MHB Quantum Computing: Change of Basis

[Ackbach]

This is Exercise 2.20 in Nielsen and Chuang's Quantum Computation and Quantum Information, on page 71.

Suppose $A'$ and $A''$ are matrix representations of an operator $A$ on a vector space $V$ with respect to two different orthonormal bases, $|v_i\rangle$ and $|w_i\rangle$. Then the elements of $A'$ and $A''$ are $A_{ij}'=\langle v_i|A|v_j\rangle$ and $A_{ij}''=\langle w_i|A|w_j\rangle$. Characterize the relationship between $A'$ and $A''$.

My solution so far: I claim that $A'$ and $A''$ are related by a unitary transformation; that is, there exists a unitary transformation $U$ such that $A'=UA''U^{-1}$. Of course, since $U^{-1}=U^{\dagger}$, this becomes $A'=UA''U^{\dagger}$, or $A'U=UA''$.

Let $\displaystyle U=\sum_i|w_i\rangle\langle v_i|$. It is easy to show that $U$ is unitary. We can show that $|w_i\rangle = U|v_i\rangle$. Here is a further list of results/known facts:
\begin{align*}
A_{ij}'&=\langle v_i|A|v_j\rangle \\
A_{ij}''&=\langle w_i|A|w_j\rangle \\
U^{\dagger}&=\sum_i|v_i\rangle\langle w_i| \\
A&=\sum_{i,j}A_{ij}'|v_i\rangle\langle v_j| \\
A&=\sum_{i,j}A_{ij}''|w_i\rangle\langle w_j| \\
U^{\dagger}AU&=\sum_{i,j}A_{ij}''|v_i\rangle\langle v_j|
\end{align*}

I need to show either that $A'=U^{\dagger}A''U$ or $A''=U^{\dagger}A'U$ (whichever is correct). But I can't seem to make further progress. Any ideas?

By the way, please don't use any fancy terms like "endomorphisms of the automorphic normal subgroup" or anything like that (sorry, Deveno!). I'm afraid I won't understand it at all.

Many thanks for your time!

 

[I like Serena]

$$A'=UA''U^{\dagger} \quad \textrm{(Hypothesis)} \tag 1$$
$$A_{ij}'=\langle v_i|A|v_j\rangle \tag 2$$
$$A=\sum_{i,j}A_{ij}''|w_i\rangle\langle w_j| \tag 3 $$

Hi Ackbach,

Changing indices, we can write $(2)$ as:
$$A_{kl}'=\langle v_k|A|v_l\rangle \tag 4 $$

When we substitute $(3)$ in $(4)$, we get:
$$A_{kl}'=\langle v_k|\sum_{i,j}A_{ij}''|w_i\rangle\langle w_j| |v_l\rangle
=\sum_{i,j}\langle v_k|w_i\rangle A_{ij}''\langle w_j|v_l\rangle \tag 5
$$

Combining $(1)$ and $(5)$:
$$U_{ij} = \langle v_i|w_j\rangle$$

 

[Ackbach]

$$U_{ij} = \langle v_i|w_j\rangle$$

Is this consistent with $\displaystyle U=\sum_i|w_i\rangle\langle v_i|$?

 

[I like Serena]

Is this consistent with $\displaystyle U=\sum_i|w_i\rangle\langle v_i|$?

I don't think so - the "dimensions" are off.
Written as $U = \sum_i|w_i\rangle\langle v_i|$ it is an operator.
Written as $U_{ij} = \langle v_i | w_j \rangle$ it is a matrix.
It should be a matrix to satisfy $A' = UA''U^\dagger$, not an operator.

 

[Ackbach]

I don't think so - the "dimensions" are off.
Written as $U = \sum_i|w_i\rangle\langle v_i|$ it is an operator.
Written as $U_{ij} = \langle v_i | w_j \rangle$ it is a matrix.
It should be a matrix to satisfy $A' = UA''U^\dagger$, not an operator.

I could be wrong, but I don't think the distinction between an operator and a matrix is important for $U$. Presumably when you write $U_{ij}$, you mean the $i,j$ component of the matrix, whereas when you write $U$, you mean the whole matrix, right?

 

[I like Serena]

I could be wrong, but I don't think the distinction between an operator and a matrix is important for $U$.

An operator is independent of a basis.
We have to pick a basis before we can multiply it with a matrix that matches it with respect to the basis.
Alternatively, we can multiply operators with each other, but only when we assign a basis to the result does it become a numerical result in the form of a matrix with respect to that basis.

Consider physics when we have lengths $l = 1\text{ m}$ and $s = 10 \text{ cm}$.
We can talk about $l + s$ leaving in the middle what the unit is.
But we cannot talk about $l + 10$ or $1 + s$ without specifying the unit.
Only when we pick for instance meters can we write it as $1 + 0.10$ with the understanding that the unit is $\textrm{m}$.

Presumably when you write $U_{ij}$, you mean the $i,j$ component of the matrix, whereas when you write $U$, you mean the whole matrix, right?

Yes.
Btw, currently the distinction between an operator and a matrix is a bit ambiguous, since the same typeface is used.
That leaves the proof that $U$ given by $U_{ij} = \langle v_i | w_j \rangle$ is a unitary matrix:
$$({UU}^\dagger)_{ij} = \sum_k \langle v_i | w_k \rangle \langle w_k | v_j \rangle
= \langle v_i | \mathcal I | v_j \rangle
= \langle v_i | v_j \rangle
= I_{ij}
$$