd_+\vert\alpha\rangle^{\otimes n_\alpha} =(n_\alpha+1)\vert\alpha\rangle^{\otimes(n_\alpha+1)}. \]

Combining this combinatorial factor with the operator normalization \(1/\sqrt{n_\alpha+1}\) gives

\[ \hat{b}_\alpha^\dagger\vert n_\alpha\rangle =\frac{1}{\sqrt{n_\alpha+1}}\cdot(n_\alpha+1)\vert\alpha\rangle^{\otimes(n_\alpha+1)} =\sqrt{n_\alpha+1}\;\vert n_\alpha+1\rangle. \]

Similarly,

\[ \hat{b}_\alpha\vert n_\alpha\rangle =\frac{1}{\sqrt{n_\alpha}}\left(\vert\alpha\rangle\lhd_+\vert\alpha\rangle^{\otimes n_\alpha}\right). \]

The recursive deletion rule sums over all removable copies of \(\vert\alpha\rangle\); now there are exactly \(n_\alpha\) deletion choices:

\[ \vert\alpha\rangle\lhd_+\vert\alpha\rangle^{\otimes n_\alpha} =n_\alpha\vert\alpha\rangle^{\otimes(n_\alpha-1)}. \]

Combining this with the normalization \(1/\sqrt{n_\alpha}\) gives

\[ \hat{b}_\alpha\vert n_\alpha\rangle =\frac{1}{\sqrt{n_\alpha}}\cdot n_\alpha\vert\alpha\rangle^{\otimes(n_\alpha-1)} =\sqrt{n_\alpha}\;\vert n_\alpha-1\rangle. \]

So the different prefactors come from different combinatorial counts: \(n_\alpha+1\) insertion slots versus \(n_\alpha\) deletion choices. For multiple modes \(\vert\ldots,n_\beta,n_\alpha,n_\gamma,\ldots\rangle\), the same coefficients follow: other modes are independent tensor factors, bosonic insertion has no sign change when passing other factors, and deletion picks only matching mode contributions via \(\langle\alpha\vert\beta\rangle=\delta_{\alpha\beta}\). Hence only \(n_\alpha\) changes.

Fermionic Operators (\(\hat{c}_\alpha\), \(\hat{c}^\dagger_\alpha\))#

For fermions, creation and annihilation operators are defined from antisymmetric insertion/deletion operators \(\rhd_-\) and \(\lhd_-\). On an \(N\)-particle first-quantized state \(\vert\Psi_N\rangle\),

(35)#\[\begin{split} \begin{split} \hat{c}_\alpha^\dagger\vert\Psi_N\rangle&=\frac{1}{\sqrt{N+1}}\left(\vert\alpha\rangle\rhd_-\vert\Psi_N\rangle\right),\\ \hat{c}_\alpha\vert\Psi_N\rangle&=\frac{1}{\sqrt{N}}\left(\vert\alpha\rangle\lhd_-\vert\Psi_N\rangle\right). \end{split} \end{split}\]

Therefore in occupation-number basis (\(n_\alpha\in\{0,1\}\)),

\[\begin{split} \hat{c}^\dagger_\alpha \vert \ldots, n_\alpha, \ldots\rangle = \begin{cases} (-1)^{P_\alpha} \vert \ldots, 1, \ldots\rangle & n_\alpha = 0 \\ 0 & n_\alpha = 1 \end{cases} \end{split}\]

\[\begin{split} \hat{c}_\alpha \vert \ldots, n_\alpha, \ldots\rangle = \begin{cases} (-1)^{P_\alpha} \vert \ldots, 0, \ldots\rangle & n_\alpha = 1 \\ 0 & n_\alpha = 0 \end{cases} \end{split}\]

where \(P_\alpha = \sum_{i<\alpha} n_i\) counts occupied modes before \(\alpha\) in the canonical mode ordering.

Derivation: Fermionic action on Fock states

We derive the formulas directly from (35) by falling back to first-quantized states.

Single mode.

Take one mode \(\alpha\), with \(\vert0_\alpha\rangle\equiv\mathbb{1}\) and \(\vert1_\alpha\rangle\equiv\vert\alpha\rangle\).

