Many-body quantum mechanics

Learning goals:

• Understanding how the many-body QM is formulated on top of the single-particle QM
• Bosons and fermions
• Fock space and operators acting on it
• Non-interacting Fermi gas
• Description of two-body interactions: the general many-body problem and the mean-field (Hartree-Fock) theory

Learning goals for this week:

- Fermions and bosons and antisymmetric/symmetric many-body wave functions
- Fock space as a new Hilbert space for many-particle systems
- Creation and annihilation operators acting on the Fock space vectors, along with their commutation relations
- Understanding how creation/annihilation operators change with transformations in the 1-particle basis
- Understanding the basic rules for calculating matrix elements; Wick's theorem to help this

The purpose of this chapter is to extend the single-particle quantum mechanics discussed in earlier quantum mechanics courses to that of many indistinguishable particles occupying the states obtained after solving the single-particle problem. At first sight this step seems quite simple, although first requiring us to write long sequences of wave functions. This step is only intermediate, as we soon find it to be much more convenient to use a simplified notation of the occupation number representation. This representation brings forth a new complete vector space, Fock space. However, it is the indistinguishability of the particles that brings forth new physics, because it requires us to consider a new symmetry, that of permuting the particles between the single-particle states. In quantum systems this has somewhat unexpected consequences: most solutions to the Schrödinger equation cannot occur in the nature! Rather, the only physically allowed solutions are those which are either completely symmetric upon permutations (bosons) or completely antisymmetric (fermions). Interestingly, this nature of the particles can be expressed through the algebra of the possible operators mapping bosonic or fermionic Fock space vectors to each other, called the creation and annihilation operators.

We start first from analysing these different possibilities for the many-body systems. Then we dwell a little more on calculating expectation values of those operators and develop a tool called Wick's theorem for this. Further, we analyse the special case of a non-interacting Fermi system (gas), which is a useful starting point for even studying interacting systems. In fact, in general the full interacting problem of a large number of particles turns out impossible to solve, which is why we need to develop approximative techniques to deal with this problem. The main technique we learn is the Hartree-Fock method, which tries to pose the interacting fermion system problem as the closest analogue noninteracting problem, but with a new definition of the fermions. The case of quantum effects in bosonic systems is deferred to the following chapter.

Let me give a reading hint for this chapter: what we do here is to overlay the many-body description on top of the single-body quantum mechanics. The latter is often denoted simply by some indices of the states. So, when reading the material, try to distinguish which part of the description is relevant for the single-particle problem, and which part for the many-particle one. In both cases we may solve a Schrödinger equation, but the details of how it is written is quite different for the two cases.

\(N\) non-identical particles

In the case of \(N\) non-identical particles, the many-body quantum mechanics is a straigthforward generalization of the one-particle case:

• Number of degrees of freedom extends from a single particle to \(N\) particles
• Observables and the related operators denote the operators separately for each particle, i.e., \[ \hat x \mapsto \hat x_1, \hat x_2, \dots \hat x_N\\ \hat p \mapsto \hat p_1, \hat p_2, \dots \hat p_N \] and so on.
• Suppose that the single-particle states have the basis \(\{|\mu\rangle\}\), where \(\mu\) are the quantum numbers of the state.
• The \(N\)-particle states then have a basis \(\{|\mu_1 \rangle |\mu_2\rangle \cdots |\mu_N\rangle\}\), which denotes a state where the \(i\)th particle occupies a state with quantum numbers \(\mu_i\). Note that the above (tensor) product between single particle states does not commute, and we cannot reorder the terms. Also note the notation here, as we will later introduce a notation \(|\mu_1 \mu_2 \cdots \mu_N\rangle\) for bosonic and fermionic states, which will in general be a different state than the above tensor product.

For example, the position-spin basis vectors for 2 particles are \[ |\vec x_1,\sigma_{1}\rangle |\vec x_2,\sigma_{2}\rangle,\quad \vec x_i \in \mathbb R^3, \sigma_{i} \in \{-s,\dots s\} \] where \(\vec x_i\) is the position of the \(i\)th particle and \(\sigma_{i}\) is the z-component of its spin. - The completeness relation includes sums and integrals over all particles. The single-particle completeness relation in position basis \[ \hat 1 = \sum_{\sigma} \int d^3 x |\vec x, \sigma\rangle \langle \vec x, \sigma |, \] generalizes to \[ \begin{aligned} \hat {\mathbb 1} &= \sum_{\sigma_{1},\dots,\sigma_{N}} \int d^3 x_1 \dots d^3 x_N |\vec x_1, \sigma_{1}\rangle \dots |\vec x_N,\sigma_{N}\rangle \langle \vec x_1, \sigma_{1}|\dots\langle\vec x_N,\sigma_{N} |\\ &= \sum_{1,2,\ldots,N} |1,2,\ldots,N\rangle \langle 1, 2,\ldots, N|, \end{aligned} \] where on the second line, we used a short-hand \[|1,2,\ldots,N\rangle = |\vec x_1, \sigma_{1}\rangle |\vec x_2, \sigma_{2}\rangle \dots |\vec x_N,\sigma_{N}\rangle\] for the position-spin basis vectors.

• The inner product between two \(N\)-particle states \(|\psi\rangle\) and \(|\phi\rangle\) is a product of the inner products between the constituent single-particle states: \[ \langle \psi | \phi \rangle = \left(\langle \mu_1 | \cdots \langle \mu_N |\right) \left( |\nu_1\rangle \cdots |\nu_N\rangle \right) = \prod_{i=1}^N \langle \mu_i| \nu_i\rangle \]
• The wave functions are written for \(N\) particles as \[ \langle 1,\ldots, N |\psi\rangle = \psi(\vec x_1,\sigma_{1};\dots;\vec x_N,\sigma_{N}) \equiv \psi(1,2,\dots,N) \] The latter short-hand notation is used also in the rest of the chapter.
• The dynamics is given by the \(N\)-particle Hamilton operator \(\hat H\), which has a position representation \[ \begin{aligned} H(1,2,\ldots,N) &\equiv H(\vec x_1,\sigma_{1};\dots;\vec x_N,\sigma_{N})\\ &= \langle 1, 2,\ldots, N | \hat H | 1, 2,\ldots, N\rangle. \end{aligned} \] More generally, the Hamiltonian could be non-local in space and non-diagonal in spin. In that case we would need to write it as a matrix \[ H(1,2,\ldots,N; 1',2',\ldots N') = \langle 1, 2,\ldots,N|\hat H | 1',2',\ldots,N'\rangle \] The matrix form of a local spin-diagonal Hamiltonian is non-zero only for the diagonal matrix elements \[ H(1,2,\dots,N; 1',2',\ldots N') = \delta(\vec x_1-\vec x_1')\delta_{\sigma_{1},\sigma_{1'}}\cdots\delta(\vec x_N-\vec x_{N'})\delta_{\sigma_{N},\sigma_{N'}} H(1,2,\dots N). \]

These steps of course increase the complication of the theory, but do not give anything qualitatively new to the description. On the other hand, when the particles are indistinguishable, we need to include this property in the description. To understand how this can be done, let us make a small detour to group theory and representations of symmetry groups in quantum mechanics.

Symmetry in quantum mechanics

Consider the unitary transformation ()

The quantum system is invariant in the transformation if the transformed states fulfill the same Schrödinger equation as the original ones:

This has the same form provided that or .

