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

                                                     enter image description here

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 ${\lvert \psi\rangle}_P$?

  • 1
    $\begingroup$ 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). $\endgroup$ – 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 $|+\rangle\equiv\frac{1}{\sqrt2}(|0\rangle+|1\rangle)$.

  • $\begingroup$ 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. $\endgroup$ – nbro Apr 24 '18 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.