# What exactly is a phase vector?

The following $2\times 2$ matrix

$$P = \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix}$$

represents a quantum gate because it's a unitary matrix.

If we multiply $P$ by the quantum state $\lvert \psi\rangle = \alpha \lvert 0\rangle + \beta \lvert 1\rangle$, we obtain ${\lvert \psi\rangle}_P = \alpha e^{i\theta} \lvert 0 \rangle + \beta e^{i\phi} \lvert 1\rangle$, which can be derived as follows

\begin{align} {\lvert \psi\rangle}_P &= \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix} \alpha \lvert 0\rangle + \beta \lvert 1\rangle \\ &= \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix} \alpha \lvert 0\rangle + \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix} \beta \lvert 1\rangle \\ &= \alpha \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{i\phi} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\ &= \alpha \begin{bmatrix} e^{i\theta} \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 0 \\ e^{i\phi} \end{bmatrix} \\ &= \alpha e^{i\theta} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta e^{i\phi} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\ &= \alpha e^{i\theta} \lvert 0 \rangle + \beta e^{i\phi} \lvert 1\rangle \end{align}

If we tried to measure ${\lvert \psi\rangle}_P$, we would obtain the computational basis state $\lvert 0 \rangle$ with probability $|\alpha|^2$ and the computational basis state $\lvert 1 \rangle$ with probability $|\beta |^2$. So, there's no difference between measuring ${\lvert \psi\rangle}_P$ or $\lvert \psi\rangle$, in terms of probabilities of obtaining one rather than the other computational basis state.

The reason to obtain same probabilities is because $e^{i\theta}$ and $e^{i\phi}$ are phase vectors, so they do not affect the probabilities.

$e^{i\theta}$ and $e^{i\phi}$ represent complex numbers, as vectors, in the complex plane. This can be easily visualized from the following picture

But what's the intuitive meaning of multiplying the "vectors" $e^{i\phi}$ by a computational basis state? In general, what is a phase and a phase vector in this context and how does it affect the mathematics and the basis vectors? What's the relation between $\lvert \psi\rangle$ and ${\lvert \psi\rangle}_P$?

• Geometrically, the $P$ gate is rotation around $Z$ axis by the angle $\phi-\theta$ on the Bloch sphere. If $\phi-\theta=\pi$ it is Pauli Z gate (up to a global phase $e^{i\theta}$ which has no physical meaning). – kludg Apr 24 '18 at 18:10

There are a few different things that you may be confusing.

Why are objects of the form $e^{i\phi}$ actually called vectors in this context?

A complex number can always be expressed as a vector in $\mathbb R^2$, because $\mathbb C$ is nothing but $\mathbb R^2$ with a particular product defined between its elements. Note that this has nothing to do with quantum mechanics or physics, it is just how complex numbers are defined.

in general what is a phase (in the context of quantum mechanics)?

You can think of a phase as a number that characterises how different modes interfere with each other. While as you noted adding a phase doesn't change the output probabilities in a fixed basis, it does change the output probabilities as soon as you measure in a different basis.

What is the relation between $|\psi\rangle$ and $|\psi\rangle_P$?

They are just two different states. As noted above, while the probabilities of measuring $|0\rangle$ or $|1\rangle$ are the same for these states, as soon as you measure in a different basis you will see that they behave differently. For example, you can easily verify that $|\psi\rangle$ and $|\psi\rangle_P$ correspond to different probabilities of measuring the outcome $|+\rangle\equiv\frac{1}{\sqrt2}(|0\rangle+|1\rangle)$.

• Sorry, I shouldn't have asked the question "Why are objects of the form $e^{i\phi}$ actually called vectors in this context". As I also say in my post, $e^{i\phi}$ can easily be seen to be a vector in the complex plane. So, clearly, that question shouldn't be part of my post and I will remove it. – nbro Apr 24 '18 at 17:16