Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer 1

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

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

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

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

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

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

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

share|improve this answer
    
It's not complete. Can we consider the exponential terms to be unity? –  Omar Shehab Apr 2 '13 at 14:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.