From the vacuum rules,

\[ \vert\alpha\rangle\rhd_-\mathbb{1}=\vert\alpha\rangle, \qquad \vert\alpha\rangle\lhd_-\mathbb{1}=0. \]

Using the recursive insertion rule on the occupied state,

\[ \vert\alpha\rangle\rhd_-\vert\alpha\rangle =\vert\alpha\alpha\rangle-\vert\alpha\alpha\rangle=0, \]

which is Pauli exclusion at the insertion-operator level.

Using the recursive deletion rule,

\[ \vert\alpha\rangle\lhd_-\vert\alpha\rangle =\langle\alpha\vert\alpha\rangle\mathbb{1}-\vert\alpha\rangle\otimes(\vert\alpha\rangle\lhd_-\mathbb{1}) =\mathbb{1}. \]

Now apply the operator definitions:

\[\begin{split} \begin{split} \hat c_\alpha^\dagger\vert0_\alpha\rangle &=\frac{1}{\sqrt{1}}(\vert\alpha\rangle\rhd_-\mathbb{1})=\vert1_\alpha\rangle,\\ \hat c_\alpha\vert0_\alpha\rangle &=0,\\ \hat c_\alpha^\dagger\vert1_\alpha\rangle &=\frac{1}{\sqrt{2}}(\vert\alpha\rangle\rhd_-\vert\alpha\rangle)=0,\\ \hat c_\alpha\vert1_\alpha\rangle &=\frac{1}{\sqrt{1}}(\vert\alpha\rangle\lhd_-\vert\alpha\rangle)=\vert0_\alpha\rangle. \end{split} \end{split}\]

So the single-mode action is fully derived from \(\rhd_-\) and \(\lhd_-\), with no extra assumptions.

Multi-mode sign \((-1)^{P_\alpha}\).

Because of Pauli exclusion, each mode can be occupied at most once. So a fermion Fock basis state is equivalent to choosing a subset of occupied modes,

\[ S=\{\alpha_1,\alpha_2,\ldots,\alpha_N\}. \]

A set is orderless, but to define states and signs unambiguously we choose the canonical ordered representative

\[ \alpha_1<\alpha_2<\cdots<\alpha_N, \qquad \vert S\rangle_-\equiv\vert\alpha_1,\alpha_2,\ldots,\alpha_N\rangle_-. \]

Adding a fermion in mode \(\alpha\) means adding a new element to the subset, \(S\to S\cup\{\alpha\}\) (if \(\alpha\notin S\)). In the antisymmetric insertion picture, this new mode must be permuted into canonical order. If exactly \(P_\alpha\) occupied modes in \(S\) are smaller than \(\alpha\), then the insertion crosses those \(P_\alpha\) modes, giving

\[ \vert\alpha\rangle\rhd_-\vert \alpha_1,\ldots,\alpha_N\rangle_- =(-1)^{P_\alpha}\vert \alpha_1,\ldots,\alpha,\ldots,\alpha_N\rangle_-. \]

Deletion is the reverse operation and carries the same crossing-count phase. Therefore in occupation notation, \(P_\alpha=\sum_{\beta<\alpha}n_\beta\), and the operator actions acquire the factor \((-1)^{P_\alpha}\).

Algebraic Relations#

The following algebras are equivalent to the operator definitions above, and in practice are often used as defining relations for boson/fermion operators.

Commutation Relations (Bosons)

(36)#\[\begin{split} \begin{split} [\hat{b}_\alpha, \hat{b}^\dagger_\beta] &= \delta_{\alpha\beta},\\ [\hat{b}_\alpha, \hat{b}_\beta] &= 0,\\ [\hat{b}^\dagger_\alpha, \hat{b}^\dagger_\beta] &= 0. \end{split} \end{split}\]

Derivation: Bosonic commutators on Fock basis

Use the bosonic Fock-action formulas above, (34) and \(\hat{b}_\alpha^\dagger\vert\ldots,n_\alpha,\ldots\rangle=\sqrt{n_\alpha+1}\,\vert\ldots,n_\alpha+1,\ldots\rangle\), \(\hat{b}_\alpha\vert\ldots,n_\alpha,\ldots\rangle=\sqrt{n_\alpha}\,\vert\ldots,n_\alpha-1,\ldots\rangle\).