Such invariance implies a constant of motion implying:

• Eigenvalues of are constants of motion
• and share the same eigenstates
• Eigenvalues are good quantum numbers (i.e., can be used to label the eigenstates of )

Consider then a set of transformations (e.g., rotations) in some group , inducing a unitary transformation for the states

If , is a symmetry transformation, is the symmetry group of the system, and the 's form an operator-valued representation of the symmetry group .

The degenerate eigenstates of \(\hat H\) generate a representation of the group \(G\). Suppose there are \(N\) degenerate mutually orthonormal eigenstates, each with energy \(E\): \[ \hat H|n\rangle = E|n\rangle, n=1,\dots,N. \] If \(G\) is a symmetry of the system, then the energy of the transformed state is the same as the energy of the original state: \[ \hat H \hat U(g) |n\rangle = \hat U(g) \hat H |n\rangle = E \hat U(g) |n\rangle. \] In other words, the symmetry operator maps the eigenvectors to each other only within the degenerate set having the same eigenvalue of \(\hat H\). We may write in general \[ \hat U(g) |n\rangle = \sum_{m=1}^N |m\rangle D_{nm}(g), \] where the coefficients are \[ D_{nm}(g) = \langle m | \hat U(g) |n\rangle. \] These \(D_{nm}(g)\) and the corresponding matrices \(\mathbb{D}(g)\) form a representation of the group \(G\) since they satisfy the group axioms. The dimension of the representation is the dimension of the matrices, e.g. \(2\times2\) matrices give a 2-dimensional representation.

The discussion here shows how a symmetry of a system can in quantum mechanics be represented in terms of certain operators or matrices acting on an \(n\)-dimensional state space. This \(n\) is called the dimension of the representation. Below, we see that this property is important in many-body quantum mechanics, because the indistinguishability of the particles imposes a symmetry between them. We hence need to classify the particles based on how their states transform in the corresponding symmetry transformations.

\(N\) identical particles

If the \(N\) particles are identical, there is a new symmetry in the system: \[ \hat H(1,2,\dots,N) = \hat H(i_1,i_2,\dots,i_N), \] where \((i_1,i_2,\dots,i_N)\) is an arbitrary permutation of \((1,2,\dots,N)\).

For example, the Hamilton operator could be of the form \[ \hat{H}(1, \ldots, N)=\sum_{i=1}^{N}\left[\frac{\hat{p}_{i}^{2}}{2 m}+V_{1}\left(x_{i}\right)\right]+\frac{1}{2} \sum_{i \neq j} V_{2}\left(\left|x_{i}-x_{j}\right|\right), \] where every particle has the same mass \(m\), feels the same potential \(V_1\) and all pairs of particles interact via the same potential \(V_2\). This is in position basis, so \(\hat{p}_i = -i\vec\nabla_{x_i}\).

We denote a permutation with the symbol \[ P=\left(\begin{array}{llll}{1} & {2} & {\ldots} & {N} \\ {i_{1}} & {i_{2}} & {\cdots} & {i_{N}}\end{array}\right). \] A permutation \(P\) is a bijective function from the set \(\{1,2,\ldots, N\}\) to itself, such that \(1\mapsto i_1\), \(2\mapsto i_2\), ... \(N\mapsto i_N\). All possible permutations of \((1,2,\ldots, N)\) form a so-called symmetric group (or permutation group) \(S_N\).

Let us define an operator \(\hat\Pi_P\), which represents the permutation \(P\). Its action on the many-particle wavefunction is \[ \hat{\Pi}_{P} \psi(1,\ldots, N)=\psi(i_{1},\ldots, i_{N}). \] The action (of the inverse permutation) on the state vectors is \[ \hat \Pi_P^{-1} | \nu_1 \rangle \cdots |\nu_N\rangle = |\nu_{i_1}\rangle \cdots |\nu_{i_N}\rangle, \] which can be verified by comparing the wavefunctions \(\langle 1,\ldots,N | \hat\Pi_P \psi\rangle\) and \(\langle i_1,\ldots,i_N| \psi \rangle\). The effect of the permutation operator is to assign the quantum numbers of particle \(k\) to particle \(i_k\).

We apply this operator on both sides of the Schrödinger equation and get \[ i \hbar \frac{\partial}{\partial t} \hat{\Pi}_{P}\psi(1, \ldots, N)=\hat{\Pi}_{P} \hat{H}\left(1 \ldots, N \right)\, \hat{\Pi}_{P}^{-1} \hat{\Pi}_{P}\, \psi(1, \ldots, N), \] where on the right hand side we inserted the identity operator \(\hat 1 = \hat{\Pi}_{P}^{-1} \hat{\Pi}_{P}\) between \(\hat H\) and \(\psi\). This can be written as \[ i \hbar \frac{\partial}{\partial t} \psi(i_1, \ldots, i_N)=\hat{H}'\left(1 \ldots, N \right) \psi(i_1, \ldots, i_N), \] where \(\hat H'\) is defined as \[\hat H'(1,\ldots,N) = \hat{\Pi}_{P} \hat{H}\left(1 \ldots, N \right)\, \hat{\Pi}_{P}^{-1}.\] To see how the permutation operator acts on the Hamiltonian, let us consider the matrix element \[ \begin{aligned} H'(1,\dots,N) &= \langle 1,\dots N| \hat H' | 1,\dots N\rangle \\ &= \langle 1,\dots N| \hat \Pi_P \hat H \hat \Pi^{-1}_P| 1,\dots N\rangle\\ &= \langle i_1,\dots i_N| \hat H| i_{1},\dots i_{N}\rangle\\ &= H(i_1,\dots,i_N). \end{aligned} \] For a Hamiltonian describing identical particles, it follows that \[ \hat H'(1,2,\dots N)= \hat H(i_1,\ldots,i_N) = \hat H(1,\ldots,N).\]

We find that \([\hat H, \hat \Pi_P] = 0\), i.e. \(\hat H\) and \(\hat\Pi_P\) have common eigenstates. As we found out above, this implies that the eigenstates of \(\hat H\) should transform under permutations according to the irreducible representations of \(S_N\): \[ \hat{\Pi}_{P} \psi_{k}=\sum_{l} \psi_{l} D_{lk}(P), \] where \(D_{lk}(P)\) is the matrix element of \(\hat\Pi_P\) and \(\psi_l\) the eigenstates of \(\hat H\) which have the same energy as \(\psi_k\), in the representation of \(S_N\) defined by the operators \(\hat\Pi_P\).

Representations of the permutation group: Bosons and fermions

What kind of matrix representation does the permutation group have? It turns out that \(S_N\) has exactly two 1-dimensional representations:

1. \[\hat{\Pi}_{P} \psi(1 \ldots, N)=\psi(1,2, \ldots, N), \quad \forall P \in S_{N}\]
   • This is the "trivial" representation, where the wavefunction \(\psi\) is a fully symmetric under any permutation of \((1,2,\ldots,N)\).
   • Particles, whose wavefunction transforms according to this representation, are called bosons.
2. \[\hat{\Pi}_{P} \psi(1 \ldots, N)=\epsilon_P \psi(1,2, \ldots, N), \quad \forall P \in S_{N}\]
   • where \(\epsilon_P = +1,\) if \(P\) is even, i.e. if \((i_1, i_2,\ldots,i_N)\) is obtained from \((1,2,\ldots,N)\) by performing an even number of exchanges (An exchange swaps the places of two elements, e.g. \(\left(\begin{array}{llll}{1} & {2} & 3 & 4 \\ 1 & 4 & 3 & 2\end{array}\right)\))
   • \(\epsilon_P = -1,\) if \(P\) is odd.
   • In this case, the wavefunction \(\psi\) is fully antisymmetric under an exchange of any two particles \(\psi(\dots,i,\dots,j,\dots) = -\psi(\dots,j,\dots,i,\dots)\).
   • Particles, whose wavefunction transforms according to this representation of \(S_N\), are called fermions.

