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+---
+title: "Second quantization"
+firstLetter: "S"
+publishDate: 2021-02-26
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-26T10:04:16+01:00
+draft: false
+markup: pandoc
+---
+
+# Second quantization
+
+The **second quantization** is a technique to deal with quantum systems
+containing a large and/or variable number of identical particles.
+Its exact formulation depends on
+whether it is fermions or bosons that are being considered
+(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)).
+
+Regardless of whether the system is fermionic or bosonic,
+the idea is to change basis to a set of certain many-particle wave functions,
+known as the **Fock states**, which are specific members of a **Fock space**,
+a special kind of [Hilbert space](/know/concept/hilbert-space/),
+with a well-defined number of particles.
+
+For a set of $N$ single-particle energy eigenstates
+$\psi_n(x)$ and $N$ identical particles $x_n$, the Fock states are
+all the wave functions which contain $n$ particles, for $n$ going from $0$ to $N$.
+
+So for $n = 0$, there is one basis vector with $0$ particles,
+for $n = 1$, there are $N$ basis vectors with $1$ particle each,
+for $n = 2$, there are $N (N \!-\! 1)$ basis vectors with $2$ particles,
+etc.
+
+In this basis, we define the **particle creation operators**
+and **particle annihilation operators**,
+which respectively add/remove a particle to/from a given state.
+In other words, these operators relate the Fock basis vectors
+to one another, and are very useful.
+
+The point is to express the system's state in such a way that the
+fermionic/bosonic constraints are automatically satisfied, and the
+formulae look the same regardless of the number of particles.
+
+
+## Fermions
+
+Fermions need to obey the Pauli exclusion principle, so each state can only
+contain one particle. In this case, the Fock states are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            n &= 0:
+            \qquad \ket{0, 0, 0, ...}
+            \\
+            n &= 1:
+            \qquad \ket{1, 0, 0, ...} \quad \ket{0, 1, 0, ...} \quad \ket{0, 0, 1, ...} \quad \cdots
+            \\
+            n &= 2:
+            \qquad \ket{1, 1, 0, ...} \quad \ket{1, 0, 1, ...} \quad \ket{0, 1, 1, ...} \quad \cdots
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The notation $\ket{N_\alpha, N_\beta, ...}$ is shorthand for
+the appropriate [Slater determinants](/know/concept/slater-determinant/).
+As an example, take $\ket{0, 1, 0, 1, 1}$,
+which contains three particles $a$, $b$ and $c$
+in states 2, 4 and 5:
+
+$$\begin{aligned}
+    \ket{0, 1, 0, 1, 1}
+    = \Psi(x_a, x_b, x_c)
+    = \frac{1}{\sqrt{3!}} \det\!
+    \begin{bmatrix}
+        \psi_2(x_a) & \psi_4(x_a) & \psi_5(x_a) \\
+        \psi_2(x_b) & \psi_4(x_b) & \psi_5(x_b) \\
+        \psi_2(x_c) & \psi_4(x_c) & \psi_5(x_c)
+    \end{bmatrix}
+\end{aligned}$$
+
+The creation operator $\hat{c}_\alpha^\dagger$ and annihilation
+operator $\hat{c}_\alpha$ are defined to live up to their name:
+they create or destroy a particle in the state $\psi_\alpha$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \hat{c}_\alpha^\dagger \ket{... (N_\alpha\!=\!0) ...}
+            &= J_\alpha \ket{... (N_\alpha\!=\!1) ...}
+            \\
+            \hat{c}_\alpha \ket{... (N_\alpha\!=\!1) ...}
+            &= J_\alpha \ket{... (N_\alpha\!=\!0) ...}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The factor $J_\alpha$ is sometimes known as the **Jordan-Wigner string**,
+and is necessary here to enforce the fermionic antisymmetry,
+when creating or destroying a particle in the $\alpha$th state:
+
+$$\begin{aligned}
+    J_\alpha = (-1)^{\sum_{j < \alpha} N_j}
+\end{aligned}$$
+
+So, for example, when creating a particle in state 4
+of $\ket{0, 1, 1, 0, 1}$, we get the following:
+
+$$\begin{aligned}
+    \hat{c}_4^\dagger \ket{0, 1, 1, 0, 1}
+    = (-1)^{0 + 1 + 1} \ket{0, 1, 1, 1, 1}
+\end{aligned}$$
+
+The point of the Jordan-Wigner string
+is that the order matters when applying the creation and annihilation operators:
+
+$$\begin{aligned}
+    \hat{c}_1^\dagger \hat{c}_2 \ket{0, 1}
+    &= \hat{c}_1^\dagger \ket{0, 0}
+    = \ket{1, 0}
+    \\
+    \hat{c}_2 \hat{c}_1^\dagger \ket{0, 1}
+    &= \hat{c}_2 \ket{1, 1}
+    = - \ket{1, 0}
+\end{aligned}$$
+
+In other words, $\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$,
+meaning that the anticommutator $\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$.
