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The state $$|\Psi \rangle = |0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle $$ is coming from a far field and incident on a double slit setup. Here j is the index of j-th atom of the detector and $\omega$ is the index of the $\omega$-th mode of the field.

On the other side there is a detector. The setup looks as follows (Prof. Shih's book).

enter image description here

So far, I have calculated the following. The field at the detector is described as follows.

The positive field arrived at the detector is: $$E^{(+)}(x, z, t) = E^{(+)}_1(x, z, t) + E^{(+)}_2(x, z, t) $$

and the negative field arrived at the detector is:

$$E^{(-)}(x, z, t) = E^{(-)}_1(x, z, t) + E^{(-)}_2(x, z, t) $$

We can derive the terms separately (assuming single polarization).

The positive field at the detector coming from the $P_1$ slit is: $$E^{(+)}_1(x, z, t) = \sum_{\overrightarrow{k_1}} g(\overrightarrow{k_1}, z-z_0, t-t_0) E^{(+)}_1(\overrightarrow{k_1}, x_0, z_0, t_0)$$

$$ = \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0}$$

, the positive field at the detector coming from the $P_2$ slit is:

$$E^{(+)}_2(x, z, t) = \sum_{\overrightarrow{k_2}} g(\overrightarrow{k_2}, z-z_0, t-t_0) E^{(+)}_2(\overrightarrow{k_2}, x_0, z_0, t_0)$$

$$ = \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}$$

, the negative field at the detector coming from the $P_1$ slit is:

$$E^{(-)}_1(x, z, t) = \sum_{\overrightarrow{k_1}} g(\overrightarrow{k_1}, z-z_0, t-t_0) E^{(-)}(\overrightarrow{k_1}, x_0, z_0, t_0)$$

$$ = \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0}$$

and the negative field at the detector coming from the $P_2$ slit is:

$$E^{(-)}_2(x, z, t) = \sum_{\overrightarrow{k_2}} g(\overrightarrow{k_2}, z-z_0, t-t_0) E^{(-)}(\overrightarrow{k_2}, x_0, z_0, t_0)$$

$$ = \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0}$$

Now we plugin the values in original equations.

$$E^{(+)}(x, z, t) = E^{(+)}_1(x, z, t) + E^{(+)}_2(x, z, t) $$

$$= \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0} $$

and

$$E^{(-)}(x, z, t) = E^{(-)}_1(x, z, t) + E^{(-)}_2(x, z, t) $$

$$ = \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} $$

So, the correlation fuction for state $|\Psi \rangle = |0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle $ is:

$$G^{(1)} = $\langle \Psi | E^{(-)}(x, z, t) E^{(+)}(x, z, t) | \Psi \rangle$$

$$ = (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \left[ \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} \int^{\frac{b}{2}}_{-\frac{b}{2}} dx_0 e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle) $$

At this point I am confused how to solved this scary looking expression.

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2  
What is it that you're trying to calculate? Usually when expressions get that big it's time to sit back and ask if your'e on the right track. – lionelbrits Nov 30 '13 at 22:44
1  
seems to me that this is just some obscure form of wroting down the scalar diffraction integral for this system accounting for coherence in the wacefield. – Numrok Feb 24 at 19:52
1  
can i ask what exactly is meant by E+ and E-? could probsbly be worked out from your formulae, but that might take a while... – Numrok Feb 24 at 19:53
2  
also: the diffraction integrals for this systems can be solved exactly in the far field limit. small amgle approximation might be worth considering for convenience but is not necessary i think. – Numrok Feb 24 at 20:00
    
@OmarShehab: would you mind responding to my questions? I would like to answer this question and I think I can if you clarify what I pointed out. – Numrok Feb 29 at 20:57

Considering the size of the slit to be negligible the correlation function becomes: $$G^{(1)} = (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \left[ \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0(x)]}}{r_0(x)} e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0(x)]}}{r_0(x)} e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

$$= (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \left[ \sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0]}}{r_0} e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0]}}{r_0} e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} \frac{e^{-i[\omega_1(t-t_0)-k_1r_0]}}{r_0} e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} \frac{e^{-i[\omega_2(t-t_0)-k_2r_0]}}{r_0} e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

$$= (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \frac{1}{r_0} \left[ \sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]} e^{i(k_{1x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]} e^{i(k_{2x_0} -k_x)x_0} \times (-i) \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]} e^{i(k_{1x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]} e^{i(k_{2x_0} -k_x)x_0} \times i \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

$$= (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \frac{1}{r_0} \left[ \sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]+ i(k_{1x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]+i(k_{2x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]+i(k_{1x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]+i(k_{2x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

$$= (\langle0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |)| \frac{1}{r_0} \left[ \sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]+ i(k_{1x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_1}}\hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]+i(k_{2x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_2}}\hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right] \left[\sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t-t_0)-k_1r_0]+i(k_{1x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_1}}\hat{a}_{\overrightarrow{k_1}} . e^{-i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-t_0)-k_2r_0]+i(k_{2x_0} -k_x)x_0} \times \mathcal{E}_{\overrightarrow{k_2}}\hat{a}_{\overrightarrow{k_2}} . e^{-i \overrightarrow{k_2} r_0}\right] | (|0\rangle + \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

$$= (\langle 0| + \sum_j \int d\omega f_j(\omega)\hat{a}_j (\omega) \langle 0 |) | \frac{e^{-i k_x x_0}}{r_0} \left[ \sum_{\overrightarrow{k_1}} e^{-i[\omega_1(t - \frac{s_1}{c}) - k_1r_0] + ik_{1x_0}x_0} \times \\ \mathcal{E}_{\overrightarrow{k_1}} \hat{a}^\dagger_{\overrightarrow{k_1}} . e^{i \overrightarrow{k_1} r_0} + \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-\frac{s_2}{c}) - k_2r_0] \\ + i(k_{2x_0} x_0} \times \\ \mathcal{E}_{\overrightarrow{k_2}} \hat{a}^\dagger_{\overrightarrow{k_2}} . e^{i \overrightarrow{k_2} r_0} \right]$$

$$ \left[\sum_{\overrightarrow{k_1}} e^{-i [\omega_1(t - \frac{s_1}{c}) - k_1r_0] \\ + i(k_{1x_0} x_0} \times \\ \mathcal{E}_{\overrightarrow{k_1}} \hat{a}_{\overrightarrow{k_1}} \\ . e^{-i \overrightarrow{k_1} r_0} + \\ \sum_{\overrightarrow{k_2}} e^{-i[\omega_2(t-\frac{s_2}{c}) \\ - k_2r_0] + \\ i (k_{2x_0}x_0} \\ \times \\ \mathcal{E}_{\overrightarrow{k_2}} \hat{a}_{\overrightarrow{k_2}}\\ . e^{-i \overrightarrow{k_2} r_0} \right] \\ | (|0\rangle + \\ \sum_j \int d\omega f_j(\omega)\hat{a}^\dagger_j (\omega) |0\rangle)$$

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It's not complete. Can we consider the exponential terms to be unity? – Omar Shehab Apr 2 '13 at 14:53
1  
I agree with lionelbrits, what is it that your trying to calculate and why does it need to be so long and complex? – Bill Alsept Jun 6 at 0:31

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