Does the matrix representation of quantum gates depend on the basis?

Question

I have read that I can chose any orthonormal basis $\{|0⟩, |1⟩\}$ of $\mathbb{C}^2$ as my basis states for a single qubit. Usually $|0⟩ = \begin{pmatrix}1\\0\end{pmatrix}$ and $|1⟩ = \begin{pmatrix}0\\1\end{pmatrix}$ is chosen, but say I chose anything else. Would the matrix representation of quantum gates still be the same or would it change (e.g. or the Hadamard gate)? I searched for a while now but didn't find any answer.

Answer 1

Say you have state $|\psi\rangle$ and you perform a change in basis by applying $V$:

$$|\phi\rangle = V |\psi\rangle$$

Now you want to apply some unknown unitary $W$ to $|\phi\rangle$ such that it has the same effect as applying the unitary $U$ to $|\psi\rangle$.

The key to finding $W$ is to remember that if you apply the conjugate transpose of $V$ to $|\phi\rangle$, you switch back to the original basis:

$$ |\psi\rangle = V^{\dagger} |\phi\rangle .$$

So applying $U$ to $|\psi\rangle$ is equivalent to:

$$ U|\psi\rangle = U \underbrace{V^{\dagger} |\phi\rangle}_{|\psi\rangle} .$$

Lastly, you can apply $V$ to the state evolved by $U$ to change the basis again to the one you want to work in:

$$ V U|\psi\rangle = V U V^{\dagger} |\phi\rangle .$$

This means that the unitary you want to apply to the state in the new basis $|\phi\rangle$ to have the same effect as what $U$ would have in the original state $|\psi\rangle$ is:

$$ W = V U V^{\dagger} .$$

A simple example:

1. You start with state $|\psi\rangle = |0\rangle$ in the computational basis $\{|0\rangle, |1\rangle\}.$
2. You want switch to the Hadamard basis $\{|+\rangle, |-\rangle\}$, so you apply a unitary $V$ equal to the Hadamard gate: $|\phi\rangle = H|0\rangle = |+\rangle$ (this is the equivalent to state in $|0\rangle$ in your new basis).
3. You want to know what gate $W$ you need to apply to $|\phi\rangle=|+\rangle$ so it has the same effect as a unitary $U$ equal to the $X$ gate on $|\psi\rangle = |0\rangle$, so you need to apply $V U V^{\dagger} = H X H$, which corresponds to a $Z$ gate.
4. To verify, you can see that applying a $Z$ gate to $|+\rangle$ gives you state $|-\rangle$, which if you transform back to the computational basis is state $|1\rangle$ (i.e., the state you get when you apply $X$ to $|0\rangle$).

Answer 2

Yes, the matrix representation changes. For example, the Pauli $X$ gate in the standard basis is $$ X\equiv \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ which you evaluate by taking your orthonormal basis, in this case $|0\rangle$ and $|1\rangle$ and evaluating terms $\langle a|X|b\rangle$ where distinct $a$ map to rows, and distinct $b$ map to columns.

If I pick another basis, perhaps the $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$ basis. Now the top-left element, for example, is $\langle +|X|+\rangle=1$. So, now you end up with a representation $$ X\equiv\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}. $$ Very different.

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More details on the procedure of representation... Pick an orthonormal basis $\{|u_i\rangle\}$. A unitary operator $U$ has a representation in that basis of a matrix $M$ with matrix elements $$ M_{ij}=\langle u_i|U|u_j\rangle. $$ After all, what this is essentially saying is that $$ U=\sum_{i,j}M_{ij}|u_i\rangle\langle u_j|, $$ which is straightforward to prove.

So, going back to the example of an $X$ operator, which we know has the effect of flipping vectors of the standard basis: $X|0\rangle=|1\rangle$ and $X|1\rangle=|0\rangle$. (You should basically take this as the definition of your operator, and forget about the matrix you're possibly used to thinking about.) As a result of these definition, and linearity, we know that $X|+\rangle=|+\rangle$ and $X|-\rangle=-|-\rangle$. So now you just work through all 4 matrix elements and find the claimed representation in the $|\pm\rangle$ basis.

Answer 3

Some other very good answers have been posted, and this answer is not really useful on it's own, but I think it adds something in conjunction with the others. Really, quantum gates should be thought of as order-2 tensors acting on the Hilbert space. Once a basis is fixed, this just means quantum gates are matrices acting on the Hilbert space, but when we change basis by a transformation $L_{ij}$, the state vectors change by $v'_i=L_{ij}v_j$ and the quantum gates $T$ transform like $T_{ij}' = L_{ip}L_{jq} T_{pq} = (LTL^T)_{ij}$. This is a particularly nice way of thinking about things.

I do not know of any examples of order-3 tensors coming up in quantum mechanics, but it's an interesting thought, I'd be curious if anyone had any examples.