## 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 $$U|1\rangle$$?

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

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.

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.