Two slit experiment - quantum theory - problem

dageki
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Hi, I'm new and I'm from Poland.
I have problem with equation(average number of photons registered behind pinhole 1 in two slit experiment):
\bar{n}_1=\langle n|a_{1}^{+}a_{1}|n\rangle=\frac{\langle 0|(a^{+})^{n}a_{1}^{+}a_{1}(a^{+})^{n}|0\rangle}{n!}
using:
a^{+}=\frac{(a_{1}^{+}+a_{2}^{+})}{ \sqrt{2}}
and
a=\frac{(a_{1}+a_{2})}{\sqrt{2}}
and
|n\rangle=\frac{1}{\sqrt{n!}}(a^{+})^{n}|0\rangle
and using usual creation and destruction oprator properties, to give finally :
\bar{n}_1=\frac{1}{2}

I have no idea how to do it...
Big thnx
 
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dageki said:
Hi, I'm new and I'm from Poland.
I have problem with equation(average number of photons registered behind pinhole 1 in two slit experiment):
\bar{n}_1=\langle n|a_{1}^{+}a_{1}|n\rangle=\frac{\langle 0|(a^{+})^{n}a_{1}^{+}a_{1}(a^{+})^{n}|0\rangle}{n!}
using:
a^{+}=\frac{(a_{1}^{+}+a_{2}^{+})}{ \sqrt{2}}
and
a=\frac{(a_{1}+a_{2})}{\sqrt{2}}
and
|n\rangle=\frac{1}{\sqrt{n!}}(a^{+})^{n}|0\rangle
and using usual creation and destruction oprator properties, to give finally :
\bar{n}_1=\frac{1}{2}

I have no idea how to do it...
Big thnx

Welcome
Hi i think that you must specify that whether there is a detector or not at any of the slits and whether the other slit is open or not.Other wise refer volume three of feynman lectures
 
We have a stream of photons incident on a pair of identical pinholes. We assume that only a single mode of the cavity (cavity formed by the lens and the first screen) is excited, with photon creation and destruction operators

a^{+}

and

a

We suppose that the two piholes provide the only means for photons in the cavity.For pinholes of equal size, the incident photons are equally likely to be registered in mode 1 or 2.

Still I don't know how to prove it.
I will be grateful for help.
 
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