Unitary acting on standard qubit basis properties

3

If we have a $U$ (unitary with all real entries) such that:

$U|0\rangle =a|0\rangle +b|1\rangle$

What is $U|1\rangle=?$

I know: the definition of what it means to be unitary ie. $U^\dagger U=UU^\dagger =I$

I've worked out: for $U|1\rangle=c|0\rangle+d|1\rangle$ must satisfy $ac+bd=0$ (by taking it's dagger and multiplying it for the constants).

Is this the only information we can derive? How can I write $U|1\rangle$?

Lock

Posted 2018-10-27T20:21:45.527

Reputation: 41

My unitary has all real entries. – Lock – 2018-10-27T20:44:45.277

I see. Well, then you still need the normalization. – Norbert Schuch – 2018-10-27T20:46:31.953

What do you mean? – Lock – 2018-10-27T20:53:10.660

@Lock, I think he means that $|a|^2 + |b|^2 = 1$ – user1271772 – 2018-10-27T21:03:58.923

Oh. Yes, I have considered this already but was not able to use this to derive anything new. – Lock – 2018-10-27T21:05:14.720

I am looking for $ab+cd$ and $a^2+c^2$ (or their counterparts) particularly. – Lock – 2018-10-27T21:06:49.547

The 2x2 real unitary (=orthogonal) matrices are parametrized by one parameter only, and are of the form [cos(x) sin(x) ; -sin(x) cos(x)], and the same with the 2nd row multiplied with -1. And you have 3 conditions for 4 variables. All good. – Norbert Schuch – 2018-10-27T22:01:23.850

Answers

4

You have $U|1\rangle=e^{i\phi}(b^*|0\rangle-a^*|1\rangle)$ (and for real entries, $e^{i\phi}=\pm1$). This condition follows automatically from $$ \langle 0|U^\dagger U |1\rangle=0 $$ -- this is exactly the condition you describe -- together with the fact that $U|0\rangle$ and $U|1\rangle$ must have the same normalization, $$ \langle k|U^\dagger U |k\rangle=1 $$ for $k=0,1$. (This also means that $|a|^2+|b|^2=1$.)

This is the only condition, since now you have ensured that all matrix elements of $U^\dagger U$ are of the correct form.

Norbert Schuch

Posted 2018-10-27T20:21:45.527

Reputation: 3 740

2

The general form of a 2x2 unitary matrix is:

$$ \begin{pmatrix} \alpha & \beta \\ -e^{i\phi}\beta^* & e^{i\phi}\alpha^* \end{pmatrix}, $$

with the constraint that $|\alpha|^2 + |\beta|^2$ = 1.

Since you say that $U|0\rangle = U\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix}$, we have that $\alpha = a$ and $-e^{i\phi}\beta^* = b$.

Therefore, the most we can say about the bottom-right corner is that $d=e^{i\phi}a^*$, and the most we can say about the top-right corner is $c=\beta = -b^* e^{i\phi}$.

So you have: $U|1\rangle = e^{i\phi}b^*|0\rangle - e^{i\phi}a^*|1\rangle$. We therefore do not have enough information to determine the phase $\phi$, but since you only ask how to write $U|1\rangle$ we don't need $\phi$ because it enters our expression for $U|1\rangle$ only as a global phase.

In conclusion: If all we know is $U|0\rangle = a|0\rangle = b|1\rangle$, then we can say that $U|1\rangle = b^*|0\rangle -a^*|1\rangle$, which is correct up to a global phase.

user1271772

Posted 2018-10-27T20:21:45.527

Reputation: 8 162