MathB.in - Share Mathematics with LaTeX and Markdown

Second Quantization

Given the non-relativistic Schrodinger equation for a system of $N$ identical particles with a one-particle interaction (e.g. no potentials of the form $V(q_i,q_j)$ etc... only $V(q_i)$)
\begin{align}
i \dfrac{\partial \psi}{\partial t} &= \hat{H} \psi \label{notexplicit} \\
&= \sum \, _{i=1}^N[-\dfrac{1}{2m} \nabla^2_i + V(q_i)] \psi \label{explicit}
\end{align}
expand the solution in terms of a complete set of quantities $n_i$ for each particle $i$, e.g. $n_1 = (\mathbf{p}_1,s_1)$, $n_2 = (\mathbf{L},s_2)$, etc..., summing over all possible values each $n_i$ can take:
\begin{align}
\psi(q_1,q_2,\dots,q_N,t) = \sum_{n_1} \sum_{n_2} \dots \sum_{n_N} c(n_1,n_2,\dots,n_N,t) \psi_{n_1}(q_1) \psi_{n_2}(q_2) \dots \psi_{n_N}(q_N)
\end{align}
and then use this in the Schrodinger equation to find
\begin{align}
i \dfrac{\partial}{\partial t} \psi  &= \sum_{n_1} \sum_{n_2} \dots \sum_{n_N} i \dfrac{\partial}{\partial t}  c(n_1,n_2,\dots,n_N,t) \psi_{n_1}(q_1) \psi_{n_2}(q_2) \dots \psi_{n_N}(q_N)  \\
&= \hat{H} \psi \\
&= \hat{H} \sum_{n_1} \sum_{n_2} \dots \sum_{n_N} c(n_1,n_2,\dots,n_N,t) \psi_{n_1}(q_1) \psi_{n_2}(q_2) \dots \psi_{n_N}(q_N) \\
&= \sum_{k=1} \sum_{n_1} \sum_{n_2} \dots \sum_{n_N} c(n_1,n_2,\dots,n_N,t) \psi_{n_1}(q_1) \dots \hat{H} \psi_{n_k}(q_k) \dots \psi_{n_N}(q_N).
\end{align}
From this we can isolate an equation for the coefficients $c$ and give the Schrodinger equation in Fourier space

\begin{align}
i \dfrac{\partial}{\partial t}  c &= i \dfrac{\partial}{\partial t}  c(m_1,m_2,\dots,m_N,t) \\
&= \int dq_1 \dots dq_N \psi_{m_1}^*(q_1) \dots \psi_{m_N}^*(q_N)  \sum_{n_1} \dots \sum_{n_N} i \dfrac{\partial}{\partial t}  c(n_1,\dots,n_N,t) \psi_{n_1}(q_1)  \dots \psi_{n_N}(q_N) \\
&= \int dq_1 \dots dq_N \psi_{m_1}^*(q_1) \dots \psi_{m_N}^*(q_N)  \sum_{k=1}^N \sum_{n_1} \dots \sum_{n_N} c \psi_{n_1}(q_1) \dots \hat{H} \psi_{n_k}(q_k) \dots \psi_{n_N}(q_N) \\
&= \sum_{k=1}^N \sum_{n_k}  \int dq_k \psi_{m_k}^*(q_k) \hat{H} \psi_{n_k}(q_k) c(m_1,\dots,n_k,\dots,m_N,t) \\
&= \sum_k \sum_{n_k} \hat{H}_{m_k;n_k} c(m_1,\dots,n_k,\dots,m_N,t) .
\end{align}

