# What is the difference between the states $i|1\rangle$ and $|+i\rangle$?

I am new to Quantum computing.

I see $$|\mbox{+}i\rangle$$ state maps to y-axis on bloch sphere ($$\theta = 90$$ degree and $$\phi = 90$$ degree) while $$i|1\rangle$$ maps on x-axis, $$i|1\rangle$$ is stated as $$|1\rangle$$ with global phase $$i$$ ($$\theta = 180$$ degree and $$\phi = 90$$ degree).

But both the notation ie. $$i|1\rangle$$ and $$|\mbox{+}i\rangle$$ looks very similar to me. How do we interpret that something is global phase while reading it.

It is just the convention that people use the notation $$|1 \rangle$$ to represent the vector $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ and $$|0 \rangle$$ to represents the matrix $$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$. Similarly, people use the notation $$|i\rangle$$ to represent the vector $$\dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$$ .

I could have very much use a different notation, say $$|k\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$$, instead. Maybe this would lessen the confusion.

So by looking at the vector itself, it would be less confusing. That is, if you look at the state $$i|1\rangle$$ you have

$$i|1\rangle = i \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

which you can see that $$i$$ here is the global phase. But if you look at the state $$|i\rangle$$ you have

$$|i \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix} = \dfrac{1}{\sqrt{2}} \bigg[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} + i \begin{pmatrix} 0 \\ 1 \end{pmatrix} \bigg] = \dfrac{1}{\sqrt{2}}\bigg[ |0 \rangle + i|1\rangle \bigg]$$

as you can see here, $$i$$ is not a global phase but rather a relative phase.

• Thanks. It clarifies...I have one more conceptual question, when we see the vector for i |1> , we say that "i" is the global phase but when we see the vector for | i > or | +>, why don't we consider 1/sqrt(2) as phase...or to put it differently given any vector, how can we say that this part is global phase... Oct 31 '20 at 20:22
• The real factor, like $\dfrac{1}{\sqrt{2}}$, is what we called the normalization constant. It is there to make sure the vector is has unit norm (normalized). Oct 31 '20 at 21:28
• global phase comes in when you have a state $|\psi \rangle$ and another state $e^{i\theta}|\psi \rangle$. Thu, if $|\psi \rangle = |1 \rangle$ then the state $e^{i \pi/2} |1 \rangle = i |1 \rangle$ is the same as the state $|\psi \rangle = |1 \rangle$ with an attached global phase of $e^{i \pi/2}$. These two states are not the same mathematically, but they are the same quantum mechanically because we cannot distinguish them under quantum measurement. Oct 31 '20 at 21:34
• @abhinav, you can imagine a global phase as a "unit" complex number which has absolute value 1. So if you write this complex number in the polar form $re^{i\theta}$, since $r=1$, you'll be left with the phase factor $e^{i\theta}$. Or in the case of $i$ as global phase your complex number has the following cartesian form $c=a+ib$, in which $a=0$ and $b=1$ so $r=\sqrt{0^2+1^2}=1$. This simple information can help you to distinguish between a global phase and a normalization constant. Nov 1 '20 at 9:59
• @abhinav, You can easily see that the absolute value of $\frac{1}{\sqrt2}$, which is $\frac{1}{2}$, is not equal to 1! And as KAJ226 explained it nicely, here $\frac{1}{\sqrt2}$ is just a normalization constant. youtube.com/watch?v=2ntBEd-bxKs Nov 1 '20 at 9:59