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Nielsen / Chuang "Quantum Computation and Quantum Information" states on p. 70:

"A matrix $ U$ is said to be unitary if $U^\dagger U = I$. Similarly, an operator $U$ is unitary if $U^\dagger U = I$. It is easily checked that an operator is unitary if and only if each of its matrix representations is unitary."

I would like to do two things: (i) confirm that I am interpreting these definitions of a unitary matrix and a unitary operator correctly and (ii) understand how to check that an operator is unitary iff each of its matrix representations is unitary.

As for (i): I think the above definition of a unitary matrix should probably be interpreted as follows:

"A matrix $U$ is unitary iff $U$ is a square matrix and $U^\dagger U = I$ (where $U^\dagger$ denotes the conjugate transpose of $ U$)."

Comparing this to the definition as stated by Nielsen / Chuang, I added the condition that $U$ must be a square matrix, which then allows the definition to be stated as a biconditional.

As for Nielsen / Chuang's definition of a unitary operator, I think this should probably be interpreted as follows:

"A linear operator $ U: V \rightarrow V $ is unitary iff $U^\dagger U = I$ (where $U^\dagger$ denotes the adjoint of $U$)."

Comparing this to the definition as stated by Nielsen / Chuang, I added the condition that the operator $U$ be a linear operator on the inner product space $V$ (i.e. that $U: V \rightarrow V$), and I also turned the definition into a biconditional.

I think that Nielsen / Chuang probably intend the definitions of a unitary matrix and a unitary operator to be interpreted as described above, but please let me know if you think that I'm mistaken.

As for (ii): I would then like to understand how it can be "checked that an operator is unitary if and only if each of its matrix representations is unitary". To again start with an interpretive question, based on some online research I think that Nielsen / Chuang probably mean that

"An operator is unitary iff each of its matrix representations with respect to an orthonormal basis is unitary."

In this restatement I added that the condition that the matrix representations must be with respect to orthnormal bases, but please let me know if you think it is not necessary to add this condition.

Interpreting the statement in this way, I then made an attempt to check the left-to-right direction. I ultimately got stuck, but here is my approach:

Suppose that the linear operator $ U: V \rightarrow V$ is unitary. Consider two arbitrary orthonormal bases $B_v = \{ |v_1\rangle , \ldots , |v_n\rangle \}$ and $B_w \{ |w_1\rangle , \ldots , |w_n\rangle \}$ of $ V $. Then we can give an outer product representation of $ U $ as follows (as described on pp.67-68 of Nielsen / Chuang):

$U = \sum_{i=1}^n \sum_{j=1}^n \langle w_j | U | v_i \rangle |w_j\rangle \langle v_i|$

From this, we can find an outer product representation of the adjoint $ U^\dagger $:

$U^\dagger = \left( \sum_{i=1}^n \sum_{j=1}^n \langle w_j | U | v_i \rangle |w_j\rangle \langle v_i| \right)^\dagger$

$=\sum_{i=1}^n \sum_{j=1}^n (\langle w_j | U | v_i \rangle |w_j\rangle \langle v_i|)^\dagger$

$=\sum_{i=1}^n \sum_{j=1}^n \langle w_j | U | v_i \rangle^* |w_j\rangle \langle v_i|^\dagger$

$=\sum_{i=1}^n \sum_{j=1}^n \langle w_j | U | v_i\rangle^* |v_i\rangle \langle w_j|$

From these outer product representations, we know that (i) the matrix representation of the operator $ U $ with respect to input basis $ B_v $ and output basis $ B_w $ will have the entry $ \langle w_j | U | v_i \rangle $ in the $i$-th column and $j$-th row; and (ii) the matrix representation of the operator $ U^\dagger $ with respect to input basis $ B_v $ and output basis $ B_w $ will have the entry $ \langle w_j | U | v_i\rangle ^* $ in the $j$-th column and $i$-th row (cf. Nielsen / Chuang p. 68).

Writing these matrix representations as $[U]_{B_v}^{B_w} $ and $[U^\dagger]_{B_v}^{B_w} $, respectively, we have that:

$[U^\dagger]_{B_v}^{B_w} [U]_{B_v}^{B_w}= \begin {bmatrix} \langle w_1 | U | v_1\rangle^* & \ldots & \langle w_n | U | v_1 \rangle^* \\ \vdots & \ddots & \vdots \\ \langle w_1 | U | v_n \rangle^* & \ldots & \langle w_n | U | v_n \rangle^* \end {bmatrix} \begin {bmatrix} \langle w_1 | U | v_1\rangle & \ldots & \langle w_1 | U | v_n \rangle \\ \vdots & \ddots & \vdots \\ \langle w_n | U | v_1 \rangle & \ldots & \langle w_n | U | v_n \rangle \end {bmatrix}$

$=\begin {bmatrix} \langle w_1 | U | v_1 \rangle^* \langle w_1 | U | v_1 \rangle + \ldots + \langle w_n | U | v_1 \rangle^* \langle w_n | U | v_1 \rangle & \ldots & \langle w_1 | U | v_1\rangle ^* \langle w_1 | U | v_n \rangle + \ldots + \langle w_n | U | v_1 \rangle^* \langle w_n | U | v_n \rangle \\ \vdots & \ddots & \vdots \\ \langle w_1 | U | v_n \rangle^* \langle w_1 | U | v_1 \rangle + \ldots + \langle w_n | U | v_n \rangle^* \langle w_n | U | v_1 \rangle & \ldots & \langle w_1 | U | v_n\rangle^* \langle w_1 | U | v_n \rangle + \ldots + \langle w_n | U | v_n \rangle^* \langle w_n | U | v_n \rangle \\ \end {bmatrix}$

I would now like to show that this last matrix is equal to the $n \times n $ identity matrix $ I_n = \begin {bmatrix} 1 & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \ldots & 1 \end {bmatrix} $ , which would give us the desired result that $[U^\dagger]_{B_v}^{B_w} [U]_{B_v}^{B_w} = I_n $.

