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diff --git a/content/know/category/numerical-methods.md b/content/know/category/numerical-methods.md new file mode 100644 index 0000000..5455fee --- /dev/null +++ b/content/know/category/numerical-methods.md @@ -0,0 +1,9 @@ +--- +title: "Numerical methods" +firstLetter: "N" +date: 2021-02-26T16:00:51+01:00 +draft: false +layout: "category" +--- + +This page will fill itself. diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc new file mode 100644 index 0000000..84c2d74 --- /dev/null +++ b/content/know/concept/density-operator/index.pdc @@ -0,0 +1,131 @@ +--- +title: "Density operator" +firstLetter: "D" +publishDate: 2021-03-03 +categories: +- Physics +- Quantum mechanics + +date: 2021-03-03T09:07:51+01:00 +draft: false +markup: pandoc +--- + +# Density operator + +In quantum mechanics, the expectation value of an observable +$\expval*{\hat{L}}$ represents the average result from measuring +$\hat{L}$ on a large number of systems (an **ensemble**) +prepared in the same state $\ket{\Psi}$, +known as a **pure ensemble** or (somewhat confusingly) **pure state**. + +But what if the systems of the ensemble are not all in the same state? +To work with such a **mixed ensemble** or **mixed state**, +the **density operator** $\hat{\rho}$ or **density matrix** (in a basis) is useful. +It is defined as follows, where $p_n$ is the probability +that the system is in state $\ket{\Psi_n}$, +i.e. the proportion of systems in the ensemble that are +in state $\ket{\Psi_n}$: + +$$\begin{aligned} + \boxed{ + \hat{\rho} + = \sum_{n} p_n \ket{\Psi_n} \bra{\Psi_n} + } +\end{aligned}$$ + +Do not let is this form fool you into thinking that $\hat{\rho}$ is diagonal: +$\ket{\Psi_n}$ need not be basis vectors. +Instead, the matrix elements of $\hat{\rho}$ are found as usual, +where $\ket{j}$ and $\ket{k}$ are basis vectors: + +$$\begin{aligned} + \matrixel{j}{\hat{\rho}}{k} + = \sum_{n} p_n \braket{j}{\Psi_n} \braket{\Psi_n}{k} +\end{aligned}$$ + +However, from the special case where $\ket{\Psi_n}$ are indeed basis vectors, +we can conclude that $\hat{\rho}$ is Hermitian, +and that its trace (i.e. the total probability) is 100%: + +$$\begin{gathered} + \boxed{ + \hat{\rho}^\dagger = \hat{\rho} + } + \qquad \qquad + \boxed{ + \mathrm{Tr}(\hat{\rho}) = 1 + } +\end{gathered}$$ + +These properties are preserved by all changes of basis. +If the ensemble is purely $\ket{\Psi}$, +then $\hat{\rho}$ is given by a single state vector: + +$$\begin{aligned} + \hat{\rho} = \ket{\Psi} \bra{\Psi} +\end{aligned}$$ + +From the special case where $\ket{\Psi}$ is a basis vector, +we can conclude that for a pure ensemble, +$\hat{\rho}$ is idempotent, which means that: + +$$\begin{aligned} + \hat{\rho}^2 = \hat{\rho} +\end{aligned}$$ + +This can be used to find out whether a given $\hat{\rho}$ +represents a pure or mixed ensemble. + +Next, we define the ensemble average $\expval*{\expval*{\hat{L}}}$ +as the mean of the expectation values for states in the ensemble, +which can be calculated like so: + +$$\begin{aligned} + \boxed{ + \expval*{\expval*{\hat{L}}} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{L}}{\Psi_n} + = \mathrm{Tr}(\hat{L} \hat{\rho}) + } +\end{aligned}$$ + +To prove the latter, +we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so: + +$$\begin{aligned} + \mathrm{Tr}(\hat{L} \hat{\rho}) + &= \sum_{j} \matrixel{j}{\hat{L} \hat{\rho}}{j} + = \sum_{j} \sum_{n} p_n \matrixel{j}{\hat{L}}{\Psi_n} \braket{\Psi_n}{j} + \\ + &= \sum_{n} \sum_{j} p_n \braket{\Psi_n}{j} \matrixel{j}{\hat{L}}{\Psi_n} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{I} \hat{L}}{\Psi_n} + = \expval*{\expval*{\hat{L}}} +\end{aligned}$$ + +In both the pure and mixed cases, +if the state probabilities $p_n$ are constant with respect to time, +then the evolution of the ensemble obeys the **Von Neumann equation**: + +$$\begin{aligned} + \boxed{ + i \hbar \dv{\hat{\rho}}{t} = [\hat{H}, \hat{\rho}] + } +\end{aligned}$$ + +This equivalent to the Schrödinger equation: +one can be derived from the other. +We differentiate $\hat{\rho}$ with the product rule, +and then substitute the opposite side of the Schrödinger equation: + +$$\begin{aligned} + i \hbar \dv{\hat{\rho}}{t} + &= i \hbar \dv{t} \sum_n p_n \ket{\Psi_n} \bra{\Psi_n} + \\ + &= \sum_n p_n \Big( i \hbar \dv{t} \ket{\Psi_n} \Big) \bra{\Psi_n} + \sum_n p_n \ket{\Psi_n} \Big( i \hbar \dv{t} \bra{\Psi_n} \Big) + \\ + &= \sum_n p_n \ket*{\hat{H} n} \bra{n} - \sum_n p_n \ket{n} \bra*{\hat{H} n} + = \hat{H} \hat{\rho} - \hat{\rho} \hat{H} + = [\hat{H}, \hat{\rho}] +\end{aligned}$$ + + diff --git a/content/know/concept/lagrange-multiplier/index.pdc b/content/know/concept/lagrange-multiplier/index.pdc new file mode 100644 index 0000000..2b14897 --- /dev/null +++ b/content/know/concept/lagrange-multiplier/index.pdc @@ -0,0 +1,103 @@ +--- +title: "Lagrange multiplier" +firstLetter: "L" +publishDate: 2021-03-02 +categories: +- Mathematics +- Physics + +date: 2021-03-02T16:28:42+01:00 +draft: false +markup: pandoc +--- + +# Lagrange multiplier + +The method of **Lagrange multipliers** or **undetermined multipliers** +is a technique for optimizing (i.e. finding the extrema of) +a function $f(x, y, z)$, +subject to a given constraint $\phi(x, y, z) = C$, +where $C$ is a constant. + +If we ignore the constraint $\phi$, +optimizing $f$ simply comes down to finding stationary points: + +$$\begin{aligned} + 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} +\end{aligned}$$ + +This problem is easy: +$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary, +so all we need to do is find the roots of +the partial derivatives $f_x$, $f_y$ and $f_z$, +which we respectively call $x_0$, $y_0$ and $z_0$, +and then the extremum is simply $(x_0, y_0, z_0)$. + +But the constraint $\phi$, over which we have no control, +adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$, +so if two are known, the third is given by $\phi = C$. +The problem is then a system of equations: + +$$\begin{aligned} + 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} + \\ + 0 &= \dd{\phi} = \phi_x \dd{x} + \phi_y \dd{y} + \phi_z \dd{z} +\end{aligned}$$ + +Solving this directly would be a delicate balancing act +of all the partial derivatives. + +To help us solve this, we introduce a "dummy" parameter $\lambda$, +the so-called **Lagrange multiplier**, and contruct a new function $L$ given by: + +$$\begin{aligned} + L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z) +\end{aligned}$$ + +Clearly, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$, +so now the problem is a single equation again: + +$$\begin{aligned} + 0 = \dd{L} + = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z} +\end{aligned}$$ + +Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$. +This choice represents satisfying the constraint, +so now the remaining $\dd{x}$ and $\dd{y}$ are independent again, +and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$. + +This generalizes nicely to multiple constraints or more variables: +suppose that we want to find the extrema of $f(x_1, ..., x_N)$ +subject to $M < N$ conditions: + +$$\begin{aligned} + \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M +\end{aligned}$$ + +This once again turns into a delicate system of $M+1$ equations to solve: + +$$\begin{aligned} + 0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N} + \\ + 0 &= \dd{\phi_1} = \phi_{1, x_1} \dd{x_1} + ... + \phi_{1, x_N} \dd{x_N} + \\ + &\vdots + \\ + 0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N} +\end{aligned}$$ + +Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$ +and define $L(x_1, ..., x_N)$: + +$$\begin{aligned} + L = f + \sum_{m = 1}^M \lambda_m \phi_m +\end{aligned}$$ + +As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$ +to eliminate $M$ of its $N$ terms: + +$$\begin{aligned} + 0 = \dd{L} + = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n} +\end{aligned}$$ diff --git a/content/know/concept/pulay-mixing/index.pdc b/content/know/concept/pulay-mixing/index.pdc new file mode 100644 index 0000000..9102c0e --- /dev/null +++ b/content/know/concept/pulay-mixing/index.pdc @@ -0,0 +1,158 @@ +--- +title: "Pulay mixing" +firstLetter: "P" +publishDate: 2021-03-02 +categories: +- Numerical methods + +date: 2021-03-02T19:11:51+01:00 +draft: false +markup: pandoc +--- + +# Pulay mixing +Some numerical problems are most easily solved *iteratively*, +by generating a series $\rho_1$, $\rho_2$, etc. +converging towards the desired solution $\rho_*$. +**Pulay mixing**, also often called +**direct inversion in the iterative subspace** (DIIS), +is an effective method to speed up convergence, +which also helps to avoid periodic divergences. + +The key concept it relies on is the **residual vector** $R_n$ +of the $n$th iteration, which in some way measures the error of the current $\rho_n$. +Its exact definition varies, +but is generally along the lines of the difference between +the input of the iteration and the raw resulting output: + +$$\begin{aligned} + R_n + = R[\rho_n] + = \rho_n^\mathrm{new}[\rho_n] - \rho_n +\end{aligned}$$ + +It is not always clear what to do with $\rho_n^\mathrm{new}$. +Directly using it as the next input ($\rho_{n+1} = \rho_n^\mathrm{new}$) +often leads to oscillation, +and linear mixing ($\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$) +can take a very long time to converge properly. +Pulay mixing offers an improvement. + +The idea is to construct the next iteration's input $\rho_{n+1}$ +as a linear combination of the previous inputs $\rho_1$, $\rho_2$, ..., $\rho_n$, +such that it is as close as possible to the optimal $\rho_*$: + +$$\begin{aligned} + \boxed{ + \rho_{n+1} + = \sum_{m = 1}^n \alpha_m \rho_m + } +\end{aligned}$$ + +To do so, we make two assumptions. +Firstly, the current $\rho_n$ is already close to $\rho_*$, +so that such a linear combination makes sense. +Secondly, the iteration is linear, +such that the raw output $\rho_{n+1}^\mathrm{new}$ +is also a linear combination with the *same coefficients*: + +$$\begin{aligned} + \rho_{n+1}^\mathrm{new} + = \sum_{m = 1}^n \alpha_m \rho_m^\mathrm{new} +\end{aligned}$$ + +We will return to these assumptions later. +The point is that $R_{n+1}$ is also a linear combination: + +$$\begin{aligned} + R_{n+1} + = \rho_{n+1}^\mathrm{new} - \rho_{n+1} + = \sum_{m = 1}^n \alpha_m \rho_m^\mathrm{new} - \sum_{m = 1}^n \alpha_m \rho_m + = \sum_{m = 1}^n \alpha_m R_m +\end{aligned}$$ + +The goal is to choose the coefficients $\alpha_m$ such that +the norm of the error $|R_{n+1}| \approx 0$, +subject to the following constraint to preserve the normalization of $\rho_{n+1}$: + +$$\begin{aligned} + \sum_{m=1}^n \alpha_m = 1 +\end{aligned}$$ + +We thus want to minimize the following quantity, +where $\lambda$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/): + +$$\begin{aligned} + \braket{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m + = \sum_{m=1}^n \alpha_m \Big( \sum_{k=1}^n \alpha_k \braket{R_m}{R_k} + \lambda \Big) +\end{aligned}$$ + +By differentiating the right-hand side with respect to $\alpha_m$, +we get a system of equations that we can write in matrix form, +which is relatively cheap to solve numerically: + +$$\begin{aligned} + \begin{bmatrix} + \braket{R_1}{R_1} & \cdots & \braket{R_1}{R_n} & 1 \\ + \vdots & \ddots & \vdots & \vdots \\ + \braket{R_n}{R_1} & \cdots & \braket{R_n}{R_n} & 1 \\ + 1 & \cdots & 1 & 0 + \end{bmatrix} + \begin{bmatrix} + \alpha_1 \\ \vdots \\ \alpha_n \\ \lambda + \end{bmatrix} + = + \begin{bmatrix} + 0 \\ \vdots \\ 0 \\ 1 + \end{bmatrix} +\end{aligned}$$ + +This method is very effective. +However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc. +are much further from $\rho_*$ than $\rho_n$, +so usually only the most recent $N$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used: + +$$\begin{aligned} + \rho_{n+1} + = \sum_{m = N}^n \alpha_m \rho_m +\end{aligned}$$ + +You might be confused by the absence of all $\rho_m^\mathrm{new}$ +in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored. +This is due to the first assumption, +which states that $\rho_n^\mathrm{new}$ are $\rho_n$ are already similar, +such that they are interchangeable. + +Speaking of which, about those assumptions: +while they will clearly become more accurate as $\rho_n$ approaches $\rho_*$, +they might be very dubious in the beginning. +A consequence of this is that the early iterations might get "trapped" +in a suboptimal subspace spanned by $\rho_1$, $\rho_2$, etc. +To say it another way, we would be varying $n$ coefficients $\alpha_m$ +to try to optimize a $D$-dimensional $\rho_{n+1}$, +where in general $D \gg n$, at least in the beginning. + +There is an easy fix to this problem: +add a small amount of the raw residual $R_m$ +to "nudge" $\rho_{n+1}$ towards the right subspace, +where $\beta \in [0,1]$ is a tunable parameter: + +$$\begin{aligned} + \boxed{ + \rho_{n+1} + = \sum_{m = N}^n \alpha_m (\rho_m + \beta R_m) + } +\end{aligned}$$ + +In other words, we end up introducing a small amount of the raw outputs $\rho_m^\mathrm{new}$, +while still giving more weight to iterations with smaller residuals. + +Pulay mixing is very effective: +it can accelerate convergence by up to one order of magnitude! + + + +## References +1. P. Pulay, + [Convergence acceleration of iterative sequences. The case of SCF iteration](https://doi.org/10.1016/0009-2614(80)80396-4), + 1980, Elsevier. diff --git a/content/know/concept/second-quantization/index.pdc b/content/know/concept/second-quantization/index.pdc new file mode 100644 index 0000000..b8d9a18 --- /dev/null +++ b/content/know/concept/second-quantization/index.pdc @@ -0,0 +1,332 @@ +--- +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. |