\input phyzzx

\def\a{\hat a^{\vphantom\dagger}}
\def\ad{\hat a^\dagger}
\def\bx{{\bf x}}
\def\hilb{{\cal H}^B}
\def\eqdef{\buildrel\rm def\over=}

\hsize=6.8in	\hoffset=-0.2in	%\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#1.\quad
Due September 18, 2000.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Bosonic creation and annihilation operators $\ad_\alpha$ and $\a_\alpha$
were defined in class in terms of their respective matrix elements in
the occupation-number basis of the bosonic Fock space ${\cal F}^B$:
$$
\eqalign{
\bra{\{n'_\beta\}_\beta} \ad_\alpha \ket{\{n_\beta\}_\beta}\ &
=\ \cases{
	\sqrt{n_\alpha+1} & provided all $n'_\beta=n_\beta+\delta_{\alpha,\beta}$ and\cr
	0 & otherwise;
    }\cr
\bra{\{n'_\beta\}_\beta}  \a_\alpha \ket{\{n_\beta\}_\beta}\ &
=\ \cases{
	\sqrt{n_\alpha} & provided all $n'_\beta=n_\beta-\delta_{\alpha,\beta}$ and\cr
	0 & otherwise.
    }\cr
}\eqn\operadef
$$
This exercise is about the way these operators acts on coordinate-space wave functions
of multi-particle states.

\spointbegin
Consider an $N$--boson state of the form
$\ket{(\alpha_1,\alpha_2,\ldots,\alpha_N)}=\ket{\{n_\beta\}_\beta}$.
Show that the coordinate-space wave function of this state has form
$$
\Psi(\bx_1,\bx_2,\ldots,\bx_N)\
=\ {1\over\sqrt{N!\,\prod_\beta n_\beta!}}
\sum \varphi_{\alpha_1}(\bx_{\nu_1})\,\varphi_{\alpha_2}(\bx_{\nu_2})\,
\cdots\,\varphi_{\alpha_N}(\bx_{\nu_N})
\eqn\WaveFunction
$$
where the sum is over all $N!$ permutations $\bx_{\nu_1},\bx_{\nu_2},\ldots,\bx_{\nu_N}$ of
the particle positions $\bx_1,\bx_2,\ldots,\bx_N$.
(In other words, $\nu_1,\nu_2,\ldots,\nu_N$ are summations indices running over all $N!$
permutations of the integers $1,2,\ldots,N$.)
\subpar
Hint: permuting the positions is equivalent to permuting the 1-particle
wave functions $\varphi_{\alpha_1},\ldots,\varphi_{\alpha_N}$, but watch for coincident terms
on the right hand side of eq.~\WaveFunction.

\spoint
Now consider a generic $N$--particle state $\ket{N,\Psi}\in\hilb_N$ with a generic
wave function $\Psi(\bx_1,\bx_2,\ldots,\bx_N)$;
more precisely, $\Psi(\bx_1,\bx_2,\ldots,\bx_N)$ must be totally symmetric with respect to the $N$
positions $\bx_1,\ldots,\bx_N$ but otherwise, it is completely generic.
\subpar
Show that a creation operator $\ad_\alpha$ acting on this state produces an $(N+1)$ particle state
$\ket{N+1,\Psi'}\in\hilb_{N+1}$ with a totally symmetric wave function
$$
\!\eqalign{
\Psi'(\bx_1,\bx_2,\ldots,\bx_{N+1})\ &
=\ {1\over\sqrt{N+1}} \sum_{i=1}^{N+1}\,
	\varphi_\alpha(\bx_i)\,\Psi(\bx_1,\ldots,\not\!\bx_i,\ldots,\bx_{N+1})\cr
{}\eqdef\ {1\over\sqrt{N+1}}\ \Bigl[ &
\varphi_\alpha(\bx_1)\,\Psi(\bx_2,\ldots,\bx_{N+1})\,
	+\,\varphi_\alpha(\bx_2)\,\Psi(\bx_1,\bx_3\ldots,\bx_{N+1})\,
	+\,\cdots\cr
&+\,\varphi_\alpha(\bx_N)\,\Psi(\bx_1,\ldots,\bx_{N-1},\bx_{N+1})\,
	+\,\varphi_\alpha(\bx_{N+1})\,\Psi(\bx_1,\ldots,\bx_N) \Bigr] .\cr
}\!\eqn\ADaction
$$
Note that while $\Psi$ is a function of $N$ positions, $\Psi'$ is a function of $N+1$ positions.
In particular, for $N=0$, $\Psi$ is simply a complex number but $\Psi'$ is a 1-particle wave function,
$\Psi'(\bx_1)=\varphi_\alpha(\bx_1)\times\Psi$.
\subpar
Hint: First prove \ADaction\ for the wave-functions $\Psi$ of the form \WaveFunction\ ---
and do not forget to verify the normalization of the resulting $\Psi'$  --- then use the fact that
the states $\ket{(\alpha_1,\alpha_2,\ldots,\alpha_N)}$ constitute a basis of the $\hilb_N$.

\spoint
Next, consider the annihilation operators $\a_\alpha$ and show that the wave function $\Psi''$
of the $N-1$ particle state $\ket{N-1,\Psi''}=\a_\alpha\ket{N,\Psi}\in\hilb_{N-1}$ can be written as
$$
\Psi''(\bx_1,\ldots,\bx_{N-1})\
=\ \sqrt{N}\int\!\!d\bx_N\,\varphi^*_\alpha(\bx_N)\,\Psi(\bx_1,\ldots,\bx_{N-1},\bx_N) .
\eqn\Aaction
$$
In particular, for $N=1$, $\Psi$ is a 1--particle wave function while $\Psi''$ is a number,
$\Psi''=\vev{\alpha|\Psi}$.
For $N=0$, formula~\Aaction\ degenerates to $\Psi''=0$ (since $\sqrt{(N=0)}=0$), which agrees with
$\a_\alpha\ket{0}=0$ (and hence $\a_\alpha\ket{N=0,\Psi}=0$ for any $\Psi$), although in this case
$\Psi''$ is rather ill-defined as a function.