Thus considering $ c(n_1,n_2,\dots,n_N,t)$ as our wave function we see
\begin{align}
| c(n_1,n_2,\dots,n_N,t)|^2
\end{align}
is the probability that the first particle is in the state $n_1$ (i.e. has some specific value e.g. $((\mathbf{p}_0)_1,s_1)$), the second particle has the state $n_2$, etc... and due to the symmetry of the wave functions of identical particles, these wave functions are symmetric functions of their arguments as well, and so the values of these functions depend only on how many of the $N$ arguments $n_1, n_2, \dots , n_N$ are equal to any of the total possible states these identical particles can acquire, where $N_1$ is the number of particles in the first possible acquirable state $n_1$, $N_2$ is the number of particles in the second possible acquirable state, $N_m$ is the number of particles in the $m$'th possible acquirable state these identical particles can attain, and so on. From this we see that the wave function is redundant using the $n_i$ variables, and should be placed in the $N_i$ variables, where
\begin{align}
|c(N_1,N_2,\dots,N_m,\dots,t)|^2
\end{align}
is the probability of finding $N_1$ particles in state $1$, the first possible acquirable state these identical particles can attain, $N_2$ particles in state $2$, $N_m$ in state $m$, and so on. We must relate this to the wave function in the $n_i$ coordinates, and we do this by noting that this probability is equal to the sum over all $|c(n_1,n_2,\dots,n_N,t)|^2$ probabilities which have $N_i$ particles in the state $n_i$ as above,
\begin{align}
|c(N_1,N_2,\dots,N_m,\dots,t)|^2 = \sum_{\begin{array}{c} n_1,n_2,\dots,n_N \ \text{s.t.} \\ N_1 \ \text{in} \ n_1, N_2 \ \text{in} \ N_2, \dots \end{array}} |c(n_1,n_2,\dots,n_N,t)|^2
\end{align}
but by symmetry, each of these terms are equal, so that the sum is equal to the sum over all $N!$ possible permutations of the variables, subject to the fact that $N_1$ are equal, i.e. dividing by $N_1!$, etc... giving
\begin{align}
|c(N_1,N_2,\dots,N_m,\dots,t)|^2 &= \sum_{\begin{array}{c} n_1,n_2,\dots,n_N \ \text{s.t.} \\ N_1 \ \text{in} \ n_1, N_2 \ \text{in} \ N_2, \dots \end{array}} |c(n_1,n_2,\dots,n_N,t)|^2 \\
&= \dfrac{N!}{N_1! N_2! \dots N_m! \dots } |c(n_1,n_2,\dots,n_N,t)|^2 \\
c(N_1,N_2,\dots,N_m,\dots ) &= \sqrt{\dfrac{N!}{N_1! N_2! \dots N_m! \dots }} c(n_1,n_2,\dots,n_N,t) \\
c(n_1,n_2,\dots,n_N,t) &= \sqrt{\dfrac{N_1! N_2! \dots N_m! \dots }{N!}} c(N_1,N_2,\dots,N_m,\dots,t) 
\end{align}
so that using this in the Schrodinger equation, where 
\begin{align}
c(m_1,\dots,n_k,\dots,m_n,t)
\end{align}
only differs from 
\begin{align}
c(m_1,\dots,m_k,\dots,m_n,t)
\end{align}
by the fact that the number of particles in the $n_k = n$ state has increased by one, and the number of particles in the $m_k = m$ state has decreased by one, so that
\begin{align}
c &= c(m_1,\dots,n_k,\dots,m_n,t) \\
&= \sqrt{\dfrac{N_1! \dots  (N_m-1)! \dots (N_n + 1)! \dots }{N!}} c(N_1,\dots,N_m-1, \dots, N_n + 1,\dots,t)