In addition, \(S_N\) (for \(N>2\)), has other unitary and irreducible higher-dimensional representations. Particles whose wavefunctions would transform according to these would be called "para-particles". However, these particles have not been detected.

However, they could occur as low-energy excitations in some condensed-matter systems, especially in cases where the particles are confined into one or two dimensions. In fact, the search of parafermions is at present an active field of study (see for example this review paper).

There is a connection between the spin of an particle and the permutation symmetry of the many-body wavefunction, known as the spin-statistics theorem:

• Particles, whose spin is an integer number (\(0,1,2,\dots\)) are bosons; they obey the Bose-Einstein statistics.
• Particles, whose spin is a half-integer number (\(\frac 1 2, \frac 3 2, \frac 5 2,\dots\)) are fermions; they obey the Fermi-Dirac statistics.

The proof of this theorem requires the tools of relativistic QFT.

Independent identical particles

Let us suppose we can write the Hamilton operator of a nonrelativistic system of \(N\) identical particles as \[ \hat{H}(1,2, \ldots, N)=\sum_{i=1}^{N} \hat{H}_{0}(i)+\frac{1}{2} \sum_{i \neq j} V_{2}(i, j)+\sum_{i \neq j \neq k} V_{3}(i, j, k)+\cdots, \] where \(V_2\) and \(V_3\) are symmetric functions of their arguments and \(\hat H_0(i) = (\hat{\vec p}_i)^2/(2m)+V_1(i)\) is a 1-particle operator.

Particles are independent (non-interacting) if the mutual interactions are so weak that we can approximate \(V_2\approx V_3\approx \ldots \approx 0\), or if the mutual interactions can be approximated by a "mean field".

The Hamilton operator for \(N\) identical, non-interacting, particles is \[ \hat H(1,2,\ldots,N) = \sum_{i=1}^N\hat H_0(i). \] Now suppose that \[ \hat{H}_{0}(i) \varphi_{\nu}\left(\vec{x}_{i}, \sigma_{i}\right)=E_\nu^{(0)} \varphi_\nu\left(\vec{x}_{i}, \sigma_{i}\right), \] i.e. that the 1-particle eigenvalues \(E_\nu^{(0)}\) and eigenstates \(\left|\nu\right\rangle\) of \(\hat H_0\) are known, and \(\nu\) is a set of suitable quantum numbers that describe the system; e.g. \(\nu=\{n,m_l,m_s\}\).

Suppose also that the eigenstates are normalized, \(\left\langle \nu | \nu^{\prime}\right\rangle=\delta_{\nu\nu'},\) i.e. \[ \sum_{\sigma} \int d^{3} \vec{x} \varphi_{\nu}^{*}\left(\vec{x}, \sigma\right) \varphi_{\nu}\left(\vec{x}, \sigma\right)=\delta_{\nu\nu'}, \] where \(\varphi_\nu(\vec x,s_z)=\left\langle\vec x,s_z|\nu\right\rangle\).

Since we now have noninteracting particles, the solution of the Schrödinger equation separates into a product of single-particle wavefunctions, i.e. \(\psi(1,2,\cdots,N) =\varphi_{\nu_1}(1)\varphi_{\nu_2}(2)\cdots \varphi_{\nu_N}(N)\). Substituting this ansatz to the SE, we find \[ E \psi(1,\cdots,N) = \hat H(1,\cdots,N)\psi(1,\cdots,N) \] \[ = \sum_{i=1}^N \hat H_0(i)\psi(1,\cdots,N) = \left(\sum_{i=1}^N E^{(0)}_{\nu_i} \right)\psi(1,\cdots,N), \] so that the energy is obtained as a sum of single-particle energies, i.e. \(E=E^{(0)}_{\nu_1} + E^{(0)}_{\nu_2} + \cdots + E^{(0)}_{\nu_N}\). Since \(E\) is the same for any permutation \(\psi(i_1,\cdots i_N)\), its degree of degeneracy is \(N!\) .

Bosons: fully symmetric wave function

The bosonic wave function is of the form \[\psi_{\nu_{1} \ldots \nu_{N}}(1,2, \ldots, N)=\mathcal{N} \sum_{P \in S_{N}} \varphi_{\nu_{1}}(i_1) \varphi_{\nu_{2}}\left(i_{2}\right) \cdots \varphi_{\nu_{N}}\left(i_{N}\right)\] where \[ \mathcal{N} = \frac{1}{\sqrt{N!} \sqrt{n_1!}\sqrt{n_2!} \cdots \sqrt{n_N!}} \] is a normalization factor and \(n_i\) is the number of particles in state \(i\), i.e., the number of times the state \(\varphi_i\) appears in the product. Obviously \(\sum_i n_i = N\).

The above is in position basis. In Dirac notation the symmetrized state vector corresponding to the above wavefunction is \[ |\nu_1\nu_2\dots \nu_N\rangle = \mathcal N \sum_{P\in S_N} |\nu_{i_1} \rangle \cdots |\nu_{i_N} \rangle. \]

The examples above show that labelling the \(N\)-particle states as \[ |\Psi\rangle = |\nu_1 \nu_1 \nu_1 \cdots \nu_1 \nu_2 \cdots \nu_2 \nu_3 \nu_4 \cdot \nu_4\rangle \] becomes soon tedious. Therefore, it is easier to use the Occupation number representation in the Fock space: \[|\Psi\rangle = |n_{\nu_1} n_{\nu_2} n_{\nu_3} \cdots \rangle,\] where \(n_{\nu_i}=\) how many times the one-particle state \(|\nu_1\rangle\) appears in \(|\Psi\rangle\). This number \(n_{\nu_i} =0,1,2,3,\dots\) for bosons.

Note that the number of particles is \(N=\sum_i n_i\) and the eigenenergy of the many-particle state \(|\Psi\rangle\) can be found by \(E=\sum_{i=1}^\infty n_{\nu_i} E_{\nu_i}^0\).

Fermions: fully antisymmetric wave function

Antisymmetrizing the product \(\varphi_{\nu_1}(1)\cdots\varphi_{\nu_N}(N)\) for an arbitrary exchange \(i\leftrightarrow j\) gives \[ \psi_{\nu_{1} \ldots \nu_{N}}(1,2, \ldots, N) = \mathcal N \sum_{P \in S_{N}} \epsilon_{P} \varphi_{\nu_{1}}(i_{1}) \varphi_{\nu_{2}}(i_{2}) \cdots \varphi_{\nu_{N}}(i_N) \] \[ = \frac{1}{\sqrt{N !}}\left|\begin{array}{ccc} \varphi_{\nu_1}(1) & \varphi_{\nu_1}(2) & \cdots & \varphi_{\nu_1}(N) \\ \varphi_{\nu_2}(1) & \varphi_{\nu_2}(2) & \cdots & \varphi_{\nu_2}(N) \\ {\vdots} & {\vdots} & & {\vdots} \\ \varphi_{\nu_N}(1) & \varphi_{\nu_N}(2) & \cdots & \varphi_{\nu_N}(N) \end{array}\right|. \] This kind of wavefunction is known as Slater determinant. The corresponding Hilbert space vector in Dirac notation is \[ |\nu_1\dots \nu_N\rangle = \mathcal N \sum_{P\in S_N} \epsilon_P |\nu_{i_1} \rangle \cdots |\nu_{i_N} \rangle. \] The normalization constant \(\mathcal N=1/\sqrt{N!}\) guarantees again \(\left\langle\nu_1\ldots\nu_N|\nu_1\ldots\nu_N\right\rangle=1\).

