# Global phases and indistinguishable quantum states, mathematical understanding

Im trying to mathematically understand this: "All four states are mathematically identical, up to a global phase, and global phases do not distinguish quantum states. " $$\displaystyle \frac{|0\rangle +|1\rangle }{\sqrt {2}}$$

$$\displaystyle 2\sin (\pi /4) |0\rangle + \sin (\pi /4)|1\rangle + \cos (3\pi /4)|0\rangle$$

$$\displaystyle \frac{-1}{\sqrt {2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$$

$$\displaystyle \frac{e^{i\pi /2}+1}{2} |0\rangle + \frac{e^{i\pi /2}-1}{2i}|1\rangle$$

I get the idea, the global phase doesn't matter, but the mathematics is above my current understanding. I will appreciate some help

## 1 Answer

$$\newcommand{\ket}{\left|#1\right>}$$ $$\newcommand{\bra}{\left<#1\right|}$$ $$\newcommand{\expect}{\left<#1\right>}$$

Let's take them one at a time, with the understanding that

$$\ket{0} = \begin{pmatrix} 1\\0\end{pmatrix}$$ and $$\ket{1} = \begin{pmatrix} 0\\1\end{pmatrix}$$

1. $$\ket{\psi_1} = \frac{1}{\sqrt{2}} \left(\ket{0} + \ket{1}\right)$$

2. \begin{align} \ket{\psi_2} &= 2\sin(\pi/4) \ket{0} + \sin(\pi/4) \ket{1} + \cos(3\pi/4)\ket{0}\\ &= \frac{2}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1} - \frac{1}{\sqrt{2}}\ket{0}\\ &= \frac{1}{\sqrt{2}} \left(\ket{0} + \ket{1}\right)\\ &= \ket{\psi_1} \end{align}

which is identical to the original state.

1. \begin{align} \ket{\psi_3} &= -\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1 \end{pmatrix}\\ &= -\frac{1}{\sqrt{2}}\left(\begin{pmatrix} 1\\0 \end{pmatrix} + \begin{pmatrix} 0\\1 \end{pmatrix}\right)\\ &= -\frac{1}{\sqrt{2}} \left(\ket{0} + \ket{1}\right)\\ &= -\ket{\psi_1} \\ &= e^{i\pi} \ket{\psi_1} \end{align}

which differs from $\ket{\psi_1}$ by an overall phase.

1. \begin{align} \ket{\psi_4} &= \frac{e^{i\pi/2}+1}{2} \ket{0} + \frac{e^{i\pi/2}-1}{2i} \ket{1} \tag{1}\\ &= \frac{i+1}{2}\ket{0} + \frac{i-1}{2i} \ket{1}\tag{2}\\ &= \frac{1+i}{2}\ket{0} - \frac{1-i}{2i} \ket{1}\tag{3}\\ &= \frac{1}{\sqrt{2}}\left(\frac{1+i}{\sqrt{2}}\ket{0} - \frac{1-i}{\sqrt{2}i} \ket{1}\right)\tag{4}\\ &= \frac{1}{\sqrt{2}}\left(e^{i\pi/4}\ket{0} - \frac{1}{i}e^{-i\pi/4} \ket{1}\right)\tag{5}\\ &= \frac{e^{i\pi/4}}{\sqrt{2}}\left(\ket{0} - \frac{1}{i}e^{-i\pi/2} \ket{1}\right)\tag{6}\\ &= \frac{e^{i\pi/4}}{\sqrt{2}}\left(\ket{0} - \frac{1}{i}(-i) \ket{1}\right)\tag{7}\\ &= \frac{e^{i\pi/4}}{\sqrt{2}}\left(\ket{0} + \ket{1}\right)\tag{8}\\ &= e^{i\pi/4} \ket{\psi_1}\tag{9} \end{align}

which again differs only by an overall phase.

## Edit:

There are a few ways you could do this one, a simpler way would have been to use

\begin{align} \frac{e^{i\pi/2}-1}{2i} &= \frac{i-1}{2i}\\ &= -i\frac{i-1}{2} \\ &= \frac{1 + i}{2} \\ &= \frac{1}{\sqrt{2}} \left(\frac{1+i}{\sqrt{2}}\right) \\ &= \frac{1}{\sqrt{2}} e^{i\pi/4} \end{align}

# Addendum: Global versus relative phases

A global phase means multiplied by a complex phase factor $e^{i\phi}$. Equivalently we can call it an "overall phase". This is to distinguish it from a relative phase. Consider the state from before, $$\ket{\psi_1} = \frac{1}{\sqrt{2}} \left(\ket{0} + \ket{1}\right)$$ Now consider the state $$\ket{\psi'} = \frac{1}{\sqrt{2}} \left(\ket{0} + e^{i\phi}\ket{1}\right)$$ This is not the same as the state $\ket{\psi_1}$, because the two components differ by a relative phase. On the other hand, the state $$\ket{\psi''} = e^{i\phi} \frac{1}{\sqrt{2}} \left(\ket{0} + \ket{1}\right) = e^{i\phi} \ket{\psi_1}$$ differs from $\ket{\psi_1}$ by an overall or global phase.