\end{align}
which means the Schrodinger equation becomes
\begin{align}
i \dfrac{\partial}{\partial t}  c &= i \dfrac{\partial}{\partial t}  c(m_1,m_2,\dots,m_N,t) \\
&= i \dfrac{\partial}{\partial t}  \sqrt{\dfrac{N_1! N_2! \dots N_m! \dots }{N!}} c(N_1,N_2,\dots,N_m,\dots,t) \\
&=  \sum_{k=1} \sum_{n_1} \sum_{n_2} \dots \sum_{n_N} c(n_1,n_2,\dots,n_N,t) \psi_{n_1}(q_1) \dots \hat{H} \psi_{n_k}(q_k) \dots \psi_{n_N}(q_N) \\
&= \sum_k \sum_{n_k} \hat{H}_{m_k;n_k} \sqrt{\dfrac{N_1! \dots  (N_m-1)! \dots (N_n + 1)! \dots }{N!}} c(N_1,\dots,N_m-1, \dots, N_n + 1,\dots,t)  \\
&= \sum_{m,n} N_m \hat{H}_{m,n} \sqrt{\dfrac{N_1! \dots  (N_m-1)! \dots (N_n + 1)! \dots }{N!}} c(N_1,\dots,N_m-1, \dots, N_n + 1,\dots,t)
\end{align} 
where the last sum comes from noting that in summing over $k$'s, this is equivalent to summing over the $m$'s where each $m_k$ occurs $N_{m_k} = N_m$ times, and then dividing the $\sqrt{\dfrac{N_1! N_2! \dots N_m! \dots }{N!}}$ coefficient on both sides leaves
\begin{align}
i \dfrac{\partial}{\partial t} c(N_1,N_2,\dots,N_m,\dots,t) &= \sum_{m,n} \hat{H}_{m,n} \sqrt{N_m} \sqrt{N_n+1} c(N_1,\dots,N_m-1, \dots, N_n + 1,\dots,t)  \\
&= \sum_{m,n} \hat{H}_{m,n} \hat{\mathbf{a}}_m \hat{\mathbf{a}}_n^* c(N_1,\dots,N_m, \dots, N_n, \dots,t) \\
&= \hat{\mathbf{H}} c(N_1, \dots , N_m , \dots , N_n , \dots ,t) 
\end{align} 
where we defined the creation and annihilation operators $\hat{\mathbf{a}}_m , \hat{\mathbf{a}}_n^* $ to act as
\begin{align}
\hat{\mathbf{a}}_m c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_m} c(N_1,\dots,N_m-1,\dots,N_n,\dots,t) \\
\hat{\mathbf{a}}_n^* c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_n+1} c(N_1,\dots,N_m,\dots,N_n+1,\dots,t) \\
\hat{\mathbf{a}}_m \hat{\mathbf{a}}_n^* &= \hat{\mathbf{a}}_n^* \hat{\mathbf{a}}_m \ \ , n \neq m,
\end{align}
so that they commute for $m \neq n$, but for $m = n$ we see they must satisfy
\begin{align}
\hat{\mathbf{a}}_m c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_m} c(N_1,\dots,N_m-1,\dots,N_n,\dots,t) \\
\hat{\mathbf{a}}_m^* \hat{\mathbf{a}}_m c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_m} \hat{\mathbf{a}}_n^* c(N_1,\dots,N_m-1,\dots,N_n,\dots,t) \\
&= \sqrt{N_m} \sqrt{(N_m-1)+1} c(N_1,\dots,N_m,\dots,N_n,\dots,t) \\
&= N_m c(N_1,\dots,N_m,\dots,N_n,\dots,t) \\
\hat{\mathbf{a}}_m^* c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_m+1} c(N_1,\dots,N_m+1,\dots,N_n,\dots,t) \\
\hat{\mathbf{a}}_m \hat{\mathbf{a}}_m^* c(N_1,\dots,N_m,\dots,N_n,\dots,t) &= \sqrt{N_m+1} \hat{\mathbf{a}}_m  c(N_1,\dots,N_m+1,\dots,N_n,\dots,t) \\
&= \sqrt{N_m+1} \sqrt{N_m+1}  c(N_1,\dots,N_m,\dots,N_n,\dots,t) \\
&= (N_m+1) c(N_1,\dots,N_m,\dots,N_n,\dots,t)
\end{align}
generating the property
\begin{align}
[\hat{\mathbf{a}}_n, \hat{\mathbf{a}}_m^* ] = \hat{\mathbf{a}}_n \hat{\mathbf{a}}_m^*  - \hat{\mathbf{a}}_n^*  \hat{\mathbf{a}}_m = \delta _{nm}.
\end{align}

