# What is the physical significance of phase difference in qubit or spin half? [duplicate]

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.

• I asked kind of the same questions a couple of years ago, I hope it helps :) . physics.stackexchange.com/questions/177588/… . TL;DR : it changes the probabilities of measuring states in other bases Commented Aug 12, 2021 at 4:54
• Thanks :) but since I’m new to QM and that one is talking about energy and spatial probability I would appreciate if someone can describe it for me in terms of qubit and spin. Commented Aug 12, 2021 at 5:15
• @GiorgioP It does in a way, but like I said above it doesn't align with the level of knowledge I have about QM, so in some sense, no it doesn't. The answer Discord Warrior gave is what I was looking for which I am studying now. Thanks though. Commented Aug 12, 2021 at 15:49

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 $$c^*c$$.

• Thanks for the great answer. So is it right to conclude (like Ofek Gillon said in above comment) that the phase difference changes the probabilities of measuring states in other basis? And if that is correct does that mean if we had measured both states in whatever basis they were originally, the probability of measuring $|0\rangle$ and $|1\rangle$ would have been 3/4 and 1/4 correspondingly? Commented Aug 12, 2021 at 17:10
• I am concluding so because I have learned that in $r_1|0\rangle+r_2 e^{I\phi}|1\rangle$, $r_1^2$ and $r_2^2$ represent the probabilities of $|0\rangle$ and $|1\rangle$. Commented Aug 12, 2021 at 17:30
• Yes, that's right. More accurately we take the probability of the state $\left|\psi\right> = r_1\left|0\right> + e^{i\phi}r_2\left|1\right>$ to be in one of the states to be $p^*p$ where $p(0/1)= \left<0/1|\psi\right>$. Here the phase goes away because of $p^*p$ cancels the $e^{i\phi}$ term. And we're just left with $r_1^2$ and $r_2^{2}$ Commented Aug 12, 2021 at 17:39