4. The Hilbert Space of Many-Body Physics#

All of the previous axioms remain valid for a composite system consisting of several quantum degrees of freedom. However, we need to know how to describe the state of the system, and thus more specifically, how to define the Hilbert space associated to such a system. It turns out that quantum mechanics forces us to distinguish two cases.

4.1. Distinguisable Particles and Tensor Products#

Consider a quantum system composed out of two subsystems, which we call $A$ and $B$, sometimes referred to as Alice and Bob in quantum information contexts. These can themselves already be many-body systems. Suppose we know the Hilbert space $\mathbb{H}^A$ in which to describe states of subsystem $A$ when considered as an isolated system on itself, and analoguously for $\mathbb{H}^B$. Now consider both systems together, but where they do not interact, so that we can still treat them independently. In particular, we can prepare subsystem $A$ in a state $\ket{\psi^A}$ and subsystem $B$ in a state $\ket{\psi^B}$. We should also be able to describe these two independent subsystems jointly, so that there must exist a map from the two arguments $(\ket{\psi^A}, \ket{\varphi^B}) \in \mathbb{H}^A \times \mathbb{H}^B$ to a single state which we denote as $\ket{ \psi^A} \otimes \ket{\varphi^B}$ and that lives in a joint Hilbert space $\mathbb{H}^{AB}$ that we have yet to determine.

Now, it makes sense that, if we build superpositions in one of the two subsystems, while keeping the other fixed, this also correspond to a superposition in the joint description of both systems together. This leads to

\[\left (a_1 \ket{\psi^A_1} + a_2 \ket{\psi^A_2}\right) \otimes \ket{\varphi^B} = a_1 \ket{\psi^A_1 }\otimes \ket{\varphi^B} + a_2 \ket{\psi^A_1 }\otimes \ket{\varphi^B}\]

and similarly

\[\ket{\psi^A} \otimes \left(b_1 \ket{\varphi^B_1} + b_2 \ket{\varphi^A_2}\right) = b_1 \ket{\psi^A }\otimes \ket{\varphi^B_1} + b_2 \ket{\psi^A }\otimes \ket{\varphi^B_2}.\]

Hence, the Hilbert space $\mathbb{H}^{AB}$ that we are trying to construct must contain all states $\ket{\psi^A} \otimes \ket{\varphi^B}$ for all $\ket{\psi^A} \in \mathbb{H}^A$ and all $\ket{\varphi^B} \in \mathbb{H}^B$, all possible linear combinations thereof (in order to be a vector space), but in such a way that the above equalities hold. This construction, which can be made mathematically precise, is known as the tensor product of vector spaces $\mathbb{H}^{AB} = \mathbb{H}^A \otimes \mathbb{H}^B$.

We have also denoted the output of the map from two states $(\ket{\psi^A}, \ket{\varphi^B}) \in \mathbb{H}^A \times \mathbb{H}^B$ to $\mathbb{H}^A \otimes \mathbb{H}^B$ using the same tensor product symbol, and refer to such a state as a (tensor) product state $\ket{ \psi^A} \otimes \ket{\varphi^B}$. Importantly, however, the tensor product space $\mathbb{H}^A \otimes \mathbb{H}^B$ certainly contains vectors which are not product states, such as

\[a_1 \ket{ \psi_1^A} \otimes \ket{\varphi_1^B} + a_2 \ket{ \psi_2^A} \otimes \ket{\varphi_2^B}.\]

This forms the basis for quantum correlations and the concept of (quantum) entanglement, which will be a fundamental property of quantum many-body systems. That the Hilbert space of a composite system is given by the tensor product of the individual Hilbert spaces is often introduced as a separate axiom. The deductive (but informal) argument just given can however be turned into a proof that depends only on the axioms given above (in fact only on the first two).

As expected (and required), it can be shown that the tensor product of two Hilbert spaces is again a Hilbert space, if we define its inner product in the following way. We first define the inner product for product states as

\[\braket{\psi_1^A \otimes \varphi_1^B | \psi_2^A \otimes \varphi_2^B } = \braket{\psi_1^A | \psi_2^A} \braket{\varphi_1^B | \varphi_2^B}\]

and then extend this definition by linearity (in the second argument and antilinearity in the first argument).