Let \(\vert\mathbf{n}\rangle\equiv\vert\ldots,n_\alpha,\ldots,n_\beta,\ldots\rangle\).

For \(\alpha\neq\beta\), both orders give the same final state with the same coefficient,

\[ \hat{b}_\alpha\hat{b}_\beta^\dagger\vert\mathbf{n}\rangle =\hat{b}_\beta^\dagger\hat{b}_\alpha\vert\mathbf{n}\rangle =\sqrt{n_\alpha}\sqrt{n_\beta+1}\,\vert\ldots,n_\alpha-1,\ldots,n_\beta+1,\ldots\rangle, \]

so \([\hat{b}_\alpha,\hat{b}_\beta^\dagger]\vert\mathbf{n}\rangle=0\).

For \(\alpha=\beta\),

\[\begin{split} \begin{split} \hat{b}_\alpha\hat{b}_\alpha^\dagger\vert\mathbf{n}\rangle&=(n_\alpha+1)\vert\mathbf{n}\rangle,\\ \hat{b}_\alpha^\dagger\hat{b}_\alpha\vert\mathbf{n}\rangle&=n_\alpha\vert\mathbf{n}\rangle, \end{split} \end{split}\]

hence \([\hat{b}_\alpha,\hat{b}_\alpha^\dagger]\vert\mathbf{n}\rangle=\vert\mathbf{n}\rangle\). Combining both cases gives \([\hat{b}_\alpha,\hat{b}_\beta^\dagger]\vert\mathbf{n}\rangle=\delta_{\alpha\beta}\vert\mathbf{n}\rangle\).

Similarly, \(\hat{b}_\alpha\hat{b}_\beta\) and \(\hat{b}_\beta\hat{b}_\alpha\) (and likewise two creation operators) produce identical states with identical coefficients, so \([\hat{b}_\alpha,\hat{b}_\beta]=0\) and \([\hat{b}_\alpha^\dagger,\hat{b}_\beta^\dagger]=0\) on every basis state.

Anticommutation Relations (Fermions)

(37)#\[\begin{split} \begin{split} \{\hat{c}_\alpha, \hat{c}^\dagger_\beta\} &= \delta_{\alpha\beta},\\ \{\hat{c}_\alpha, \hat{c}_\beta\} &= 0,\\ \{\hat{c}^\dagger_\alpha, \hat{c}^\dagger_\beta\} &= 0. \end{split} \end{split}\]

Derivation: Fermionic anticommutators on Fock basis

Use the fermionic Fock-action formulas above from (35): \(\hat{c}_\alpha^\dagger\vert\ldots,n_\alpha,\ldots\rangle\) and \(\hat{c}_\alpha\vert\ldots,n_\alpha,\ldots\rangle\) with the phase \((-1)^{P_\alpha}\).

For \(\alpha=\beta\), evaluate on \(\vert\mathbf{n}\rangle\) by cases:

if \(n_\alpha=0\): \(\hat{c}_\alpha\hat{c}_\alpha^\dagger\vert\mathbf{n}\rangle=\vert\mathbf{n}\rangle\), while \(\hat{c}_\alpha^\dagger\hat{c}_\alpha\vert\mathbf{n}\rangle=0\);
if \(n_\alpha=1\): \(\hat{c}_\alpha\hat{c}_\alpha^\dagger\vert\mathbf{n}\rangle=0\), while \(\hat{c}_\alpha^\dagger\hat{c}_\alpha\vert\mathbf{n}\rangle=\vert\mathbf{n}\rangle\).

So in either case, \(\{\hat{c}_\alpha,\hat{c}_\alpha^\dagger\}\vert\mathbf{n}\rangle=\vert\mathbf{n}\rangle\).