\bigskip
\point
Formul\ae\ \Aaction\ and \ADaction\ allow for straightforward translation between first-quantized and
second-quantized forms of various operators.
In particular, consider an additive one-particle operator of the form
$$
\hat A_{\rm tot}(N\ {\rm particles})\
= \sum_{i=1}^N\, \hat A_1(i\coeff{\rm th}{}\ {\rm particle}).
\eqn\AddOPO
$$
As argued in class, the second-quantized form of such an operator is
$$
\hat A\ = \sum_{\alpha,\beta}\bra{\alpha}\hat A_1\ket\beta\,\ad_\alpha\a_\beta\,.
\eqn\SecondOPO
$$

\spointbegin
Use formul\ae\ \Aaction\ and \ADaction\ to explicitly calculate the wave function
$\Psi'(\bx_1,\ldots,\bx_N)$ of the $N$ particle state $\ket{N,\Psi'}=\hat A\ket{N,\Psi}$
(assume generic totally symmetric $\Psi(\bx_1,\ldots,\bx_N)$)
and show that in the first-quantized formalism, $\hat A_{\rm tot}$ acting on $\Psi$ yields
exactly same $\Psi'$.
\subpar
Hint: Prove and use
$$
\displaylines{
\hat A_1(i\coeff{\rm th}{}\ {\rm particle})\Psi(\bx_1,\ldots,\bx_i,\ldots,\bx_N)
\hfill\eqname\ithOPO\cr
{}\hfill = \sum_{\alpha,\beta}\bra{\alpha}\hat A_1\ket\beta\,
    \varphi_\alpha(\bx_i)\!\int\!\!d\bx'_i\,\varphi^*_\beta(\bx'_i)\,
    \Psi(\bx_1,\ldots,\bx'_i,\ldots,\bx_N) .\qquad
}$$

\pointcon
Now consider an additive two-particle interaction operator such as
$$
\hat V_{\rm tot}\ =\ \half\sum_{i\neq j} V(\hat\bx_i-\hat\bx_j)
$$
for some two-body potential $V(\bx_i-\bx_j)$.
More generally, one has a two-particle operator $\hat A_2$ involving positions or other quantum
numbers of two particles and the total $\hat A$ of an $N$ particle system is given by
$$
\hat A_{\rm total}\
=\ \half\sum_{i,j=1,\ldots,N\atop i\neq j}
\hat A_2(i\coeff{\rm th}{}\ {\rm and}\ j\coeff{\rm th}{}\ {\rm particles}).
\eqn\AddTPO
$$
The second quantized form of such an additive two-particle operator is given by
$$
\hat A\ =\ \half\sum_{\alpha,\beta,\gamma,\delta} \bra{\alpha\otimes\beta}\hat A_2\ket{\gamma\otimes\delta}\,
\ad_\alpha\ad_\beta\a_\gamma\a_\delta
\eqn\SecondTPO
$$
where $\bra{\alpha\otimes\beta}$ and $\ket{\gamma\otimes\delta}$ are non-symmetrized
two-distinct-particles states
whose respective wave functions are simply $\varphi^*_\alpha(\bx_1)\varphi^*_\beta(\bx_2)$
and $\varphi_\gamma(\bx_1)\varphi_\delta(\bx_2)$.

\spoint
Verify by an explicit wave-function calculation
that the operators \AddTPO\ and \SecondTPO\ indeed produce identical results when acting
on any $N$ particle state $\ket{N,\Psi}$.
\subpar
Note special cases of $N=0$ or $N=1$ where both \AddTPO\ and \SecondTPO\ yield 0.
For $N\ge2$, use formul\ae\ \Aaction\ and \ADaction\ and a suitable analog of eq.~\ithOPO.


\bigskip
\point %3
Finally, an exercise in using the bosonic commutation relations
$$
[\a_\alpha,\a_\beta]\ =\ [\ad_\alpha,\ad_\beta]\ =\ 0,\quad
[\a_\alpha,\ad_\beta]\ =\ \delta_{\alpha,\beta}\,.
\eqn\CCR
$$

\spointbegin
Calculate the commutators $[\ad_\alpha\a_\beta,\ad_\gamma]$, $[\ad_\alpha\a_\beta,\a_\delta]$
and $[\ad_\alpha\a_\beta,\ad_\gamma\a_\delta]$.

\spoint
Consider to one-particle operators $\hat A_1$ and $\hat B_1$ and let $\hat C_1$ be their
commutator, $\hat C_1=[\hat A_1,\hat B_1]$.
Let $\hat A$, $\hat B$ and $\hat C$ be the second-quantized forms of the respective additive operators,
\cf\ eq.~\SecondOPO.
\subpar
Show that $[\hat A,\hat B]=\hat C$.

\spoint
Next, calculate the commutator $[\ad_\mu\a_\nu,\ad_\alpha\ad_\beta\a_\gamma\a_\delta]$.

\spoint
Finally, consider a one-particle operator $\hat A_1$, a two-particle operator $\hat B_2$
and a two-particle operator $\hat C_2=\left[\bigl(\hat A_1(1\coeff{\rm st}{})
 +\hat A_1(2\coeff{\rm nd}{})\bigr),\hat B_2\right]$.
Show that in this case the second-quantized $\hat C$ is again the commutator of the second-quantized
$\hat A$ with the second-quantized $\hat B$.


\bye