In practice, given two finite-dimensional Hilbert spaces $\mathbb{H}^A \cong \mathbb{C}^{d^A}$ and $\mathbb{H}^B \cong \mathbb{C}^{d^B}$ with a basis $\{ \ket{j}, j=1,\dots, d^A\}$ and $\{\ket{k}, k=1,\dots, d^B\}$, the tensor product space is spanned by a basis composed of all products

\[\{ \ket{j,k} = \ket{j} \otimes \ket{k}, j=1,\dots, d^A, k=1,\dots, d^B\}\]

and thus has dimension $d^A \cdot d^B$. A general state $\ket{\Psi} \in \mathbb{H}^{A}\otimes \mathbb{H}^B$ can then be expanded as

\[\ket{\Psi} = \sum_{j=1}^{d^A }\sum_{k=1}^{d^B} \Psi_{jk} \ket{j,k}\]

The expansion coefficients $\Psi_{jk}$ thus have two indices, and it is often useful to think of them as a matrix. Note that we will almost always use this product basis, also referred to as the computational basis, for working with tensor product spaces. However, one can certainly also use more complicated basis choices, where the basis vectors are not simple product states. One well known choice that you might remember from your quantum mechanics course is in the case of two spin-1/2 systems. If we denote the basis for a single spin-1/2 system as $\{\ket{\uparrow},\ket{\downarrow}\}$, then the product basis for a system consisting of two spin-1/2 systems is given by $\{\ket{\uparrow,\uparrow}, \ket{\downarrow,\uparrow}, \ket{\uparrow,\downarrow}, \ket{\downarrow,\downarrow}, \}$. However, in the context of spin coupling (see Section on Symmetries), one also uses the coupled basis

\[\begin{split}\ket{0,0} &= \frac{1}{\sqrt{2}} \left(\ket{\uparrow,\downarrow} - \ket{\downarrow,\uparrow}\right)\\ \ket{1,+1} &= \ket{\uparrow,\uparrow}\\ \ket{1,0} &= \frac{1}{\sqrt{2}} \left(\ket{\uparrow,\downarrow} + \ket{\downarrow,\uparrow}\right)\\ \ket{1,-1} &= \ket{\downarrow,\downarrow}\end{split}\]

Note that we also use the same tensor product notation as an operation to map operators from the subsystems into operators acting on the full tensor product Hilbert space. In particular, the process of measuring operator $\hat{A}$ in subsystem $A$ and simultaneously operator $\hat{B}$ in subsystem $B$ is associated with an operator $\hat{A}\otimes \hat{B}$ acting on $\mathbb{H}^A \otimes \mathbb{H}^B$, the action of which is first defined on the product states as

\[\left(\hat{A} \otimes \hat{B}\right) \left(\ket{\psi^A}\otimes \ket{\varphi^B}\right) = \left(\hat{A}\ket{\psi^A}\right) \otimes \left(\hat{B}\ket{\varphi^B}\right)\]

and then extended by linearity. It furthermore holds that

\[(\hat{A}_1 \otimes \hat{B}_1) (\hat{A}_2 \otimes \hat{B}_2) = (\hat{A}_1 \hat{A}_2) \otimes (\hat{B}_1 \hat{B}_2).\]

With respect to a product basis, the matrix representation of $\left(\hat{A} \otimes \hat{B}\right)$ is given by the Kronecker product.

When we are only interested in an operator $\hat{O}$ acting on subsystem $A$ without doing anything on subsystem $B$, we should create the operator $\hat{O} \otimes \hat{1}_B$, with $\hat{1}_B$ the identity operator of the Hilbert space $\mathbb{H}^B$. Often, we will omit this explicit tensor product with the identity operator, and simply use some notation which indicates that an operator acts on a certain subsystem, such as $\hat{O}^{(A)} = \hat{O} \otimes \hat{1}_B$. This also makes it explicit that operators defined on different subsystems, when lifted to act on the full Hilbert space, commute, i.e.

\[\left[\hat{O}_1^{(A)} , \hat{O}_2^{(B)}\right] = \left[ \hat{O}_1 \otimes \hat{1}_B, \hat{1}_A \otimes \hat{O}_2\right] = 0.\]

The tensor product construction extends readily to systems with multiple subsystems. Consider for example a system consisting of qubits, where every individual qubit has an associated Hilbert space $\mathbb{C}^2$ with basis denoted as $\{\ket{0},\ket{1}\}$. The Hilbert space $\mathbb{H}^N$ of $N$ qubits is then spanned by a computational basis which we can denote as

\[\{\ket{s_1, s_2, \ldots, s_N} = \ket{s_1} \otimes \ket{s_2} \otimes \cdots \otimes \ket{s_N}; s_1 =0,1; s_2 =0,1; \ldots; s_n =0,1\}.\]