+You can verify for youself that
+the general anticommutators of these operators are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \{\hat{c}_\alpha, \hat{c}_\beta\} = \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\} = 0
+        \qquad \quad
+        \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\} = \delta_{\alpha\beta}
+    }
+\end{aligned}$$
+
+Each single-particle state can only contain 0 or 1 fermions,
+so these operators **quench** states that would violate this rule.
+Note that these are *scalar* zeros:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{c}_\alpha^\dagger \ket{... (N_\alpha\!=\!1) ...} = 0
+        \qquad \quad
+        \hat{c}_\alpha \ket{... (N_\alpha\!=\!0) ...} = 0
+    }
+\end{aligned}$$
+
+Finally, as has already been suggested by the notation, they are each other's adjoint:
+
+$$\begin{aligned}
+    \matrixel{... (N_\alpha\!=\!1) ...}{\hat{c}_\alpha^\dagger}{... (N_\alpha\!=\!0) ...}
+    = \matrixel{...(N_\alpha\!=\!0) ...}{\hat{c}_\alpha}{... (N_\alpha\!=\!1) ...}
+\end{aligned}$$
+
+Let us now use these operators to define the **number operator** $\hat{N}_\alpha$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+    }
+\end{aligned}$$
+
+Its eigenvalue is the number of particles residing in state $\psi_\alpha$
+(look at the hats):
+
+$$\begin{aligned}
+    \hat{N}_\alpha \ket{... N_\alpha ...}
+    = N_\alpha \ket{... N_\alpha ...}
+\end{aligned}$$
+
+
+## Bosons
+
+Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state.
+The Fock states are therefore as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            n &= 0:
+            \qquad \ket{0, 0, 0, ...}
+            \\
+            n &= 1:
+            \qquad \ket{1, 0, 0, ...} \quad \ket{0, 1, 0, ...} \quad \ket{0, 0, 1, ...} \quad \cdots
+            \\
+            n &= 2:
+            \qquad \ket{1, 1, 0, ...} \quad \ket{1, 0, 1, ...} \quad \ket{0, 1, 1, ...} \quad \cdots
+            \\
+            &\qquad\:\:\:
+            \qquad \ket{2, 0, 0, ...} \quad \ket{0, 2, 0, ...} \quad \ket{0, 0, 2, ...} \quad \cdots
+        \end{aligned}
+    }
+\end{aligned}$$
+
+They must be symmetric under the exchange of two bosons.
+To achieve this, the Fock states are represented by Slater *permanents*
+rather than determinants.
+
+The boson creation and annihilation operators $\hat{c}_\alpha^\dagger$ and
+$\hat{c}_\alpha$ are straightforward:
+
+$$\begin{gathered}
+    \boxed{
+        \begin{aligned}
+            \hat{c}_\alpha^\dagger \ket{... N_\alpha ...}
+            &= \sqrt{N_\alpha + 1} \: \ket{... (N_\alpha \!+\! 1) ...}
+            \\
+            \hat{c}_\alpha \ket{... N_\alpha ...}
+            &= \sqrt{N_\alpha} \: \ket{... (N_\alpha \!-\! 1) ...}
+        \end{aligned}
+}\end{gathered}$$
+
+Applying the annihilation operator $\hat{c}_\alpha$ when there are zero
+particles in $\alpha$ will quench the state:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{c}_\alpha \ket{... (N_\alpha\!=\!0) ...} = 0
+    }
+\end{aligned}$$
+
+There is no Jordan-Wigner string, and therefore no sign change when commuting.