To do this, I have been trying to manipulate the inner products that appear in the entries of the last matrix and possibly use the assumption that $U^\dagger U = I $ somewhere along the way. However, I have not been able to get anywhere towards showing that this last matrix is equal to the identity matrix.

Can the proof that I started be completed? Or have I made any mistakes in my reasoning so far? Or am I taking the wrong approach towards showing that any matrix representation of $ U $ with respect to an orthonormal basis is unitary?

Any help would be greatly appreciated.

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  • $\begingroup$ MathJax does not completely incorporate LaTeX. In particular, \( and \) are not recognized, and you must use dollar signs to go in and out of math mode. $\endgroup$ Apr 2 at 16:48
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    $\begingroup$ (and \braket is not a standard command in any case) $\endgroup$ Apr 2 at 16:54
  • $\begingroup$ Thank you, I've fixed these issues. My apologies for the misformatted initial post - I'm still learning how to use this forum. $\endgroup$
    – mchk
    Apr 2 at 17:41
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    $\begingroup$ For $U : V \to V$ where $V$ is an infinite-dimensional space you will need both $U^\dagger U = I$ and $UU^\dagger = I$. For finite-dimensional $V$, one of these implies the other. $\endgroup$
    – GEdgar
    Apr 3 at 14:27

2 Answers 2

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All your restatements look correct and necessary to me.

To proceed further, use $\langle w|U|v\rangle^\ast=\langle v|U^\dagger|w\rangle$ (note $U^\dagger=U^{-1}$ when unitary) and

$$ |w_1\rangle\langle w_1|+\cdots+|w_n\rangle\langle w_n|=I $$

for any unitary basis $\{w_1,\cdots,w_n\}$.

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I will try to fill in the missing steps to complete the proof from left to right, and then attempt the proof from right to left.

Left-to-right: Note that, in the matrix above that results from multiplying out $[U^\dagger]_{B_v}^{B_w} [U]_{B_v}^{B_w}$, the entry in the $j$-th row and $i$-th column is given by

$\langle w_1 | U | v_j \rangle^* \langle w_1 | U | v_i \rangle + \ldots + \langle w_n | U | v_j \rangle^* \langle w_n | U | v_i \rangle $

We then have that

$\langle w_1 | U | v_j \rangle^* \langle w_1 | U | v_i \rangle + \ldots + \langle w_n | U | v_j \rangle^* \langle w_n | U | v_i \rangle $

$=\langle \langle w_1 | U | v_j \rangle^{**} | w_1 \rangle | U | v_i \rangle + \ldots + \langle \langle w_n | U | v_j \rangle^{**} | w_n \rangle | U | v_i \rangle $

$=\langle \langle w_1 | U | v_j \rangle | w_1 \rangle | U | v_i \rangle + \ldots + \langle \langle w_n | U | v_j \rangle | w_n \rangle | U | v_i \rangle $

$=\langle \left( \langle w_1 | U | v_j \rangle | w_1 \rangle + \ldots + \langle w_n | U | v_j \rangle | w_n \rangle \right) | U | v_i \rangle $

$=\langle \left( | w_1 \rangle \langle w_1 | U | v_j \rangle + \ldots + | w_n \rangle \langle w_n | U | v_j \rangle \right) | U | v_i \rangle $

$=\langle U | v_j \rangle | U | v_i \rangle $

$=\langle v_j | U^\dagger U | v_i \rangle $

$=\langle v_j | v_i \rangle $

$=\delta_{ij}$

where the last equality holds due to the orthonormality of the basis $ B_v$. This means that the entries on the diagonal of the matrix (where $i=j $) will be $1$ and all other entries will be $0$. Thus we have shown, for any arbitrary orthonormal input and output bases $B_v$ and $B_w$ of $V$, that $[U^\dagger]_{B_v}^{B_w} [U]_{B_v}^{B_w}=I_n$, completing the left-to-right direction of the proof.

Right-to-left: For the right-to-left direction, assume that, for arbitrary orthonormal input and output bases $B_v$ and $B_w$ of $V$, we have that $[U^\dagger]_{B_v}^{B_w} [U]_{B_v}^{B_w}=I_n$.

As we saw above, the product of $[U^\dagger]_{B_v}^{B_w}$ and $[U]_{B_v}^{B_w}$ is a matrix whose entry in the $j$-th row and $i$-th column is equal to

$\langle v_j | U^\dagger U | v_i \rangle$

Since the entry in the $j$-th row and $i$-th column of the identity matrix is equal to $\delta_{ij}$, we have that

$\delta_{ij} = \langle v_j | v_i \rangle = \langle v_j | U^\dagger U | v_i \rangle $

Since this is true for all orthonormal bases, we know that $U^\dagger U | v_i \rangle = | v_i \rangle$ for all unit vectors $| v_i \rangle \in V$. But since, for any vector $|v \rangle \in V$, there is a unit vector $|v_u \rangle$ such that $|v \rangle = || |v \rangle || |v_u \rangle$, we have that

$U^\dagger U | v \rangle = U^\dagger U || |v \rangle || |v_u \rangle = || |v \rangle || U^\dagger U |v_u \rangle = || |v \rangle || |v_u \rangle = | v \rangle$

Therefore $U^\dagger U = I$.

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