For \(\alpha\neq\beta\), both orderings lead to the same occupation pattern, but one path has exactly one extra fermionic swap relative to the other; therefore the two amplitudes are negatives and cancel:

\[ \hat{c}_\alpha\hat{c}_\beta^\dagger\vert\mathbf{n}\rangle =-\hat{c}_\beta^\dagger\hat{c}_\alpha\vert\mathbf{n}\rangle, \]

including occupancy-forbidden cases where both sides are zero. Hence \(\{\hat{c}_\alpha,\hat{c}_\beta^\dagger\}\vert\mathbf{n}\rangle=0\) for \(\alpha\neq\beta\).

The same sign-cancellation argument gives \(\{\hat{c}_\alpha,\hat{c}_\beta\}=0\) and \(\{\hat{c}_\alpha^\dagger,\hat{c}_\beta^\dagger\}=0\) on the full Fock basis.

Number Operator#

The number operator \(\hat{n}_\alpha = \hat{a}^\dagger_\alpha \hat{a}_\alpha\) counts particles in mode \(\alpha\):

\[\hat{n}_\alpha \vert \ldots, n_\alpha, \ldots\rangle = n_\alpha \vert \ldots, n_\alpha, \ldots\rangle\]

Fock states are eigenstates of every \(\hat{n}_\alpha\) simultaneously.

Derivation: Number operator for bosons and fermions

Use the operator actions already derived above.

For bosons,

\[ \hat{b}_\alpha^\dagger\hat{b}_\alpha\vert\ldots,n_\alpha,\ldots\rangle =\hat{b}_\alpha^\dagger\left(\sqrt{n_\alpha}\vert\ldots,n_\alpha-1,\ldots\rangle\right) =n_\alpha\vert\ldots,n_\alpha,\ldots\rangle. \]

So the bosonic number operator in mode \(\alpha\) is \(\hat{n}_\alpha^{(B)}=\hat{b}_\alpha^\dagger\hat{b}_\alpha\).

For fermions (\(n_\alpha\in\{0,1\}\)), check both cases:

\[\begin{split} \begin{split} \hat{c}_\alpha^\dagger\hat{c}_\alpha\vert\ldots,0,\ldots\rangle&=0,\\ \hat{c}_\alpha^\dagger\hat{c}_\alpha\vert\ldots,1,\ldots\rangle&=\vert\ldots,1,\ldots\rangle. \end{split} \end{split}\]

Hence \(\hat{c}_\alpha^\dagger\hat{c}_\alpha\vert\ldots,n_\alpha,\ldots\rangle=n_\alpha\vert\ldots,n_\alpha,\ldots\rangle\), so \(\hat{n}_\alpha^{(F)}=\hat{c}_\alpha^\dagger\hat{c}_\alpha\).

Proof: Quantization of Boson Number

For one bosonic mode, let

\[ \hat{n}\vert n\rangle=n\vert n\rangle. \]

Using positivity, for every integer \(m\ge 0\),

\[ \langle n\vert(\hat{b}^\dagger)^m(\hat{b})^m\vert n\rangle\ge 0. \]

To evaluate this matrix element, expand only the lowering action:

\[\begin{split} \begin{split} \hat b\vert n\rangle&=\sqrt n\,\vert n-1\rangle,\\ \hat b^2\vert n\rangle&=\sqrt{n(n-1)}\,\vert n-2\rangle,\\ &\ \vdots\\ \hat b^m\vert n\rangle&=\sqrt{n(n-1)\cdots(n-m+1)}\,\vert n-m\rangle. \end{split} \end{split}\]

Now use \(\langle n\vert(\hat b^\dagger)^m\hat b^m\vert n\rangle=\|\hat b^m\vert n\rangle\|^2\). Squaring the coefficient gives

\[ \langle n\vert(\hat b^\dagger)^m\hat b^m\vert n\rangle =n(n-1)\cdots(n-m+1) =\prod_{\ell=0}^{m-1}(n-\ell)\ge 0, \qquad (\forall m\in\mathbb{N}). \]

The first few constraints are

\[\begin{split} \begin{split} m=1&:\quad n\ge 0,\\ m=2&:\quad n(n-1)\ge 0,\\ m=3&:\quad n(n-1)(n-2)\ge 0,\ \ldots \end{split} \end{split}\]

The only values satisfying all of them simultaneously are

\[ n\in\{0,1,2,\ldots\}=\mathbb{N}. \]

So bosonic number-operator eigenvalues are natural numbers.