Unlike in the bosonic case, here the order of \(\nu_i\)'s does matter. \(\ket{\nu_1 \nu_2}\) and \(\ket{\nu_2 \nu_1}\) represent the same state, but with a phase difference of \(\pi\): \[ |\nu_1 \nu_2\rangle = - |\nu_2\nu_1\rangle. \]

Recall here the definition of a determinant of an \(N\times N\) matrix M: \[\det M = \sum_{P\in S_N} \epsilon_P(i_1,\ldots,i_N) M_{1,i_1} M_{2,i_2} \cdots M_{N,i_N} = \sum_{P\in S_N} \epsilon_P(i_1,\ldots,i_N) M_{i_1,1} M_{i_2,2} \cdots M_{i_N,N}\]

From this definition it follows that \(\det(\varphi_{\nu_i}(j))=0\) if two rows are identical, i.e. the wavefunction vanishes, \(\psi_{\nu_1\ldots\nu_N}(1,\ldots,N)=0\), if two fermions share the same state (\(\nu_i=\nu_j\) for some pair \((i,j)\)).

This property of the fermions is commonly stated as Pauli exclusion principle
Two identical fermions cannot have identical quantum numbers, i.e. two identical fermions do not appear in the same quantum state.

\(\det(\varphi_{\nu_i}(j))=0\) also if two columns are identical, in other words \(\psi_{\nu_1\ldots\nu_N}(1,\ldots,N)=0\) if two identical fermions in different 1-particle states \(\nu_i\neq \nu_j\) have the same coordinates \((i)=(j)\), e.g. \(\vec x_i=\vec x_j\) and \(s_{iz}=s_{jz}.\)

This means that there is an effective repulsion force between identical fermions, even without interactions (\(V_2=V_3=\ldots=0\))!

Also in the identical-fermion systems, we use the occupation number representation in the Fock space: \[|\psi\rangle=\left|n_{\nu_1}, n_{\nu_2}, n_{\nu_3}, \cdots\right\rangle,\] where \(n_{\nu_i}=0\) or \(1\) for fermions. The total number of particles and the energy are \[ \sum_{i=1}^\infty n_{\nu_i} = N \quad \text{and}\quad \sum_{i=1}^\infty n_{\nu_i} E^{(0)}_{\nu_i} = E. \]

Examples of antisymmetric wavefunctions of \(N\) identical fermions

Ground state of helium

A crude model for helium is given by the Hamiltonian \[ \hat H(1,2) = \hat H_0(1) + \hat H_0(2), \] where the 1-particle Hamiltonians are \[ \hat H_0(i) = \frac{(\hat{\vec p}_i)^2}{2m_e}-\frac{Ze^2}{4\pi\varepsilon_0 r_i}, \] where \(Z=2\), or less if we use a variational approach (see QMI). The single-particle Hamiltonians are familiar from the hydrogen atom.

1-particle solutions are known: \[ \begin{array}{l} {\hat{H}_{0}(1) \varphi_{nlm}(\vec{r}_1)=\varepsilon_{n}^{(0)} \varphi_{nlm}(\vec{r}_1)},\\ {\hat{H}_{0}(2) \varphi_{nlm}(\vec{r}_2)=\varepsilon_{n}^{(0)} \varphi_{nlm}(\vec{r}_2)}. \end{array} \] The spatial part of the many-particle wavefunction consists of \(\varphi_{nlm}(\vec r_1)\varphi_{n'l'm'}(\vec r_2)\), and the energy is \(E=\varepsilon_n^{(0)}+\varepsilon_{n'}^{(0)}\). There is no spin in \(\hat H_0(i)\), so the different spin states are degenerate.

Consider then the lowest possible energy (the ground state) of the system, and set \(n=n'=1, l=l'=0, m=m'=0\). We see that the spatial part \[ \psi(\vec r_1,\vec r_2)=\varphi_{100}(\vec r_1)\varphi_{100}(\vec r_2) \] is symmetric in the exchange \(1\leftrightarrow 2\). The full wavefunction \[ \Psi_{\rm GS}(1,2) = \psi(\vec r_1,\vec r_2)\chi_S(1,2) \] on the other hand is antisymmetric. Thus the spin-part \(\chi_S\) of the wavefunction must be antisymmetric. As we find out in the box below, only the singlet state qualifies.

We denote the singlet state by where .

The ground state wavefunction is thus \[ \Psi^{\rm GS}_{\nu_1,\nu_2} (1,2) = \psi(\vec r_1,\vec r_2) \chi_{S=0}(1,2) \\%= \varphi_{100}(\vec r_1)\varphi_{100}(\vec r_2) \chi_{S=0} \] \[ =\frac{1}{\sqrt{2}} \underbrace{\varphi_{100}(\vec r_1) \chi_+(1)}_{\varphi_{\nu_1}(1)} \;\cdot\; \underbrace{\varphi_{100}(\vec r_2) \chi_-(2)}_{\varphi_{\nu_2}(2)} \] \[ +\frac{1}{\sqrt{2}} \underbrace{\varphi_{100}(\vec r_1) \chi_+(1)}_{\varphi_{\nu_1}(2)} \;\cdot\; \underbrace{\varphi_{100}(\vec r_2) \chi_-(2)}_{\varphi_{\nu_2}(1)} \] \[ = \frac{1}{\sqrt{2}}\left|\begin{array}{ll}{\varphi_{\nu_{1}}(1)} & {\varphi_{\nu_{1}}(2)} \\ {\varphi_{\nu_{2}}(1)} & {\varphi_{\nu_{2}}(2)}\end{array}\right|, \] which is a Slater determinant, and \(\nu_1\) and \(\nu_2\) are single-particle states with quantum numbers \(n=1\), \(l=m=0\) and \(s_z = \pm\frac 1 2\).

Fock space

In many-particle systems we might want to consider cases where the particle number \(N\) can change --- such as creation and annihilation of particles in relativistic quantum field theory, particle-hole creation or phonons in condensed-matter physics, or quantization of electromagnetic field (with a different number of photons), one should consider all \(N\)-particle sectors simultaneously.

This means that we should generalize the above considerations and introduce the Fock space \(\mathcal{H}\): \[ \mathcal{H} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \mathcal{H}_2 \oplus \cdots, \] where \(\mathcal{H}_N\) denotes the \(N\)-particle Hilbert space.

In the occupation number representation these spaces can be specified as follows

• \(N=0\): \(\mathcal H_0\) consists of the single vector \(|0000\dots \rangle \equiv |0\rangle\), i.e., the vacuum state, with no particles. This vacuum state is normalized: \(\langle 0 | 0\rangle = 1\).

• \(N=1\): \(\mathcal H_1\) consists of an infinite number of vectors \(|10000\dots\rangle\), \(|01000\dots\rangle\), \(|00100\dots\rangle\), and so on. In other words, there is one particle on one of the possible one-particle eigenstates, and no particles on the others.

• \(N=2\): \(\mathcal H_2\) consists of the vectors \[ \begin{aligned} |110000\dots\rangle, |101000\dots\rangle, |011000\dots\rangle, \dots\\ |200000\dots\rangle, |02000\dots\rangle, |002000\dots\rangle, \dots, \end{aligned} \] where the states of the type in the second line is possible only for bosons.

