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diff --git a/source/know/concept/second-quantization/index.md b/source/know/concept/second-quantization/index.md index e446557..605ffd1 100644 --- a/source/know/concept/second-quantization/index.md +++ b/source/know/concept/second-quantization/index.md @@ -15,29 +15,26 @@ 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/), +the idea is to change basis to a set of many-particle wavefunctions +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. +$$\psi_k(x)$$ and $$N$$ identical particles $$x_k$$, +the Fock states are all the wavefunctions which contain $$n$$ particles, +for $$n$$ going from $$0$$ to $$N$$. 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 +In other words, these operators relate the Fock basis states 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. +The idea is to express states in such a way +that the fermionic/bosonic constraints are automatically satisfied, +and that the formulas look the same regardless of the number of particles. + ## Fermions @@ -56,6 +53,8 @@ $$\begin{aligned} \\ n &= 2: \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots + \\ + &\:\:\vdots \qquad \qquad \qquad \vdots \end{aligned} } \end{aligned}$$ @@ -79,16 +78,17 @@ $$\begin{aligned} The creation operator $$\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$$: +they create or destroy a particle in the state $$\psi_\alpha$$. +Formally, this means: $$\begin{aligned} \boxed{ \begin{aligned} - \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!0) ...} - &= J_\alpha \Ket{... (N_\alpha\!=\!1) ...} + \hat{c}_\alpha^\dagger \Ket{...0_\alpha...} + &= J_\alpha \Ket{...1_\alpha...} \\ - \hat{c}_\alpha \Ket{... (N_\alpha\!=\!1) ...} - &= J_\alpha \Ket{... (N_\alpha\!=\!0) ...} + \hat{c}_\alpha \Ket{...1_\alpha...} + &= J_\alpha \Ket{...0_\alpha...} \end{aligned} } \end{aligned}$$ @@ -98,7 +98,8 @@ and is necessary here to enforce the fermionic antisymmetry, when creating or destroying a particle in the $$\alpha$$th state: $$\begin{aligned} - J_\alpha = (-1)^{\sum_{j < \alpha} N_j} + J_\alpha + = (-1)^{\sum_{j < \alpha} N_j} \end{aligned}$$ So, for example, when creating a particle in state 4 @@ -110,7 +111,8 @@ $$\begin{aligned} \end{aligned}$$ The point of the Jordan-Wigner string -is that the order matters when applying the creation and annihilation operators: +is that the order matters when applying the creation and annihilation operators, +so, for example: $$\begin{aligned} \hat{c}_1^\dagger \hat{c}_2 \Ket{0, 1} @@ -124,14 +126,21 @@ $$\begin{aligned} In other words, $$\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$$. -You can verify for youself that +You can verify for yourself that the general anticommutators of these operators are given by: $$\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} + \begin{aligned} + \{\hat{c}_\alpha, \hat{c}_\beta\} + &= 0 + \\ + \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\} + &= 0 + \\ + \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\} + &= \delta_{\alpha\beta} + \end{aligned} } \end{aligned}$$ @@ -141,24 +150,29 @@ Note that these are *scalar* zeros: $$\begin{aligned} \boxed{ - \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!1) ...} = 0 - \qquad \quad - \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0 + \begin{aligned} + \hat{c}_\alpha^\dagger \Ket{...1_\alpha...} + &= 0 + \\ + \hat{c}_\alpha \Ket{...0_\alpha...} + &= 0 + \end{aligned} } \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) ...} + \matrixel{...1_\alpha...}{\hat{c}_\alpha^\dagger}{...0_\alpha...} + = \matrixel{...0_\alpha...}{\hat{c}_\alpha}{...1_\alpha...}^{*} \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 + \hat{N}_\alpha + = \hat{c}_\alpha^\dagger \hat{c}_\alpha } \end{aligned}$$ @@ -171,6 +185,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Bosons Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state. @@ -188,8 +203,10 @@ $$\begin{aligned} n &= 2: \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots \\ - &\qquad\:\:\: + &\qquad\:\,\, \qquad \Ket{2, 0, 0, ...} \quad \Ket{0, 2, 0, ...} \quad \Ket{0, 0, 2, ...} \quad \cdots + \\ + &\:\:\vdots \qquad \qquad \qquad \vdots \end{aligned} } \end{aligned}$$ @@ -212,23 +229,31 @@ $$\begin{gathered} \end{aligned} }\end{gathered}$$ -Applying the annihilation operator $$\hat{c}_\alpha$$ when there are zero -particles in $$\alpha$$ will quench the state: +Applying the annihilation operator $$\hat{c}_\alpha$$ +when there are zero particles in $$\alpha$$ quenches the state: $$\begin{aligned} \boxed{ - \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0 + \hat{c}_\alpha \Ket{...0_\alpha...