Bosons vs Fermions: Complete Comparison#

Aspect	Bosons	Fermions
Statistics	Bose-Einstein	Fermi-Dirac
Wavefunction	Symmetric (permanent)	Antisymmetric (determinant)
Occupation	\(n_\alpha \in \{0,1,2,\ldots\}\)	\(n_\alpha \in \{0,1\}\)
Creation	\(\hat{b}^\dagger \vert n\rangle = \sqrt{n+1}\,\vert n{+}1\rangle\)	\(\hat{c}^\dagger \vert 0\rangle = \vert 1\rangle\), \(\hat{c}^\dagger \vert 1\rangle = 0\)
Annihilation	\(\hat{b} \vert n\rangle = \sqrt{n}\,\vert n{-}1\rangle\)	\(\hat{c} \vert 1\rangle = \vert 0\rangle\), \(\hat{c} \vert 0\rangle = 0\)
Algebra	\([\hat{b},\hat{b}^\dagger]=1\)	\(\{\hat{c},\hat{c}^\dagger\}=1\)
Phase	None	\((-1)^{P_\alpha}\) (parity sign)
Repeated creation	\((\hat{b}^\dagger)^n \vert 0\rangle = \sqrt{n!}\,\vert n\rangle\)	\((\hat{c}^\dagger)^2 = 0\)
Key consequence	Stimulated emission, BEC	Pauli exclusion
Examples	Photons, phonons, \(^4\text{He}\)	Electrons, protons, neutrons

Boson Enhancement#

Apply \(\hat{b}^\dagger\) to a superposition of Fock states:

\[\hat{b}^\dagger \vert\psi\rangle = \hat{b}^\dagger \sum_n c_n \vert n\rangle = \sum_n c_n \sqrt{n+1}\;\vert n+1\rangle\]

The \(\sqrt{n+1}\) factor means it is more probable to add a boson to a mode that already contains many bosons. This “rich get richer” effect is the origin of stimulated emission—the transition rate into a mode with \(n\) bosons is proportional to \(n+1\) (stimulated + spontaneous). It also drives Bose-Einstein condensation.

Example: Stimulated Emission

Problem. A cavity mode contains \(n\) photons. A two-level atom in the excited state can emit a photon into the mode. Show that the emission matrix element is proportional to \(\sqrt{n+1}\).

Solution. The relevant operator is \(\hat{b}^\dagger\) acting on the photon Fock state \(\vert n\rangle\):

\[\hat{b}^\dagger \vert n\rangle = \sqrt{n+1}\;\vert n+1\rangle\]

The emission amplitude is \(\langle n+1\vert \hat{b}^\dagger \vert n\rangle = \sqrt{n+1}\), so the emission rate \(\propto n+1\). The “\(+1\)” corresponds to spontaneous emission (present even at \(n=0\)), while the “\(n\)” part is stimulated emission. This is why lasers work: a large photon population exponentially amplifies further emission.

Pauli Exclusion in Operator Language#

For fermions, the anticommutation relation \(\{\hat{c}^\dagger_\alpha, \hat{c}^\dagger_\alpha\} = 0\) implies:

\[(\hat{c}^\dagger_\alpha)^2 = 0\]

Attempting to create a second fermion in an occupied mode gives zero—the state is annihilated. This is the Pauli exclusion principle encoded directly in the algebra, not imposed by hand.

Discussion: Why Commutators vs Anticommutators?

Bosons obey commutation relations, fermions obey anticommutation relations. This is not an arbitrary choice—it is dictated by the symmetry of the wavefunction. Symmetric states require commuting creation operators; antisymmetric states require anticommuting ones. The spin-statistics theorem (a consequence of relativistic quantum field theory) links integer spin to bosonic statistics and half-integer spin to fermionic statistics. Can you see why \([\hat{b}^\dagger_1, \hat{b}^\dagger_2] = 0\) produces symmetric states while \(\{\hat{c}^\dagger_1, \hat{c}^\dagger_2\} = 0\) produces antisymmetric ones?

