What is $\sum_{i}\langle i \vert U \vert j\rangle$ for unitary $U$?

2

The question is basically the title but given a unitary operator $U$ and a computational basis, can we say anything about the complex number below?

$$c = \sum_{i}\langle i \vert U \vert j\rangle$$

I expected that it would be $|c| = 1$ but this does not seem to generally hold.

Oli

Posted 2020-11-24T21:50:11.403

Reputation: 23

Are you interested in just the inner products or their squares? – Hasan Iqbal – 2020-11-24T22:04:52.813

@HasanIqbal I am only interested in the inner product given above, not the squares – Oli – 2020-11-25T11:08:50.813

Answers

2

Welcome to the quantum computing stack exchange.

If you view $\sum_{i} \langle i |$ as a (non-properly normalized) (bra) state $\sqrt{d}\langle \psi |$, where $|\psi\rangle = \frac{1}{\sqrt{d}}\sum_{x \in \{0,1\}^{n}}|x\rangle$, the quantity $c$ becomes just the inner product of $\sqrt{d}\langle \psi |U|j\rangle$. Here, $d =2^{n}$ is the proper normalization constant.

Without imposing any other constraints on $U$, $U|j\rangle$ just becomes some random state $|\phi\rangle$. Thus, $c = \sqrt{d}\langle \psi|\phi\rangle$ and the only thing we can conclude is:

$$ 0 \leq |c| \leq \sqrt{d} $$

JSdJ

Posted 2020-11-24T21:50:11.403

Reputation: 3 248

No, in my answer $|i\rangle$ and $|j\rangle$ are also normalized; only the 'state' $\sum_{i}\langle i |$ is not, so that's why there's the factor of $\sqrt{d}$. – JSdJ – 2020-11-25T11:31:49.140

1Sorry realized what you meant after I commented, so I removed it! – Oli – 2020-11-25T11:53:07.840

1@JSdJ if I may interject, I am a little confused as to your equation $|c|\leq\sqrt{d}$. If $c=\sqrt{d}\langle\psi|\phi\rangle$ and $|\phi\rangle$ is normalised because $|j\rangle$ is, then $|c|=\sqrt{d}$ would only happen if $\langle\psi|\phi\rangle =1$. But that cannot happen as $|\psi\rangle$ isn't normalised, and $|\phi\rangle$ is. $|\phi\rangle$ would need to be $\sqrt{d}|11...11\rangle$, which isn't normalised. – GaussStrife – 2020-11-25T15:15:47.297

$|\psi\rangle = \frac{1}{\sqrt{d}}|11\ldots 11\rangle$ so definitely is properly normalized. The state $\sum_{i}|i\rangle = \sqrt{d}|\psi\rangle$ is not properly normalized, so therefore $|c| \leq \sqrt{d}$ and not $1$. – JSdJ – 2020-11-25T15:19:43.170

$\langle\psi|\psi\rangle=\frac{1}{d}$, so $|\psi\rangle$ is not normalized. $\sqrt{d}\sqrt{d}\langle\psi|\psi\rangle=\frac{d}{d}=1$, so it is normalized. You even state in your response that $\sqrt{d}$ is the normalization factor. If $|\psi\rangle$ was normalized, it would be $|\psi\rangle = |11...11\rangle$ – GaussStrife – 2020-11-25T15:37:33.103

1Ah, I now see your point, and my sloppy notation. I've updated the text. – JSdJ – 2020-11-25T16:08:13.597

Ah yes, apologies I see what you were getting at. – GaussStrife – 2020-11-25T17:12:11.350

1

You know that the columns of a unitary matrix are orthonormal: https://math.stackexchange.com/questions/1688950/why-do-the-columns-of-a-unitary-matrix-form-an-orthonormal-basis. This means that the squares of the elements in each column should add up to 1, but that doesn't mean the sum of the elements in each column (what you have above) will add up to a value of magnitude 1.

For example, you could have $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},$$ which is clearly unitary.

The squared elements of the first half column add up to $\frac{1}{2}+\frac{1}{2}=1$ as do the squared elements of the second column. However, the elements of the first column itself add up to $\frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}}$ while the elements of the second column add up to $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0.$

nosuchthingasmagic

Posted 2020-11-24T21:50:11.403

Reputation: 61

Is there a maximum value that the sum of elements can take? It seems like it is $\sqrt{d}$ from the other answer – Oli – 2020-11-25T11:14:47.423

Yes, that is what @JSdJ showed in their answer. In the example I included of $H$, $\langle \psi | \phi \rangle$ takes its maximum value of 1 for the first column, and since $d=2$, the sum comes out be $\sqrt{2}$, as predicted. – nosuchthingasmagic – 2020-11-27T03:58:36.320