} + = 0 } \end{aligned}$$ There is no Jordan-Wigner string, and therefore no sign change when commuting. -Consequently, these operators therefore satisfy the following: +Consequently, these operators satisfy the following commutators: $$\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} + \begin{aligned} + [\hat{c}_\alpha, \hat{c}_\beta] + &= 0 + \\ + [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] + &= 0 + \\ + [\hat{c}_\alpha, \hat{c}_\beta^\dagger] + &= \delta_{\alpha\beta} + \end{aligned} } \end{aligned}$$ @@ -237,90 +262,93 @@ ensure that $$\hat{N}_\alpha$$ keeps the same nice form: $$\begin{aligned} \boxed{ - \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha + \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$$th particle: +In the second quantization, +changing between different bases of single-particle states +is done in the usual way, where $$\alpha$$ and $$b$$ need not be in the same basis. +Note that $$\Ket{0}$$ is the zero-particle Fock state, +and $$\Ket{\alpha}$$ etc. are one-particle Fock states: $$\begin{aligned} - \hat{V} - = \sum_{n = 1}^N \hat{V}_1 + \hat{c}_b^\dagger \Ket{0} + = \Ket{b} + = \sum_{\alpha} \Ket{\alpha} \inprod{\alpha}{b} + = \sum_{\alpha} \inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0} \end{aligned}$$ -This can be rewritten using the second quantization operators as follows: +With this, we define the **field operators**, +which create or destroy a particle at a position $$\vb{r}$$: $$\begin{aligned} \boxed{ - \hat{V} - = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}_1}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta + \hat{\Psi}^\dagger(\vb{r}) + = \sum_{\alpha} \inprod{\alpha}{\vb{r}} \hat{c}_\alpha^\dagger + \qquad \qquad + \hat{\Psi}(\vb{r}) + = \sum_{\alpha} \inprod{\vb{r}}{\alpha} \hat{c}_\alpha } \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}$$ in first-quantized form: +By the same basis-changing principle, +any single-particle (non-interacting) operator $$\hat{V}$$ can be translated +to its second-quantized $$N$$-particle version as follows: $$\begin{aligned} \hat{V} - = \sum_{n \neq m} v(\vec{r}_n, \vec{r}_m) + &= \sum_{\alpha, \beta} \ket{\alpha} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\beta} + = \sum_{\alpha, \beta} \ket{\hat{c}_\alpha^\dagger 0} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\hat{c}_\beta^\dagger 0} \end{aligned}$$ -We can rewrite this in second-quantized form as follows. -Note the ordering of the subscripts: +We take out the creation operators, +which allows us to generalize to multi-particle states: $$\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 + = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta } \end{aligned}$$ -Where the constant $$v_{\alpha \beta \gamma \delta}$$ is defined from the -single-particle wave functions: +Where the matrix element $$\matrixel{\alpha}{\hat{V}}{\beta}$$ +is to be evaluated in the normal way: $$\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} + \matrixel{\alpha}{\hat{V}}{\beta} + = \int \psi_\alpha^*(\vb{r}) \: \hat{V}(\vb{r}) \: \psi_\beta(\vb{r}) \dd{\vb{r}} \end{aligned}$$ -Finally, in the second quantization, changing basis is done in the usual way: +In the same way, a two-particle interaction operator $$\hat{W}$$ +can be rewritten in the form below. +Note the ordering of the operators' subscripts: $$\begin{aligned} - \hat{c}_b^\dagger \Ket{0} - = \Ket{b} - = \sum_{\alpha} \Ket{\alpha} \Inprod{\alpha}{b} - = \sum_{\alpha} \Inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0} + \boxed{ + \hat{W} + = \sum_{\alpha, \beta, \gamma, \delta} + W_{\alpha \beta \gamma \delta} \: \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma + } \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}$$: +Where the constant $$W_{\alpha \beta \gamma \delta}$$ +is defined from the single-particle wavefunctions like so: $$\begin{aligned} - \boxed{ - \hat{\Psi}^\dagger(\vec{r}) - = \sum_{\alpha} \Inprod{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger - \qquad \quad - \hat{\Psi}(\vec{r}) - = \sum_{\alpha} \Inprod{\vec{r}}{\alpha} \hat{c}_\alpha - } + W_{\alpha \beta \gamma \delta} + \equiv \iint \psi_\alpha^*(\vb{r}_1) \: \psi_\beta^*(\vb{r}_2) + \: W(\vb{r}_1, \vb{r}_2) \: \psi_\gamma(\vb{r}_1) + \: \psi_\delta(\vb{r}_2) \dd{\vb{r}_1} \dd{\vb{r}_2} \end{aligned}$$ + ## References 1. L.E. Ballentine, *Quantum mechanics: a modern development*, 2nd edition, |