Hence, the Hilbert space thus has dimension $2^N$, and a general state $\ket{\Psi}$ has expansion coefficients

\[\Psi_{s_1,s_2, \ldots, s_N}\]

which can be interpreted as a single vector of length $2^N$, or as a $N$-dimensional tensor, where every tensor index ranges over the two values 0 and 1. This exponential increase of the Hilbert space dimension with the number of particles is exactly why the quantum many-body problem is so difficult, but also essential for providing a quantum computer with its speed-up. It is exactly these type of quantum states living in a many-body Hilbert space, which is thus composed of many tensor product factors, that we will represent as a tensor network.

Finally, we also have to specify the Hamiltonian of a many-body system. It typically takes the form of a sum of terms, where every individual term acts nontrivially on only a few subsystems. One important example that will reappear throughout these tutorials is the “Quantum Ising Model with transverse magnetic field”, which acts on a system composed of qubits or spin-1/2 particles, and is defined as

\[\hat{H} = - J \sum_{\langle i, j \rangle} \sigma^z_i \otimes \sigma^z_j - h \sum_i \sigma^x_i\]

Here, the summation variables $i$ and $j$ correspond to the sites of a lattice. The notation $\sum_{\langle i,j \rangle}$ denotes a sum over pairs of neighbouring lattice sites $i$ and $j$. The second sum contains terms $\sigma^x_i$ which act nontrivially only on the site $i$, and as the identity operator elsewhere. If, for example, we enumerate the sites from $1$ to $N$, it would act as

\[\sigma^x_i = \underbrace{1 \otimes 1 \otimes \ldots \otimes 1}_{\text{$i-1$ factors}} \otimes \sigma^x \otimes \underbrace{1 \otimes \ldots \otimes 1}_{\text{$N-i-1$ factors}}\]

with $\sigma^x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ the Pauli x matrix, and $1$ the $2 x 2$ unit matrix. The first set of terms in $\hat{H}$ acts nontrivially on two sites, and is defined analoguously, using the Pauli z matrices $\sigma^z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.

4.2. Identical Particles and Pauli’s Exclusion Principle#

The tensor product construction needs to be revised when discussing the Hilbert space of a system composed of identical particles. Consider for example a system made out of $N$ identical particles. To every individual particle we can associate a particular Hilbert space, which we denote as $\mathbb{H}^{(1)}$, for example $\mathbb{H}^{(1)} = L^2(\mathbb{R})$ for a particle moving on the real line, or $\mathbb{H}^{(1)} = \mathbb{C}^L$ for a particle living on the sites of a chain of length $L$.

If we temporarily assign each of the $N$ particles a label $n=1, \dots, N$, then the Hilbert space of the composite system would be given by the $N$-fold tensor product $\widetilde{\mathbb{H}}^{(N)} = \left(\mathbb{H}^{(1)}\right)^{\otimes N}$. However, for identical particles, our labeling is completely arbitrary. For the case of $N=2$ particles on a chain of $L$ sites, we cannot distinguish between the state $\ket{j_1, j_2}$ where particle $1$ is on site $j_1$ and particle $2$ is on site $j_2$ versus the state $\ket{j_2, j_1}$ where site $j_1$ is occupied by the particle that we gave label $2$ and site $j_2$ is occupied by the particle with label $1$. A general redefinition of the particle labels amounts to a permutation, and we have to require that no physical measurement can distinguish between such permutations. Hence, this permutation invariance does not behave like a regular symmetry (like e.g. rotation symmetry, one can still construct observables along preferred directions such that they can detect rotations).

We are forced to restrict our tensor product Hilbert space $\left(\mathbb{H}^{(1)}\right)^{\otimes N}$ to the subspace $\mathbb{H}^{(N)}$ of physical states which are not affected by acting with such permutations. Note that, due to the fact that quantum states actually correspond to rays of vectors, it is still allowed that the vectors in $\mathbb{H}^{(N)}$ pick up a phase factor when applying certain permutations. It is a result in the representation theory of the permutation group that there are only two possibilities. Either the phase factor is always absent (or thus 1), or the phase factor is (-1) for odd permutations and (+1) for even permutations, i.e. the phase factor equals the sign(ature) of the permutation. Identical particles for which the phase factor is always one are known as bosons, whereas those with the nontrival phase factor choice correspond to fermions. Indeed, the nontrivial phase factor automatically gives rise to Pauli’s exclusion principle: two fermions cannot be in the same quantum state, since $P_{12} \ket{j_1,j_2} = \ket{j_2,j_1} = -\ket{j_1,j_2}$ and for $j_1=j_2$ we would thus find $\ket{j,j} = -\ket{j,j}$.