In the case of \(\mathcal H_N\) we hence have the states of the form \[ |n_{\nu_1} n_{\nu_2} n_{\nu_3} \dots\rangle \in \mathcal H_N, \sum_i n_{\nu_i} = N \]

For a generic Fock space vector \(|n_{\nu_1} n_{\nu_2} n_{\nu_3}\dots\rangle \in \mathcal H\), we can however lift the requirement of the number of particles, since it can vary.

We can choose the basis \(\{|\nu\rangle\}\) for the one-particle states so that it is orthonormal and complete, i.e., \[ \langle \nu |\nu'\rangle = \delta_{\nu \nu'}, \quad \hat 1 = \sum_\nu |\nu\rangle \langle \nu|. \] Similarly we can write for the Fock states \[ \begin{aligned} &\left\langle n_{\nu}^{\prime} n_{\nu_{2}}^{\prime} \cdots | n_{\nu_{1}} n_{\nu_{2}} \cdots\right\rangle=\delta_{n_{\nu_1}, n_{\nu_2}} \delta_{n_{\nu_1}, n_{\nu_2}} \cdots &\text{ orthonormality}\\ &\hat{1}=\sum_{n_{\nu_1} n_{\nu_2} \cdots}\left|n_{\nu_1}, n_{\nu_{2}} \cdots\right\rangle\left\langle n_{\nu_1}, n_{\nu_{2}} \cdots\right| &\text{ completeness} \end{aligned} \] In addition, the states corresponding to a different total number of particles are orthogonal, \[ {}_{N'}\langle \{n_\nu'\}|\{n_\nu\}\rangle_N = 0 \text{ if } N\neq N' \] This means that an inner product of Fock space states \(|\Phi\rangle = \sum_N c_N |\phi\rangle_N \in \mathcal H\) and \(|\Psi\rangle = \sum_N d_N |\psi\rangle_N \in \mathcal H\), where \(|\phi\rangle_N, |\psi\rangle_N \in \mathcal H_N\), becomes \[ \langle \Phi|\Psi\rangle = \sum_{N,M} c_N^* d_M \enspace {}_N\langle \phi|\psi\rangle_M = \sum_N c_N^* d_N \enspace {}_N\langle \phi|\psi\rangle_N. \]

The occupation number states of the form \(|\{n_\nu\}\rangle\) hence form an orthonormal basis for the Fock space \(\mathcal H\).

Bosonic system

Let us now discuss the operators acting on the Fock space vectors. In fact, it is enough to define two types of operators: the annihilation operator removing a particle from a certain one-particle state, and its conjugate operator, creation operator, that creates a particle in a given one-particle state.

Bosonic annihilation operator
\[ b_\nu |n_{\nu_1} n_{\nu_2} \dots n_\nu \dots\rangle = \sqrt{n_\nu} |n_{\nu_1} n_{\nu_2} \dots n_\nu-1 \dots \rangle \] In other words, \(b_\nu\) removes one particle from state \(\nu\). There are as many such operators as there are one-particle states considered in the problem.

The above then also implies that \[ b_\nu |0\rangle = 0, \] since there is nothing to be annihilated in a vacuum state.

Note that you have encountered this type of an operator already before: in the discussion of the harmonic oscillator, where the states \(|n\rangle\) correspond to its excitation states, and the annihilation operator thus couples the state \(|n\rangle\) with state \(|n-1\rangle\).

In quantum mechanics, operators are typically defined via their matrix elements. From the above definition, the matrix element of \(b_\nu\) is \[ \left\langle n_{\nu_1}^{\prime} n_{\nu_2}^{\prime} \dots n_{\nu}^{\prime} \dots\left|b_{0}\right| n_{\nu_1}, n_{\nu_2}, \dots, n_{\nu}, \dots\right\rangle=\sqrt{n_{\nu}} \delta_{n_{\nu}^{\prime}, n_{\nu}-1} \prod_{\nu_k \neq \nu} \delta_{n_{\nu_{k}^{\prime}}n_{\nu_{k}}} \]

This is a complex number (or in fact a real number in this case). Let us take its complex conjugate. We get \[ \langle n_{\nu_1} n_{\nu_2}\dots n_\nu \dots |b_\nu^\dagger|n_{\nu_1}' n_{\nu_2}' \dots n_\nu'\dots\rangle = \sqrt{n_{\nu'}+1} \delta_{n_{\nu}'+1,n_\nu} \prod_{\nu_k \neq \nu} \delta_{n_{\nu_k'} n_{\nu_k}} \] This defines the Bosonic creation operator
\[ b_\nu^\dagger |n_{\nu_1} n_{\nu_2} \dots n_\nu \dots\rangle = \sqrt{n_{\nu}+1}|n_{\nu_1} n_{\nu_2} \dots n_{nu}+1\dots\rangle. \] which adds the occupation of state \(|\nu\rangle\) by one.

Let us check what happens if we first create a particle in state \(\mu\) and then annihilate one from state \(\nu\neq \mu\).

We find that the bosonic operators satisfy the Bosonic commutation relations \[ [b_\mu,b_\nu^\dagger]=b_\mu b_\nu^\dagger-b_\nu^\dagger b_\mu = \delta_{\mu\nu} \] and \[ [b_\mu,b_\nu]=[b_\mu^\dagger,b_\nu^\dagger]=0. \] The first of these is proven in the above collapsible region. The second is straightforward to show with a similar manner.

Note that we can now start from the vacuum and generate all states \(|\{n_\nu\}\rangle\) by operating with the creation operator multiple times: \[ \begin{aligned} b_\nu^\dagger |0\rangle &= \sqrt{1}|0\dots 1_\nu \dots\rangle\\ (b_\nu^\dagger)^2 |0\rangle &= \sqrt{1}\sqrt{2} |0\dots 2_\nu \dots \rangle\\ \dots\\ (b_\nu^\dagger)^{n_\nu} |0\rangle &= \sqrt{n_\nu!} |0\dots n_\nu \dots \rangle\\ b_\mu^\dagger |0\dots n_\nu \dots 0_\mu \dots\rangle &= \sqrt{1} |0\dots n_\nu \dots 1_\mu \dots \rangle\\ \dots\\ (b_\mu^\dagger)^{n_\mu} |0\dots n_\nu \dots 0_\mu \dots\rangle &= \sqrt{n_\mu!} |0\dots n_\nu \dots n_\mu \dots \rangle \end{aligned} \] Continuing this, we find that a given Fock state specified by the set of occupation numbers \(|\{n_\nu\}\rangle\) can be defined by \[ \boxed{|n_{\nu_1} n_{\nu_2} \dots n_\nu \dots \rangle = \frac{1}{\sqrt{n_{\nu_1}! n_{\nu_2}! \cdots n_{\nu}!}} (b_{\nu_1}^\dagger)^{n_{\nu_1}} (b_{\nu_2}^\dagger)^{n_{\nu_2}} \cdots (b_{\nu}^\dagger)^{n_{\nu}} \dots |0\rangle}. \]

