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


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

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.


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 $\hat{O}$.

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

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.