# How to measure relative phases of quantum states

If I have a large number of identical systems in identical quantum state $\Psi$ and an observable $A$ whose eigenstates are $\alpha_n$:

$$\Psi = \sum_n c_n \alpha_n$$

I can get absolute values of $|c_n|^2$ by measuring $A$, but with which experiment I can get relative phases of $c_n$?

• arxiv.org/pdf/1410.0916v1.pdf Commented Jun 9, 2016 at 13:30
• @valerio92 could you maybe summarize the article? I'm not a professional physicist.
– xaxa
Commented Jun 9, 2016 at 16:17

The phase ambiguity is a bit worse than you think. There is a global phase ambiguity in $|\Psi⟩$, for sure, but if the state $$|\Psi⟩=\sum_n c_n |\alpha_n⟩ \tag 1$$ is all you have around, then there is also a phase ambiguity in the phase of each individual $c_n$. This is because if you change $|\alpha_n⟩$ to $|\alpha'_n⟩=e^{i\theta_n}|\alpha_n⟩$, the transformed state is also an eigenstate of $A$ with eigenstate $|\alpha_n⟩$, so the two have exactly equal standing. If your universe only ever contains $|\Psi⟩$ (and a measurement device for $A$), then you only ever care about the absolute values $|c_n|^2$ of the coefficients.

The phase of these coefficients comes into play if you have two or more states, i.e. if you introduce some second state $$|\Phi⟩=\sum_n b_n |\alpha_n⟩ \tag 2$$ into the mix. Here transforming the eigenstates by some phase, as before, will still change the phases of the individual coefficients, but it will leave their differences $\arg(c_n)-\arg(b_n)$ unchanged. These are the phase differences that can be detected by experiment.

So, to come to the question, how do you detect them? There are a bunch of ways, but they all involve interference in some way. As a simple example, take your two states $|\Phi⟩$ and $|\Psi⟩$, and a superposition of them, $$\frac{|\Phi⟩+|\Psi⟩}{\sqrt 2} = \sum_n \frac{c_n+b_n}{\sqrt 2} |\alpha_n⟩,\tag 3$$ and measure $A$. Then you will measure the eigenvalue $\alpha_n$ with a probability $$P_n=\frac12 |c_n+b_n|^2$$ which is sensitive to the relative phase of the two coefficients.

• That's a nice catch about phases of eigenstates... Now I'm a bit confused: to which one of them does a system collapse after measurement according to QM? Of a set of systems in identical state, will each one collapse to the same $e^{i\theta_n}\alpha_n$ or $\theta_n$ may be different?
– xaxa
Commented Jun 9, 2016 at 15:24
• They're the same physical state, because they only differ by a phase. The phase would matter if this was one branch of an interference experiment, but because you've done a projective measurement, all the coherence gets destroyed and you won't see any interference. Commented Jun 9, 2016 at 15:50
• Seems like I'm missing something... Can't they be used as inputs in interference experiment later - after I've done the measurement?
– xaxa
Commented Jun 9, 2016 at 16:13
• Interferometers always measure phase differences, and those are always reset during a projective measurement. For more details, consult your favourite QM textbook. Commented Jun 9, 2016 at 16:50

## Relative Phase Determination with Quantum Circuit (Approach)

$$| \psi \rangle = \cos\left(\frac{\theta}{2}\right) |{0}\rangle + e^{-i\phi}\sin\left(\frac{\theta}{2}\right) |{1}\rangle$$

In this case, the operator $$A=Z$$ the $$\sigma^z$$ Pauli matrix whose eigen-states are indeed $$|0\rangle$$ and $$|1\rangle$$. The relative phase $$\phi$$ between components is computationally important (Represent the state in the Bloch sphere). However, if you have only $$Z$$ operator in the world, you can only measure the angle $$\theta$$ without knowing any information about $$\phi$$. To determine $$\phi$$, it is necessary to rotate the state in the Bloch sphere or simply mix the amplitudes of the state. Namely, we need both $$R_y\left(\frac{\pi}{2}\right)$$ and $$R_x\left(\frac{\pi}{2}\right)$$: $$R_y\left(\frac{\pi}{2}\right)| \psi \rangle = \frac{1}{2} \left( \cos\left(\frac{\theta}{2}\right) + e^{-i\phi}\sin\left(\frac{\theta}{2}\right) \right) |{0}\rangle + \frac{1}{2} \left( \cos\left(\frac{\theta}{2}\right) - e^{-i\phi}\sin\left(\frac{\theta}{2}\right) \right) |{1}\rangle$$

In this case, the probability of obtaining $$|0\rangle$$ by measuring $$Z$$ is : $$p_0 = \frac{1}{2}+\cos(\phi)\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$$ and the probability of obtaining $$|1\rangle$$ by measurering $$Z$$ is : $$p_1 = \frac{1}{2}-\cos(\phi)\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$$

Determining $$\cos(\phi)$$ is not enough. Hence, it is necessary to apply $$R_x\left(\frac{\pi}{2}\right)$$:

$$R_x\left(\frac{\pi}{2}\right)| \psi \rangle = \frac{1}{2} \left( \cos\left(\frac{\theta}{2}\right) + ie^{-i\phi}\sin\left(\frac{\theta}{2}\right) \right) |{0}\rangle + \frac{1}{2} \left( \cos\left(\frac{\theta}{2}\right) - ie^{-i\phi}\sin\left(\frac{\theta}{2}\right) \right) |{1}\rangle$$.

In this case, the probability of obtaining $$|0\rangle$$ by measuring $$Z$$ is : $$p_0 = \frac{1}{2}+\sin(\phi)\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$$ and the probability of obtaining $$|1\rangle$$ by measurering $$Z$$ is : $$p_1 = \frac{1}{2}-\sin(\phi)\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$$

By this way, the relative phase angle $$\phi$$ is determined.

In general, the quantum state is multi-dimensional in Hilbert space: $$|\Psi\rangle = \sum_{i=0}^{2^N - 1} \alpha_i |i\rangle$$
But, it is possible to write that state in the form of two ortho-normal states: For any eigenstate $$|k\rangle$$,
$$|\Psi\rangle = \alpha_k |k\rangle + \bar{\alpha}_k |\bar{k}\rangle$$ such that: $$|\bar{k}\rangle = \frac{1}{\sum_{i\neq k} |\alpha_i|^2} \sum_{i\neq k} \alpha_i |i\rangle$$ $$\bar{\alpha}_k = \sum_{i\neq k} |\alpha_i|^2$$
Now, we can redo the previous protocol by writing: $$\bar{\alpha}_k = \cos\left(\frac{\theta}{2}\right)$$ $${\alpha}_k = e^{-i\phi}\sin\left(\frac{\theta}{2}\right)$$
We repeat this protocol iteratively for $$k=0,1,...,2^N-1$$.