During the derivation of the bosonic commutation relations, we also found that the Fock states are eigenstates of the operator \(N_\nu = b_\nu^\dagger b_\nu\) with the eigenvalue \(n_\nu\). In other words, this operator counts the number of particles in the one-particle state \(|\nu\rangle\): Particle number operator \(\hat n_\nu=b_\nu^\dagger b_\nu\) \[ \hat n_\nu |n_{\nu_1} n_{\nu_2} \dots n_\nu \dots \rangle = n_\nu |n_{\nu_1} n_{\nu_2} \dots n_\nu \dots \rangle. \] The total particle number operator is then \[ \hat N=\sum_\nu \hat n_\nu = \sum_\nu b_\nu^\dagger b_\nu. \]

Fermionic system

The state vector in a fermionic system can be presented in terms of the Slater determinant, \[ \psi_{\nu_{1} \ldots \nu_{N}}(1,2, \ldots, N) = \mathcal N \sum_{P \in S_{N}} \epsilon_{P} \varphi_{\nu_{1}}(i_{1}) \varphi_{\nu_{2}}(i_{2}) \cdots \varphi_{\nu_{N}}(i_N) \] \[ = \frac{1}{\sqrt{N !}}\left|\begin{array}{ccc} \varphi_{\nu_1}(1) & \varphi_{\nu_1}(2) & \cdots & \varphi_{\nu_1}(N) \\ \varphi_{\nu_2}(1) & \varphi_{\nu_2}(2) & \cdots & \varphi_{\nu_2}(N) \\ {\vdots} & {\vdots} & & {\vdots} \\ \varphi_{\nu_N}(1) & \varphi_{\nu_N}(2) & \cdots & \varphi_{\nu_N}(N) \end{array}\right|\\ = \langle \vec x_1,\sigma_1;\dots \vec x_N,\sigma_N|1_{\nu_1} 1_{\nu_2} \dots 1_{\nu_N}\rangle, \] so that the occupation number representation (the last line) contains all those states involved in the Slater determinant.

The state vector has the property that it changes sign if the indices of any pair of states is changed, i.e., if for example \(\nu_1 \leftrightarrow \nu_2\). This means that we have to fix the order in which the one-particle states or their occupation number appear. The precise way of fixing the order does not matter, as long as it stays consistent throughout the description.

Let us define the Fermionic creation operator \[ a_\nu^\dagger |n_1 n_2 \dots 0_\nu\dots \rangle = (-1)^{\sum_{\mu=1}^{\nu-1} n_\mu} |n_1 n_2 \dots 1_\nu \dots \rangle. \] and \[ a_\nu^\dagger |n_1 n_2 \dots 1_\nu \dots \rangle = 0 \] The latter property is needed to make sure that we cannot create two fermions in the same single-particle state. It also implies \[ (a_\nu^\dagger)^2 = 0. \]

Let us create now two particles in different single-particle states \(\mu\) and \(\nu\) so that \(\nu\) is further right from \(\mu\): \[ a_\mu^\dagger a_\nu^\dagger |n_1 n_2 \dots 0_\mu \dots 0_\nu \dots \rangle = a_\mu^\dagger (-1)^{\sum_{\alpha=1}^{\nu-1} n_\alpha}|n_1 n_2 \dots 0_\mu \dots 1_\nu \dots \rangle \\ = (-1)^{2\sum_{\alpha=1}^{\mu-1} n_\alpha + \sum_{\alpha=\mu+1}^{\nu-1} n_\alpha} |n_1 n_2 \dots 1_\mu \dots 1_\nu \dots \rangle. \] Note that this uses the fact that in the first sum \(n_\mu=0\). We can do it also in the opposite order: \[ a_\nu^\dagger a_\mu^\dagger |n_1 n_2 \dots 0_\mu \dots 0_\nu \dots \rangle = a_\nu^\dagger (-1)^{\sum_{\alpha=1}^{\mu-1} n_\alpha}|n_1 n_2 \dots 1_\mu \dots 0_\nu \dots \rangle \\ = (-1)^{2\sum_{\alpha=1}^{\mu-1} n_\alpha +1+ \sum_{\alpha=\mu+1}^{\nu-1} n_\alpha} |n_1 n_2 \dots 1_\mu \dots 1_\nu \dots \rangle. \] We hence get an extra "1" in the exponent, because the latter expression assumes \(n_\mu=1\). Because of that, the latter result is \((-1)\) times the previous.

Adding these results together hence yields \[ (a_\mu^\dagger a_\nu^\dagger + a_\nu^\dagger a_\mu^\dagger)|n_1 n_2 \dots 0_\mu \dots 0_\nu\rangle = 0. \] If either of the states is already occupied, the result is zero no matter which order these operators operate. The result is hence valid for an arbitrary Fock state. Since we required also that \((a_\nu^\dagger)^2=0\), we can write \[ \boxed{\{a_\mu^\dagger,a_\nu^\dagger\} \equiv (a_\mu^\dagger a_\nu^\dagger + a_\nu^\dagger a_\mu^\dagger) = 0} \] Here we defined the anticommutator between two operators \(A\) and \(B\): \(\{A,B\} \equiv AB+BA\).

Correspondingly, we can define the Fermion annihilation operator \(a_\nu\) \[ \begin{aligned} a_\nu |n_1 n_2 \dots 1_\nu \dots \rangle &= (-1)^{\sum_{\mu=1}^{\nu-1} n_\mu} |n_1 n_2 \dots 0_\nu \dots |\rangle\\ a_\nu |n_1 n_2 \dots 0_\nu \dots \rangle &= 0, \text{ i.e., } a_\nu^2=0 \end{aligned} \] In the same manner as for the creation operators, we can get the anticommutation relations for \(a_\nu\) and \(a_\mu\): \[ \boxed{\{a_\mu,a_\nu\}=0\}} \] Alternatively, this could be obtained by taking the conjugate of the matrix elements of \(\{a_\nu^\dagger,a\mu^\dagger\}\).

What about the anticommutator \(\{a_\mu^\dagger,a_\nu\}\)? Let us do it explicitely. First, with \(\mu < \nu\): \[ a_\mu^\dagger a_\nu |n_1 n_2 \dots 0_\mu \dots 1_\nu \dots \rangle = (-1)^{2\sum_{\alpha=1}^{\mu-1} n_\alpha + \sum_{\alpha=\mu+1}^{\nu-1} n_\alpha} |n_1 n_2 \dots 1_\mu \dots 0_\nu\dots \rangle \] and \[ a_\nu a_\mu^\dagger |n_1 n_2 \dots 0_\mu \dots 1_\nu \dots \rangle = (-1)^{2\sum_{\alpha=1}^{\mu-1} n_\alpha + 1+\sum_{\alpha=\mu+1}^{\nu-1} n_\alpha} |n_1 n_2 \dots 1_\mu \dots 0_\nu\dots \rangle. \] In other words, we get the same anticommutation relation as above, and it is easy to see (for example by taking the conjugate matrix element) that this holds also for \(\mu > \nu:\) \[ \boxed{\{a_\nu,a_\mu^\dagger\}=0, \text{ for }\nu \neq \mu}. \] On the other hand, if \(\nu=\mu\), we get \[ \begin{aligned} a_\mu^\dagger a_\mu |n_1 n_2 \dots 1_\mu \dots \rangle &= \underbrace{(-1)^{2\sum_{\alpha=1}^{\mu-1} n_\alpha}}_{=+1 = n_\mu} |n_1 n_2 \dots 1_\mu \dots \rangle = 1 |n_1 n_2 \dots 1_\mu \dots \rangle\\ a_\mu a_\mu^\dagger |n_1 n_2 \dots 1_\mu \dots \rangle &= 0 = n_\mu |n_1 n_2 \dots 1_\mu \dots \rangle. \end{aligned} \] We thus found the Fermion number operator \[ \hat n_\mu = a_\mu^\dagger a_\mu \] for which \(\hat n_\mu |n_1 n_2 \dots n_\mu \dots \rangle = n_\mu |n_1 n_2 \dots n_\mu \dots \rangle\).