Bosons are thus described by states which are symmetric under permutations, whereas fermions are described by states which are called antisymmetric. We can define an operator on $\tilde{\mathbb{H}}^{(N)} = \left(\mathbb{H}^{(1)}\right)^{\otimes N}$ that maps any given state onto such a (anti)symmeric state, namely by first defining its action on product states as

\[\hat{S}^{\pm} \ket{\psi_1} \otimes \ket{\psi_2} \otimes \cdots \otimes \ket{\psi_N} = \frac{1}{\sqrt{N!}} \sum_{\sigma \in S_N} \epsilon_\sigma \ket{\psi_{\sigma(1)}} \otimes \ket{\psi_{\sigma(2)}} \otimes \cdots \otimes \ket{\psi_{\sigma(N)}}\]

and then extending it by linearity. Here, $S_N$ is the symmetric group containing all permutations $\sigma$ of $N$ elements, where the permutation $\sigma$ is a bijective map from integers $j \in \{1,\dots,N\}$ to a new number $\sigma(j) \in \{1,\dots,N\}$. The sign(ature) $\epsilon_\sigma$ of the permutation takes the value $+1$ or $-1$, depending on whether the permutation $\sigma$ can be obtained by composing an even or odd number of elementary transpositions. An elementary transposition $\tau_{i,j}$ is a permutation which only interchanges the two numbers $i$ and $j \neq i$:

\[\tau_{i,j}(i) =j, \tau_{i,j}(j) = i, \tau_{i,j}(k) =k, \forall k\neq i \land k \neq j\]

Note that $\hat{S}^{\pm}$ does not necessarily yield a normalised state, and can indeed even map a state to zero, in order to give rise to Pauli’s exclusion principle: $\hat{S}^-\ket{j,j} = 0$. The image of $\hat{S}^{\pm}$ contains all states with the proper behaviour under relabeling permutations, and thus correspond to the physical Hilbert space for bosons or fermions:

\[\mathbb{H}^{(N)} = \hat{S}^{\pm} \widetilde{\mathbb{H}}^{(N)} = \hat{S}^{\pm} \left(\mathbb{H}^{(1)}\right)^{\otimes N}\]

Note that in this case, the physical Hilbert space is not a tensor product. However, we can think of it as a subspace of an auxiliary Hilbert space, $ \widetilde{\mathbb{H}}^{(N)}$, which is a tensor product. The restriction to this subspace can thus be thought of as a constraint, and the same scenario happens in other constrained quantum systems. The most notable example is that of quantum gauge theories, where there is an extensive set of constraints, namely that physical quantum states need to be gauge invariant.

Now consider a single particle Hilbert space $\mathbb{H}^{(1)}$ with an orthonormal basis $\{\ket{j}, j=1,\ldots,L\}$, for example where $\ket{j}$ corresponds to the particle being positioned on site $j$ of a lattice with $L$ sites. We also refer to these single particle states as modes. To construct a basis for $\mathbb{H}^{(N)}$, we can start from the tensor product basis of $\widetilde{\mathbb{H}}^{(N)}$ and apply $\hat{S}^{\pm}$ to each of its $L^N$ elements. Let us henceforth denote these states as

\[\ket{j_1,j_2,\ldots ,j_N} = \hat{S}^{\pm} \left(\ket{j_1} \otimes \ket{j_2} \otimes \cdots \otimes \ket{j_N}\right)\]

The application of $\hat{S}^{\pm}$ will create certain linear dependences. In particular, states $ \ket{j_1,j_2, \ldots, j_N}$ that contain the same set of modes $j_k$, i.e. for which the $j_k$’s are related by a permutation, are equal (up to a sign in the case of $\hat{S}^-$). We can thus select a single state by ordering the $j_k$ arguments. Furthermore, in the case of $\hat{S}^{-}$, the state is mapped to zero as soon as two $j_k$ values coincide, so we can eliminate such states. If we thus restrict the set to states $\ket{j_1,j_2,\ldots ,j_N}$ which are such that the modes are ordered as $j_1 < j_2 < \ldots < j_N$ (for fermions) or $j_1 \leq j_2 \leq \ldots \leq j_N$ (for bosons), then we have a linearly independent set of states. For fermions, this implies in particular that we need to have $N \leq L$, there cannot be more fermions in the system then there are linearly independent modes (single particle states).

