Question

Q1 let = Jeno R(6) 147 (41 Rio) e to do SREEJICHIR Where 14>= alo> tbli> e 20/ and Rlo) = le 20/2 e Compute the integral and show that it can be written as lap? abtē lb 1² 2 こ a bēr

Answer 1

1) To perform the calculation first we need to write the state as a column vector, therefore let

$|0\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix},\; |1\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix} \quad (1)$

Hence

$|\psi\rangle = a|0\rangle + b|1\rangle = \begin{pmatrix} a\\ b \end{pmatrix} \quad (2)$

$|\psi\rangle\langle\psi | = \begin{pmatrix} a\\ b \end{pmatrix}\begin{pmatrix} a^{*} & b^{*} \end{pmatrix} = \begin{pmatrix} |a|^{2} & ab^{*}\\ a^{*} b& |b|^{2} \end{pmatrix} \quad (3)$

Finally

-1/2 0 e i0/2 R(O)(v|R* (0) Talab atb 1612 0 ei /2 0 e-10/2

e-i8/21a12 e-io/2ab* el/2a*b ei8/21612 :) (6.52 |a12 eWat b ei ab* 1612 (4)

The integral becomes

$\rho = \frac{1}{\sqrt{4\pi\sigma}}\int_{-\infty}^{\infty}R(\theta)|\psi\rangle\langle\psi |R^{*}\left(\theta \right )e^{-\theta^{2}/4\sigma}d\theta = \frac{1}{\sqrt{4\pi\sigma}}\int_{-\infty}^{\infty} \begin{pmatrix} |a|^{2} & e^{-i\theta}ab^{*}\\ e^{i\theta}a^{*}b & |b|^{2} \end{pmatrix}e^{-\theta^{2}/4\sigma}d\theta$

$= \frac{1}{\sqrt{4\pi\sigma}} \begin{pmatrix} \int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}|a|^{2}d\theta & \int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}e^{-i\theta}ab^{*}d\theta\\ \int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}e^{i\theta}a^{*}bd\theta & \int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}|b|^{2}d\theta \end{pmatrix} \quad (5)$

Consider the element in first row and first column of the matrix

$\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}|a|^{2}d\theta = |a|^{2}\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}d\theta = |a|^{2}\sqrt{\frac{\pi}{\frac{1}{4\sigma}}} = |a|^{2}\sqrt{4\pi\sigma} \quad (6)$

I have used the identity for Gaussian integral

$\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx = \sqrt{\frac{\pi}{\alpha}} \quad (7)$

Using the same identity, we get

$\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}|b|^{2}d\theta = |b|^{2}\sqrt{4\pi\sigma} \quad (8)$

Now consider the element in first row and second column of the matrix

$\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}e^{-i\theta}ab^{*}d\theta = ab^{*}\int_{-\infty}^{\infty}\exp\left(-\frac{1}{4\sigma}\left(\theta^{2}+i4\sigma\theta \right )\right)d\theta = ab^{*}\int_{-\infty}^{\infty}\exp\left(-\frac{1}{4\sigma}\left(\theta^{2}+2*i2\sigma*\theta+(i2\sigma)^{2}-(i2\sigma)^{2} \right )\right)d\theta$

$= ab^{*}e^{-\sigma}\int_{-\infty}^{\infty}\exp\left(-\frac{1}{4\sigma}\left(\theta+i2\sigma \right )^{2}\right)d\theta \quad (9)$

Substituting $\theta\to \theta+i2\sigma$ in last integral of equation (9) ,we get

$\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}e^{-i\theta}ab^{*}d\theta = ab^{*}e^{-\sigma}\int_{-\infty}^{\infty}\exp\left(-\frac{\theta^{2}}{4\sigma} \right )d\theta = ab^{*}e^{-\sigma}\sqrt{4\pi\sigma} \quad (10)$

Similarly we can show that

$\int_{-\infty}^{\infty}e^{-\theta^{2}/4\sigma}e^{-i\theta}a^{*}bd\theta = a^{*}be^{-\sigma}\sqrt{4\pi\sigma} \quad (11)$

Hence

$\rho = \frac{1}{\sqrt{4\pi\sigma}}\begin{pmatrix} \sqrt{4\pi\sigma}|a|^{2} & \sqrt{4\pi\sigma}e^{-\sigma}ab^{*}\\ \sqrt{4\pi\sigma}e^{-\sigma}a^{*}b & \sqrt{4\pi\sigma}|b|^{2} \end{pmatrix} = \begin{pmatrix} |a|^{2} & e^{-\sigma}ab^{*}\\ e^{-\sigma}a^{*}b & |b|^{2} \end{pmatrix} \quad (12)$