+Consequently, these operators therefore satisfy the following:
+
+$$\begin{aligned}
+    \boxed{
+        [\hat{c}_\alpha, \hat{c}_\beta] = [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] = 0
+        \qquad
+        [\hat{c}_\alpha, \hat{c}_\beta^\dagger] = \delta_{\alpha\beta}
+    }
+\end{aligned}$$
+
+The constant factors applied by $\hat{c}_\alpha^\dagger$ and $\hat{c}_\alpha$
+ensure that $\hat{N}_\alpha$ keeps the same nice form:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+    }
+\end{aligned}$$
+
+
+## Operators
+
+Traditionally, an operator $\hat{V}$ simultaneously acting on $N$ indentical particles
+is the sum of the individual single-particle operators $\hat{V}_1$ acting on the $n$th particle:
+
+$$\begin{aligned}
+    \hat{V}
+    = \sum_{n = 1}^N \hat{V}_1
+\end{aligned}$$
+
+This can be rewritten using the second quantization operators as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{V}
+        = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}_1}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
+    }
+\end{aligned}$$
+
+Where the matrix element $\matrixel{\alpha}{\hat{V}_1}{\beta}$ is to be
+evaluated in the normal way:
+
+$$\begin{aligned}
+    \matrixel{\alpha}{\hat{V}_1}{\beta}
+    = \int \psi_\alpha^*(\vec{r}) \: \hat{V}_1(\vec{r}) \: \psi_\beta(\vec{r}) \dd{\vec{r}}
+\end{aligned}$$
+
+Similarly, given some two-particle operator $\hat{V}$ in first-quantized form:
+
+$$\begin{aligned}
+    \hat{V}
+    = \sum_{n \neq m} v(\vec{r}_n, \vec{r}_m)
+\end{aligned}$$
+
+We can rewrite this in second-quantized form as follows.
+Note the ordering of the subscripts:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{V}
+        = \sum_{\alpha, \beta, \gamma, \delta}
+        v_{\alpha \beta \gamma \delta} \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
+    }
+\end{aligned}$$
+
+Where the constant $v_{\alpha \beta \gamma \delta}$ is defined from the
+single-particle wave functions:
+
+$$\begin{aligned}
+    v_{\alpha \beta \gamma \delta}
+    = \iint \psi_\alpha^*(\vec{r}_1) \: \psi_\beta^*(\vec{r}_2)
+    \: v(\vec{r}_1, \vec{r}_2) \: \psi_\gamma(\vec{r}_1)
+    \: \psi_\delta(\vec{r}_2) \dd{\vec{r}_1} \dd{\vec{r}_2}
+\end{aligned}$$
+
+Finally, in the second quantization, changing basis is done in the usual way:
+
+$$\begin{aligned}
+    \hat{c}_b^\dagger \ket{0}
+    = \ket{b}
+    = \sum_{\alpha} \ket{\alpha} \braket{\alpha}{b}
+    = \sum_{\alpha} \braket{\alpha}{b} \hat{c}_\alpha^\dagger \ket{0}
+\end{aligned}$$
+
+Where $\alpha$ and $b$ need not be in the same basis.
+With this, we can define the **field operators**,
+which create or destroy a particle at a given position $\vec{r}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\psi}^\dagger(\vec{r})
+        = \sum_{\alpha} \braket{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger
+        \qquad \quad
+        \hat{\psi}(\vec{r})
+        = \sum_{\alpha} \braket{\vec{r}}{\alpha} \hat{c}_\alpha
+    }
+\end{aligned}$$
+
+
+## References
+1.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.