On the other hand, \[ a_\mu a_\mu^\dagger |n_1 n_2 \dots 0_\mu \dots \rangle = \underbrace{(+1)}_{n_\mu} |n_1 n_2 \dots 0_\mu \dots \rangle\\ a_\mu a_\mu^\dagger |n_1 n_2 \dots 1_\mu \dots \rangle = 0 = 1-n_\mu. \] Since the operator \(a_\mu a_\mu^\dagger\) connects the number states to themselves (i.e., it has only diagonal matrix elements), and has the eigenvalue \(1-n_\mu\), we can identify it with \[ a_\mu a_\mu^\dagger = 1-n_\mu. \] Putting this together with the definition of the fermion number operator, we get the remaining Fermion anticommutation relations \[ \{a_\nu,a_\mu^\dagger\}= a_\nu a_\mu^\dagger +a_\mu^\dagger a_\nu = \delta_{\nu \mu}. \]

Let us then create the fermionic state \(|n_1 n_2 n_3 \dots\rangle\) from the vacuum state \(|0\rangle\) by operating multiple times with the creation operators of the different single-particle states. Consider for example the state \(|1_1 0_2 1_3 1_4 0_5 0_6 \dots\rangle\), i.e., where all states above the 4th state are empty. To get there, we start from the rightmost occupied single-particle state: \[ \begin{aligned} a_4^\dagger|0\rangle &= (-1)^{0+0+0} |0_1 0_2 0_3 1_4 0_5 \dots\rangle\\ a_3^\dagger |0_1 0_2 0_3 1_4 0_5 \dots\rangle &= (-1)^{0+0} |0_1 0_2 1_3 1_4 0_5 \dots\rangle\\ a_1^\dagger |0_1 0_2 1_3 1_4 0_5 \dots\rangle &= |1_1 0_2 1_3 1_4 0_5 \dots\rangle\\ &= a_1^\dagger a_3^\dagger a_4^\dagger |0\rangle. \end{aligned} \] This we can generalize to an arbitrary fermion state Constructing a fermion state from vacuum \[ |n_1 n_2 n_3 \dots\rangle = (a_1^\dagger)^{n_1} (a_2^\dagger)^{n_2} (a_3^\dagger)^{n_3}\cdots |0\rangle \] with \(n_i \in \{0,1\}\). The order of the operators is important!

Analogous to the bosonic state, the total fermion number operator is \[ \hat N=\sum_\mu \hat n_\mu \] Notice how the fermionic nature of the particles (states) lies in the definition of the creation and annihilation operators, or equivalently in their algebra, originating from their anticommutation relations!

Let us then create the fermionic state \(|n_1 n_2 n_3 \dots\rangle\) from the vacuum state \(|0\rangle\) by operating multiple times with the creation operators of the different single-particle states. Consider for example the state \(|1_1 0_2 1_3 1_4 0_5 0_6 \dots\rangle\), i.e., where all states above the 4th state are empty. To get there, we start from the rightmost occupied single-particle state: \[ \begin{aligned} a_4^\dagger|0\rangle &= (-1)^{0+0+0} |0_1 0_2 0_3 1_4 0_5 \dots\rangle\\ a_3^\dagger |0_1 0_2 0_3 1_4 0_5 \dots\rangle &= (-1)^{0+0} |0_1 0_2 1_3 1_4 0_5 \dots\rangle\\ a_1^\dagger |0_1 0_2 1_3 1_4 0_5 \dots\rangle &= |1_1 0_2 1_3 1_4 0_5 \dots\rangle\\ &= a_1^\dagger a_3^\dagger a_4^\dagger |0\rangle. \end{aligned} \] This we can generalize to an arbitrary fermion state Constructing a fermion state from vacuum \[ |n_1 n_2 n_3 \dots\rangle = (a_1^\dagger)^{n_1} (a_2^\dagger)^{n_2} (a_3^\dagger)^{n_3}\cdots |0\rangle \] with \(n_i \in \{0,1\}\). The order of the operators is important!

Analogous to the bosonic state, the total fermion number operator is \[ \hat N=\sum_\mu \hat n_\mu \] Notice how the fermionic nature of the particles (states) lies in the definition of the creation and annihilation operators, or equivalently in their algebra, originating from their anticommutation relations!

Second quantization and field operators

Consider a change of basis for 1-particle states from one basis to another basis . For fermions, we define and .

Now, which is equivalent to Therefore, we have This and the following holds similarly for bosonic operators.

Take then \(\ket{\lambda} = \ket{\vec x, \sigma}\) with \(\sigma \equiv S_z\) for brevity and interpret \(a^\dagger_{\vec x, \sigma}\) as an operator which creates a fermion at \(\vec x\) with \(S_z = \sigma\).

We now denote \[ a^\dagger_{\vec x, \sigma} = \sum_\mu \underbrace{\langle\mu|\vec x, \sigma\rangle}_{\varphi^*_\mu(\vec x, \sigma) \in \mathbb{C}; \text{WF}} a_\mu^\dagger = \sum_\mu \varphi_\mu^*(\vec x,\sigma) a_\mu^\dagger\equiv \Psi^\dagger(\vec x, \sigma) \] and say that \(\Psi^\dagger(\vec x,\sigma)\) is a field operator.

Correspondingly, for the annihilation operator \[ a_{\vec x, \sigma} = \sum_\mu \langle\vec x, \sigma|\mu\rangle a_\mu = \sum_\mu \varphi_\mu(\vec x, \sigma) a_\mu \equiv \Psi(\vec x, \sigma). \] \(\Psi(\vec x, \sigma)\) is a field operator that destroys a fermion at \(\vec x\) with \(S_z = \sigma\).

This looks as if we have replaced the wave function by an operator, by interpreting the coefficients of the basis-WF \(\varphi_\mu\)'s as operators \(a_\mu\). Such a procedure is of fundamental importance in e.g. relativistic QM and the theory of open many-particle Q-systems. This procedure is called the second quantization, in reference to "the first quantization" where one relates physical observables to Hermitian operators.

Notice, however, that now, for nonrelativistic theory, we have not done anything new but basically just rewritten \(a^\dagger_{\vec x, \sigma} = \Psi^\dagger(\vec x, \sigma)\) and its Hermitian conjugate.

Using the orthogonality of \(\varphi_\mu\)'s, \(\sum_\sigma \int d^3 x \varphi_\mu^*(\vec x,\sigma)\varphi_v(\vec x,\sigma) = \delta_{\mu v}\) we obtain \[ a^\dagger_\mu = \sum_\sigma \int d^3 x \varphi_\mu(\vec x,\sigma)\Psi^\dagger(\vec x,\sigma),\\ a_\mu = \sum_\sigma \int d^3 x \varphi^*_\mu(\vec x,\sigma)\Psi(\vec x,\sigma). \]

The anticommutation rules remain as before \[ \begin{aligned} \{ a_\lambda, a^\dagger_{\lambda'} \} &= \{\sum_\mu \langle\lambda|\mu\rangle a_\mu, \sum_{\mu'} \langle\mu'|\lambda'\rangle a^\dagger_{\mu'} \} \\ &= \sum_{\mu,\mu'} \langle\lambda|\mu\rangle\langle\mu'|\lambda'\rangle \underbrace{\{ a_\mu, a^\dagger_{\mu'}\}}_{=\delta_{\mu \mu'}} \,= \bra{\lambda} \underbrace{(\sum_\mu \ket{\mu} \bra{\mu})}_{= \hat 1}\ket{\lambda} \\ &= \delta_{\lambda \lambda'} \end{aligned} \] which is expected; the same rules hold in the new basis.

