**Background**

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So basically there is a dictionary/correspondence between first and second quantised quantum mechanical theories. When I suddenly turn on a potential in potential in first quantised theories I can use the sudden approximation and find the probability of the final state. I can use the dictionary to find a result in the second quantised theory as well. My question is what happens when the potential turned on is such that the particle number is not conserved? Is there a way to extend the approximation?

**Calculations**

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Let us restrict our discussion to bosons and adopt the convention First Quantised $\leftrightarrow $ Second Quantised Theory (We are following these Ashok Sen's [Quantum Field Theory I][1] of HRI institute also on my [google drive][2]). Consider a single particle system's hamiltonian $\hat h$ and energy operator $\hat E$:

$$ \hat h \psi = \hat E \psi $$

with energy eigenstates $u_i$ and eigenvalues $e_n$

$$ \hat h u_n = e_n u_n$$

Now moving to an assembly of quantum mechanical Hamiltonians (many body system) $\sum_i h_i$ with an interacting potential $\hat v_{ik}$ corresponds to a second quantised version (Page 16):

$$ \hat H_N = \sum_{i=1}^N \hat h_i + \frac{1}{2}\sum_{\substack i\neq j

\\ i,j=1 }^N \hat v_{i,j} \leftrightarrow \sum_{n=1}^\infty e_n a_n^\dagger a_n + \frac{1}{2} \sum_{m,n,p,q=1}^\infty \Big(\int \int d^3 r_1 d^3 r_2 u_{m}(\vec r_1)^* u_{n}(\vec r_2)^* \hat v_{12} u_{m}(\vec r_1) u_{p}(\vec r_2) \Big) a^\dagger_m a^\dagger_n a_p a_q$$

where $e_n$ i the $n$th energy eigevalue, $u_i$ are the one-particle eigenstates and $a_i^\dagger$ is the creation operator of the $i$'th particle and they obey the energy eigenvalue system $h_i u_n = e_n u_n$. The symmetric wave function for $H_N$ corresponds as a second quantised version (Page 6):

$$ u_{n_1,n_2,\dots n_N} \equiv \frac{1}{\sqrt{N!}} \sum_{\text{Permutations of $r_1,\dots,r_N$}}u_{n_1} (\vec r_1) \dots u_{n_N} (\vec r_N) \leftrightarrow (a_1^\dagger)^{n_1} (a_2^\dagger)^{n_2} \dots (a_N^\dagger)^{n_N} |0 \rangle$$

The potential corresponds for one body operator (Page 14):

$$ \sum_{i=1}^N \hat B_{i} \leftrightarrow \sum_{n,p = 1}^\infty \Big( \int d^3 r_1 u_n^* (\vec r_1) B_{1} u_p (\vec r_1) \Big) a^\dagger_n a_p$$

Now, let's say I suddenly turn on the one body operator (potential) in the assembly of quantum mechanical systems. Then:

$$ H'_N = \sum_{i=1}^N \hat h_i + \frac{1}{2}\sum_{\substack i\neq j

\\ i,j=1 }^N \hat v_{i,j} + \sum_i^N \hat B_{i} $$

Let the new energy eigenstates be $\tilde u_n$ with the original state be $u_{n_1,n_2,\dots,n_N}$. Then the final state being in $\tilde u_{n'_1,n'_2,\dots,n'_N}$ has probability $| \int \tilde u_{n'_1,n'_2,\dots,n'_N}u^*_{n_1,n_2,\dots,n_N} d^3 r|^2 $. Now using the dictionary we can find the corresponding $2$nd quantised state. This is nothing but the **sudden approximation**.

**Question**

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Is there a way to extend the sudden approximation to potentials which do not conserve particle number?