Finally, one can wonder about the normalisation of these states. For fermions, the superposition created by $\hat{S}^-$ contains $N!$ terms, which are mutually orthogonal, so that the resulting state is normalised, because of the $1/\sqrt{N!}$ prefactor in the definition of $\hat{S}^{-}$. More generally, one then finds

\[\braket{i_1 < i_2 < \ldots < i_N | j_1 < j_2 < \ldots < j_N} = \delta_{i_1,j_1} \delta_{i_2,j_2} \cdots \delta_{i_N,j_N}\]

For bosons, the situation is more complicated in the case that some $j_k$ values coincide. Some of the $N!$ terms created by $\hat{S}^+$ are then equal and contribute differently to the norm. If we denote with $n_1, n_2, \ldots, n_L$ the number of $j$ values that equal the value $1, 2, \ldots, L$, i.e. the number of particles in mode $1, 2, \ldots, L$, then we find

\[\braket{i_1 \leq i_2 \leq \ldots \leq i_N | j_1 \leq j_2 \leq \ldots \leq j_N} = (n_1! n_2! \cdots n_L!) \delta_{i_1,j_1} \delta_{i_2,j_2} \cdots \delta_{i_N,j_N}\]

This more general exprression is also valid for fermions, where every $n_j$ is restricted to be zero or one. In fact, the values $n_j$ for $j=1,\ldots,L$ completely characterise the state, and can thus be used to relabel the basis. Instead of specifying the mode $j_k$ that each particle $k=1,\ldots,N$ occupies (where the labeling of the particles is arbitrary because they are identical), we can move to a mode-based description and thus specify the number of particles in each mode, also known as the mode occupation number. We can then refer to the basis vectors as

\[\ket{n_1, n_2, \ldots, n_L}\]

where $n_j = 0, 1$ (fermions) or $n_j = 0,1,2, \ldots $ (bosons) and furthermore $\sum_{j=1}^{L} n_j = N$. Furthermore, we define these states to be normalised to 1, i.e. we absorb a suitable normalisation factor when defining $\ket{n_1, n_2, \ldots, n_L}$ in terms of the construction above.

This way of labelling the basis states now is again reminiscent of a tensor product structure, i.e. we could think of $\ket{n_1, n_2, \ldots, n_L}$ as the tensor product of states $\ket{n_j}$ associated to every mode, and where the Hilbert space associated with such a mode is two-dimensional in the case of fermions, or infinite-dimensional in the case of bosons. However, there is still a global constraint $\sum_{j=1}^{L} n_j = N$ so that we cannot let the different $n_j$ values vary completely independently from each other. Furthermore, some caution is now needed as to what it means to have operators acting on these different “mode Hilbert spaces”. The correct formalism is that of second quantisation, which we introduce next.

Note

In many applications, people do still work with the framework of first quantisation, and consider $N$-particle states constructed by symmetrising or antisymmetrising the tensor product of $N$ single-particle states, in a so-called independent particle model or approximation. Such states are quite cumbersome to work with. As can already be seen, the antisymmetric case is slightly easier and is known as a Slater determinant. Indeed, the antisymmetrisation formula is reminiscent of the Leibniz formula of a determinant, and for example the inner product between two Slater determinants constructed from $\{\ket{\psi_n},n=1,\ldots,N\}$ and $\{\ket{\varphi_n},n=1,\ldots,N\}$ is given by the determinant of the matrix containing all overlaps $\braket{\varphi_m \vert \psi_n}$. Slater determinants form the basis of Hartree-Fock theory for approximating the state of electrons in an atom or molecule.

The bosonic version occurs in the context of Bose-Einstein condensation and cold atom systems more generally. In that case, the inner product between two such states gives rise to a determinant-like formula, but without the minus signs. This construction is known as the permenant, but unlike the determinant it is very hard to compute in general and really requires to explicitly sum up all $N!$ terms.