For field operators, we get similarly (using \(\varphi_\mu(\vec x, \sigma) = \langle \vec x,\sigma |\mu \rangle\)) \[ \begin{aligned} \{ \Psi(\vec x, \sigma), \Psi^\dagger(\vec x', \sigma') \} &= \sum_{\mu,\mu'} \varphi_\mu(\vec x, \sigma) \varphi_{\mu'}^*(\vec x', \sigma') \{ a_\mu, a^\dagger_{\mu'} \} \\ &= \langle \vec x, \sigma |\vec x', \sigma' \rangle = \delta^{(3)}(\vec x - \vec x')\delta_{\sigma \sigma'}, \end{aligned} \] which leads to: Anticommutation rules for fermionic field operators \[ \begin{aligned} \{ \Psi(\vec x, \sigma), \Psi^\dagger(\vec x', \sigma') \} &= \delta^{(3)}(\vec x - \vec x')\delta_{\sigma \sigma'}\\ \{ \Psi(\vec x, \sigma), \Psi(\vec x', \sigma') \} &= 0\\ \{ \Psi^\dagger(\vec x, \sigma), \Psi^\dagger(\vec x', \sigma') \} &= 0\\ \end{aligned} \]

For bosonic operators, just replace \(\{,\} \rightarrow [,]\), i.e., use the commutation rules \([b_\mu,b^\dagger_{\mu'}] = \delta_{\mu \mu'}, [b_\mu,b_{\mu'}] = [b^\dagger_\mu,b^\dagger_{\mu'}] = 0\).

Now, we can write the many-particle states with the help of the field operators. We again take the examples from fermionic systems.

For a 1-particle state, we have simply \[ \begin{aligned} \Psi^\dagger(\vec x, \sigma)\ket{0} &= \sum_\mu \varphi^*_\mu(\vec x, \sigma) a^\dagger_\mu\ket{0} = \sum_\mu \varphi^*_\mu(\vec x, \sigma) \ket{\mu} = \ket{\vec x, \sigma} \sum_\mu \ket{\mu}\bra{\mu}\\ &= \ket{\vec x, \sigma} = a^\dagger_{\vec x, \sigma}\ket{0} \end{aligned} \] and for 2-particle states \[ \Psi^\dagger(\vec x_1,\sigma_1) \Psi^\dagger(\vec x_2,\sigma_2) \ket{0} = \ket{\vec x_1,\sigma_1 ; \vec x_2,\sigma_2}. \] This idea generalizes to N-particle states as \[ \Psi^\dagger(\vec x_1, \sigma_1) \Psi^\dagger(\vec x_2, \sigma_2) \dots \Psi^\dagger(\vec x_N, \sigma_N)\ket{0} = \ket{\vec x_1, \sigma_1; \vec x_2, \sigma_2; \dots; \vec x_N, \sigma_N} \]

Wick's theorem

Often one needs to compute matrix elements in the vacuum, vacuum expectation values, of an operator \(O=O(a,a^\dagger)\); such as \(a_{\mu'} a_{\mu} a_\nu^\dagger a_{\nu'}^\dagger\) above. In general, \(O\) can be quite complicated (long product), and the calculation becomes tedious. Wick's theorem helps to cope with this in a systematic manner.

First we need to define some notation. Let \(A_i \in \{a_\mu,a^\dagger_\mu\}\) (fermions) or \(A_i \in \{b_\mu,b^\dagger_\mu\}\) (bosons). We denote a normal-ordered product of operators as \[ :\! A_1 A_2\cdots A_n\!:\; \] and define it as a product where all the creation operators \(a^\dag_\mu\) (or \(b^\dag_\mu\)) have been brought to the left and the annihilation operators \(a_\mu\) (\(b_\mu\)) to the right. If \(A_i\)'s are fermionic, the product is multiplied with \(\epsilon_P = (-1)^{n_P}\), where \(n_P\) is the number of exchanges needed for bringing the \(A_1 A_2 \cdots A_n\) to the normal order.

Examples

• \(:\!b_\mu b_\nu^\dagger\!:\; = b_\nu^\dagger b_\mu\) (bosonic)
• \(:\!a_\mu a_\nu^\dagger\!:\; = -a_\nu^\dagger a_\mu\) (fermionic)
• Let \(A_1=c_1 a_1 + c_2 a_2^\dagger\) and \(A_2=c_3 a_3 + c_4 a_4^\dagger\), where \(c_i\)'s are complex numbers. Then \[ \begin{aligned} :\!A_1 A_2\!: &=\; :\!(c_1 a_1 + c_2 a_2^\dagger)(c_3 a_3 + c_4 a_4^\dagger)\!:\\ &= c_1 c_3 :\!a_1a_3\!: \;+\; c_1 c_4 :\!a_1 a_4^\dagger\!: \;+\; c_2 c_3:\!a_2^\dagger a_3\!: \;+\; c_2c_4:\!a_3^\dagger a_4^\dagger\!:\\ &= c_1 c_3 a_1a_3 - c_1 c_4 a_4^\dagger a_1 + c_2 c_3a_2^\dagger a_3 + c_2c_4 a_3^\dagger a_4^\dagger. \end{aligned} \]
• For bosonic operators \(A_1=c_1 b_1 + c_2 b_2^\dagger\) and \(A_2=c_3 b_3 + c_4 b_4^\dagger\) we get no sign-changes: \[ \begin{aligned} :\!A_1 A_2\!: &=\; :\!(c_1 b_1 + c_2 b_2^\dagger)(c_3 a_3 + c_4 b_4^\dagger)\!:\\ &= c_1 c_3 b_1b_3 + c_1 c_4 b_4^\dagger b_1 + c_2 c_3 b_2^\dagger b_3 + c_2c_4 b_3^\dagger b_4^\dagger. \end{aligned} \]

Properties of the normal ordered product

Under normal ordering, we can change the order of the operators:

The vacuum expectation value of the normal ordered product is zero: since . (and ).

Contractions

Next we define the contraction of two operators as As an expectation value, the contraction is not an operator anymore, but a complex number.

Table of contractions

Next we define the contracted normal-ordered product: where in the product means that is absent. is the number of exchanges necessary to change from to .

An example:

For an operator , Wick's theorem tells us that it can be written as

Which can be checked by using the commutation relations. The general proof of Wick's theorem can be done by mathematical induction.

The Wick's theorem and (its generalizations in field theory) is used when calculating expectation values. For a vacuum expectation value all the terms except those in which all the operators are contracted, disappear thanks to and . For an odd number of operators , the vacuum expectation value is thus always 0.

Let us calculate an earlier example using Wick's theorem: Wick's theorem removed the need to commute the operators by hand. When using the theorem, one typically does not want to write all the possible contractions, but only those which are not obviously zero. One can immediately discard e.g. the terms containing a contraction of two annihilation or two creation operators, and the terms which contain a contraction in the wrong order (see the table of contractions above). This reduces the number of terms that need to be evaluated drastically.

The sign of a single term in the contraction expansions can be determined from the number of intersections of contraction lines (first line of the above equation). For odd number of crossings we get a minus sign (e.g. first term above) and for even number a plus sign (the second term).