When we compute the expectation of any observable $\hat{O}$ in state $\ket{\psi}$, we are computing $$\expect{O}_\psi = \bra{\psi}\hat{O}\ket{\psi}$$ where the subscript denotes we are measuring the expect of $\hat{O}$ in the state $\ket{\psi}$.

Consider the case where $\ket{\psi} = \ket{\psi''} = e^{i\phi}\ket{\psi_1}$. Then the expectation is \begin{align} \expect{O}_{\psi''} &= \bra{\psi''}\hat{O}\ket{\psi''}\\ &= \bra{\psi_1}e^{-i\phi} \hat{O} e^{i\phi}\ket{\psi_1}\\ &= \bra{\psi_1}e^{-i\phi} e^{i\phi} \hat{O}\ket{\psi_1}\\ &= \bra{\psi_1} \hat{O}\ket{\psi_1}\\ &= \expect{O}_{\psi_1} \end{align} which says that the expectation value of $\hat{O}$ in the state $\ket{\psi} = e^{i\phi} \ket{\psi_1}$ is the same as in the state $\ket{\psi_1}.$ The global phase has no influence on the observed value, meaning global phases cannot be detected in any experiment, and therefore are unphysical.

I want to be complete, so I'm going to write out the full expectation value, continuing we have \begin{align} \expect{O}_{\psi''} = \expect{O}_{\psi_1} &= \bra{\psi_1} \hat{O}\ket{\psi_1}\\ &= \frac{1}{\sqrt{2}} \left(\bra{0} + \bra{1}\right) \hat{O} \frac{1}{\sqrt{2}}\left(\ket{0} + \ket{1}\right) \\ &= \frac{1}{2} \left(\bra{0} + \bra{1}\right)\left(\hat{O}\ket{0} + \hat{O}\ket{1}\right) \\ &= \frac{1}{2}\left(\bra{0} \hat{O}\ket{0} + \bra{0}\hat{O}\ket{1} + \bra{1}\hat{O}\ket{0} + \bra{1}\hat{O}\ket{1}\right)\\ &= \frac{1}{2}\left(\bra{0} \hat{O}\ket{0} + 2\Re(\bra{0}\hat{O}\ket{1}) + \bra{1}\hat{O}\ket{1}\right) \tag{10} \end{align} where $\Re$ means "the real part", and I have used the fact that $\hat{O}$ is hermitian: $\bra{0}\hat{O}\ket{1} = \left(\bra{1}\hat{O}\ket{0}\right)^*$, and for a complex number $z$, we have $z + z^* = 2\Re(z)$.

Now let's consider the case where we measure $\hat{O}$ in the state $\ket{\psi'} = \frac{1}{\sqrt{2}} \left(\ket{0} + e^{i\phi}\ket{1}\right)$ with a relative phase. We have \begin{align} \expect{O}_{\psi'} &= \bra{\psi'} \hat{O}\ket{\psi'}\\ &= \frac{1}{\sqrt{2}} \left(\bra{0} + \bra{1}e^{-i\phi}\right) \hat{O} \frac{1}{\sqrt{2}}\left(\ket{0} + e^{i\phi}\ket{1}\right) \\ &= \frac{1}{2} \left(\bra{0} + \bra{1}e^{-i\phi}\right)\left(\hat{O}\ket{0} + \hat{O}e^{i\phi}\ket{1}\right) \\ &= \frac{1}{2}\left(\bra{0} \hat{O}\ket{0} + e^{i\phi}\bra{0}\hat{O}\ket{1} + e^{-i\phi}\bra{1}\hat{O}\ket{0} + \bra{1}\hat{O}\ket{1}\right)\\ &= \frac{1}{2}\left(\bra{0} \hat{O}\ket{0} + 2\Re\left(e^{i\phi}\bra{0}\hat{O}\ket{1}\right) + \bra{1}\hat{O}\ket{1}\right) \tag{11} \end{align}

Compare equation (10) and (11). They are not the same due to the relative phase. Therefore the states $\ket{\psi'}$ and $\ket{\psi''}$ are distinguishable. We can do an experiment and conclude which of the two states the system is in, because $\expect{O}_{\psi'} \neq \expect{O}_{\psi''}$. On the other hand, $\ket{\psi''}$ and $\ket{\psi_1}$ are indistinguishable. There is no experiment which can conclude whether the system was in one or the other, because $\expect{O}_{\psi''} = \expect{O}_{\psi_1}$ for any observable $\hat{O}$.

• thank you!!! I will look carefully! in (2) you wrote 2sin(π/2) instead of 2sin(π/4), just a typo, right? – Juan Manuel Jones Volonté Apr 23 '18 at 5:53
• Oops yes that was a typo – Kai Apr 23 '18 at 6:08
• Hi, why $$e^{i\pi/4}$$ is an overall factor? I got that $$e^{i\pi}$$ is -1 – Juan Manuel Jones Volonté Apr 23 '18 at 15:54
• I added equation numbers, I assume you are referring to the last one, could you point to what line is confusing you? – Kai Apr 23 '18 at 17:18
• the last line, some confusion about what a global phase is, and what is not. – Juan Manuel Jones Volonté Apr 23 '18 at 22:47