Summary#

Second quantization replaces particle labels with occupation numbers \(\{n_\alpha\}\), making identical-particle statistics automatic.
Fock space \(\mathcal{F} = \bigoplus_N \mathcal{H}_N\) spans all particle-number sectors; Fock states \(\vert n_1, n_2, \ldots\rangle\) form its orthonormal basis.
Creation/annihilation operators add and remove particles; commutators \([\hat{b},\hat{b}^\dagger]=1\) enforce bosonic statistics, anticommutators \(\{\hat{c},\hat{c}^\dagger\}=1\) enforce fermionic statistics.
Boson enhancement: the \(\sqrt{n+1}\) factor in \(\hat{b}^\dagger\vert n\rangle\) drives stimulated emission and Bose-Einstein condensation.
Pauli exclusion: \((\hat{c}^\dagger_\alpha)^2 = 0\) forbids double occupation, following automatically from anticommutation.
The number operator \(\hat{n}_\alpha = \hat{a}^\dagger_\alpha \hat{a}_\alpha\) counts particles per mode.

Homework#

1. Using the bosonic commutation relation \([\hat{b}_\alpha, \hat{b}^\dagger_\alpha] = 1\), show that \(\hat{n}_\alpha = \hat{b}^\dagger_\alpha \hat{b}_\alpha\) satisfies \(\hat{n}_\alpha \vert n_\alpha\rangle = n_\alpha \vert n_\alpha\rangle\), with \(n_\alpha = 0, 1, 2, \ldots\) Verify explicitly for \(n_\alpha = 0\) and \(n_\alpha = 1\).

2. Compute the commutators \([\hat{n}_\alpha, \hat{b}^\dagger_\beta]\) and \([\hat{n}_\alpha, \hat{b}_\beta]\) using \([\hat{b}_\alpha, \hat{b}^\dagger_\beta] = \delta_{\alpha\beta}\). Interpret: how does \(\hat{b}^\dagger_\beta\) change the eigenvalue of \(\hat{n}_\alpha\)?

3. Show directly from \(\{\hat{c}^\dagger_\alpha, \hat{c}^\dagger_\alpha\} = 0\) that \((\hat{c}^\dagger_\alpha)^2 = 0\). Explain why this is the Pauli exclusion principle in operator language.

4. Compute \(\langle 0 \vert \hat{b}_\alpha \hat{b}^\dagger_\alpha \vert 0\rangle\) and \(\langle 0 \vert \hat{b}^\dagger_\alpha \hat{b}_\alpha \vert 0\rangle\). What is the physical meaning of their difference?

5. For a two-mode bosonic system, list all Fock states with total particle number \(N = 2\). Do the same for a two-mode fermionic system. How many states exist in each case?

6. For non-interacting particles with single-particle energies \(\epsilon_\alpha\), the Hamiltonian is \(\hat{H} = \sum_\alpha \epsilon_\alpha \hat{n}_\alpha\). Compute \(\langle n_1, n_2, \ldots \vert \hat{H} \vert n_1, n_2, \ldots\rangle\).

7. Show that \([\hat{H}, \hat{N}] = 0\) for the Hamiltonian in problem 6, where \(\hat{N} = \sum_\alpha \hat{n}_\alpha\) is the total number operator. What conservation law does this express?

8. The single-particle kinetic energy eigenvalue in a plane-wave basis is \(\epsilon_{\boldsymbol{k}} = \hbar^2 k^2 / 2m\). Write the second-quantized kinetic energy operator \(\hat{T} = \sum_{\boldsymbol{k}} \epsilon_{\boldsymbol{k}} \hat{b}^\dagger_{\boldsymbol{k}} \hat{b}_{\boldsymbol{k}}\) for bosons. Compare this to the first-quantized form \(\sum_{i=1}^N \hat{\boldsymbol{p}}_i^2 / 2m\): which is simpler and why?