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Whenever it is talked about phase difference or lack of it, double slit example is used as the example. But I want to know what is the physical manifestation of phase difference for a qubit (or spin half)? If it affects the interference then what is interference for qubit and how would phase difference affect it?
As a sample let's take $|\Psi\rangle=\frac{\sqrt{3}}{2}|u\rangle+\frac{1}{2}e^{\frac{-i\Pi}{3}}|d\rangle$ in which $e^{\frac{-i\Pi}{3}}$ is the phase difference.

Qmechanic
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al pal
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1 Answers1

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I will answer this by using the example you gave in the context of qubits. Consider, $$\left|\Psi_1\right> = \frac{\sqrt{3}}{2}\left|0\right> + e^{i\pi/3} \frac{1}{2}\left|1\right>$$ and the same state without the phase, $$\left|\Psi_2\right> = \frac{\sqrt{3}}{2}\left|0\right> + \frac{1}{2}\left|1\right>$$

The corresponding density matrices are,

$$\rho_1 = \begin{pmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4}e^{i\pi/3}\\ \frac{\sqrt{3}}{4}e^{-i\pi/3} & \frac{1}{4} \end{pmatrix} \quad \rho_2 = \begin{pmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{4} \end{pmatrix} $$ As you can see, the off-diagonal elements differ because of the difference in relative phase.

Now let's perform a measurement in $\left|+\right> =\frac{1}{\sqrt{2}} (\left|0\right> +\left|1\right>)$. The projector for this is,

$$P = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ For the first state $\left|\Psi_1\right>$, the measurement probability is, $$Tr(P\rho_1) = \frac{1}{2} + \frac{\sqrt{3}}{4}(\cos(\pi/3))$$

But for the second state $\left|\Psi_2\right>$,the measurement probability is, $$Tr(P\rho_2) = \frac{1}{2} + \frac{\sqrt{3}}{4}$$

The change in relative phase affects the statistics of this measurement.

Alternatively, you can check using c = $\left<+|\Psi\right>$ too followed by $c^*c$.

Discord Warrior
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