How to find the matrix representation of an operator from its action on a basis?

Question

First, I apologize if something is poorly written but English is not my first language.

I know that these exercises have been solved in this question. But I do not agree. Inner product and concrete vectors are used and I think that this question has to be solved only with information we have from the beginning of chapter 2 until the exercise.

So I think we only have to use eq. 2.12 in this way:

if we write $\left|0\right>=\left|v_{0}\right>$ and $\left|1\right>=\left|v_{1}\right>$ and we use them as input and output basis, we can write (2.12) as $A\left|v_{j}\right> = \sum_{i} A_{ij} \left|v_{i}\right>$; so,

$A\left|v_{0}\right> = A_{00}\left|v_{0}\right> + A_{10}\left|v_{1}\right> = \left|v_{1}\right> \Rightarrow A_{00}=0; A_{10}=1$

$A\left|v_{1}\right> = A_{01}\left|v_{0}\right> + A_{11}\left|v_{1}\right> = \left|v_{0}\right> \Rightarrow A_{01}=1; A_{11}=0$

$$ A = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} $$

As we can see I don't use concrete basis vectors and this works for basis $\begin{pmatrix}0 \\1 \end{pmatrix}$ and $\begin{pmatrix}1 \\0 \end{pmatrix}$.

But not with basis like $\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\1 \end{pmatrix}$ and $\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\-1 \end{pmatrix}$.

My feeling was that this solution doesn't depend on the basis but it does. So why?

Answer 1

Ok I think I've got it. The first part of my question is correct; the second one no. The big mistake is that I've tried to use eq 2.12 with the elements of vectors and not with the vectors themselves.

As I've done before, taking any basis, say $|v_{0}\rangle$ and $|v_{1}\rangle$, I'm looking for the matrix representation of the lineal operator $A$ such that $A |v_{0}\rangle = |v_{1}\rangle$ and $A |v_{1}\rangle = |v_{0}\rangle$. Applying eq. 2.12 we obtain:

$A |v_{0}\rangle = A_{00} |v_{0}\rangle + A_{10} |v_{1}\rangle = |v_{1}\rangle \Rightarrow A_{00} = 0; A_{10} = 1$

$A |v_{1}\rangle = A_{01} |v_{0}\rangle + A_{11} |v_{1}\rangle = |v_{0}\rangle \Rightarrow A_{01} = 0; A_{11} = 1$

$$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

And now my mistake:

I said this works for computational basis $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ but not for $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ basis.

And I said that because I did these wrong calculations:

• With computational basis:

$ A |v_{0}\rangle = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \times 0 + 1 \times 1 \\ 1 \times 0 + 0 \times 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = |v_{1}\rangle $

• With the other basis:

$ A |v_{0}\rangle = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \times 1 + 1 \times 1 \\ 1 \times 1 + 0 \times 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \ne |v_{1}\rangle $

I used elements of the vectors and not vectors, so all these calculations are wrong.

Let's see how is the correct way:

• With computational basis:

$A |v_{0}\rangle = 0 \times |v_{0}\rangle + 1 \times |v_{1}\rangle = 0 \times \begin{pmatrix} 0 \\ 1 \end{pmatrix} + 1 \times \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = |v_{1}\rangle $

$A |v_{1}\rangle = 1 \times |v_{0}\rangle + 0 \times |v_{1}\rangle = 1 \times \begin{pmatrix} 0 \\ 1 \end{pmatrix} + 0 \times \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = |v_{0}\rangle $

• With the other basis:

$A |v_{0}\rangle = 0 \times |v_{0}\rangle + 1 \times |v_{1}\rangle = 0 \times \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 1 \times \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = |v_{1}\rangle $

$A |v_{1}\rangle = 1 \times |v_{0}\rangle + 0 \times |v_{1}\rangle = 1 \times \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 0 \times \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = |v_{0}\rangle $

I hope all become clear now. I couldn't explain my problem correctly before. I'm really sorry.

I have to thanks a lot to teclado from another forum web page.

Answer 2

A simple explanation is if we look geometrically at what $A$ is doing, which is a reflection. For orthogonal basis $|v_1\rangle, |v_2\rangle$ we want to find unitary transform $A$, where

$A|v_1\rangle = |v_2\rangle$,

and

$A|v_2\rangle = |v_1\rangle$,

i.e. we want to find a reflection matrix. We want to find the vector, $r$, that reflects $|v_1\rangle$ to $|v_2\rangle$, and this is dependant on the choice of $|v_1\rangle$ to $|v_2\rangle$. The axis of reflection is the vector inbetween so

$r = \frac{1}{2}\left( |v_1\rangle - |v_2\rangle \right)$.

Using the Householder identity for the reflection matrix $R = I - 2rr^T = A$, gives us $A$ which is dependent on the basis vectors.

As an example for the computational basis we get:

$r = 0.5\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0.5 \\ -0.5 \end{pmatrix}$

plugging this into the Householder identity:

$A = I - 2\begin{pmatrix} 0.5 \\ -0.5 \end{pmatrix}\begin{pmatrix} 0.5 & -0.5 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Repeating this for your second basis vectors we can find that we are reflecting about the x-axis, if you can't immediately see this try plotting the two basis vectors. We then get,

$A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$