Using creation and annihilation operators, the Hamiltonian can be re-expressed by expanding $H_{m,n}$

\begin{align}
\hat{\mathbf{H}} &= \sum _{m,n} \hat{\mathbf{a}}_m H _{m,n} \hat{\mathbf{a}}^* _n  \\
&= \sum _{m,n} \hat{\mathbf{a}} _m \int dq  \psi_m^*(q)\hat{H} \psi _n (q) \hat{\mathbf{a}}^* _n \\ 
&= \int dq  (\sum_m \hat{\mathbf{a}}_m \psi_m^*(q)) \hat{H}(\sum_n \psi_n(q) \hat{\mathbf{a}}_n^{\dagger}) \\
&= \int dq \, \hat{\psi}^*(q) \hat{H} \hat{\psi}(q)
\end{align}

generating the wave function operators, referred to as quantum fields,

\begin{align}
\hat{\psi} = \sum_n  \hat{\mathbf{a}}_n \psi_n(q), \ \ \ \ \ \ \ \ \ \ \ \ \hat{\psi}^* = \sum_n  \hat{\mathbf{a}}_n^* \psi_n^*(q),
\end{align}
From the fact that $[\hat{\mathbf{a}} _m , \hat{\mathbf{a}}^* _n ] = \delta _{m,n}$, coupled with the completeness property of eigenfunctions, we note

\begin{align}
\delta(q' - q) &= \sum _n \psi^* _n (q') \psi _n (q) \\
&= \sum _{m,n} \delta _{m,n} \psi^*  _n (q') \psi _m (q) \\
&= \sum _{m,n} [\hat{\mathbf{a}}_m, \hat{\mathbf{a}}_n^* ] \psi_n^*(q')\psi_m(q) \\
&= \sum _{m,n} (\hat{\mathbf{a}}_m \hat{\mathbf{a}}_n^* - \hat{\mathbf{a}}_n^* \hat{\mathbf{a}}_m  ) \psi_n^*(q')\psi_m(q) \\
&= \sum _{m,n} \hat{\mathbf{a}}_m \hat{\mathbf{a}}_n^* \psi_n^*(q')\psi_m(q) - \sum _{m,n} \hat{\mathbf{a}}_n^*\hat{\mathbf{a}}_m  \psi_n^*(q')\psi_m(q)  \\
&= (\sum _m \hat{\mathbf{a}}_m \psi_m(q) )(\sum _n \hat{\mathbf{a}}_n^* \psi_n^*(q')) - (\sum _n \hat{\mathbf{a}}_n^* \psi_n^*(q')) (\sum _m \hat{\mathbf{a}}_m   \psi_m(q))  \\
&= [\hat{\psi}(q), \hat{\psi}^*(q') ] 
\end{align}

From the Lagrangian for the Schrodinger equation, formed by simply multiplying the Schrodinger equation by

\begin{align}
\delta \psi^*(q)
\end{align}

and treating it as the extremum of a Lagrangian varying w.r.t. $\psi^*$, we can derive the conjugate momentum:

\begin{align}
0 &= (\delta \psi^* )(i\frac{\partial \psi}{\partial t} - \hat{H}\psi) = \delta [ \psi^* (i\frac{\partial }{\partial t} - \hat{H})\psi ] = \delta \mathcal{L} = \delta_{\psi^*} \mathcal{L}, \\
\pi(q') &= \dfrac{\partial \mathcal{L}}{\partial (\partial_t \psi(q'))} = i \psi^*(q')
\end{align}

implying the above commutation relations, on multiplying across by $i$, give
\begin{align}
[\hat{\psi}(q), \hat{\pi}(q')] = i \delta(q' - q).
\end{align}

Tue, 12 Dec 2017 10:01 GMT