Is it possible to derive matrix representation of quantum operators in any complete orthogonal bases?

Question

We can find matrix representation of operator $\hat{A}$ by following:

$$\hat{A}=I\hat{A}I=\sum_n \left| \phi_n \right>\left< \phi_n \right| \hat{A} \sum_m \left| \phi_m \right>\left< \phi_m \right|=\sum_n\sum_m \left< \phi_n \right| \hat{A} \left| \phi_m \right>\left| \phi_n \right>\left< \phi_m \right| \text{ (*)}$$ Now if we assume that $$\phi_1=\begin{bmatrix}1 \\0\end{bmatrix} \text{and }\phi_2=\begin{bmatrix}0 \\1\end{bmatrix}$$ We can see that $$\sum_n\sum_m\left| \phi_n \right>\left< \phi_m \right|= \begin{bmatrix}1 &1 \\1 & 1\end{bmatrix} $$

And thus $$\hat{A}=\begin{bmatrix}A_{11} & A_{12} \\A_{21} & A_{22}\end{bmatrix}$$

The question is, are orthogonal bases $\{ \left| \phi_n \right>\text{}\}$ arbitrary? Obviously, the only limition I can think of is completeness of $\{ \left| \phi_n \right>\text{}\}$. Because after all equation (*) needs nothing more than that. But if it's indeed arbitrary, then I will assume the following bases: $$\phi_1'=\frac{1}{\sqrt{2}}\begin{bmatrix}1 \\1\end{bmatrix} \text{and }\phi_2'=\frac{1}{\sqrt{2}}\begin{bmatrix}1 \\-1\end{bmatrix}$$ We can verify completeness of $\{ \left| \phi_n' \right>\text{}\}$ simply by

$$\sum_n\left| \phi_n' \right>\left< \phi_n' \right| =I \text{ and } \left< \phi_1' \middle| \phi_2' \right> = 0 $$

However, the problem arises when I try to compute the following:

$$\sum_n\sum_m\left| \phi_n' \right>\left< \phi_m' \right|= \frac{1}{2}\begin{bmatrix}1 \\1 \end{bmatrix} \begin{bmatrix}1 &1 \end{bmatrix} + \frac{1}{2}\begin{bmatrix}1 \\1 \end{bmatrix} \begin{bmatrix}1 &-1 \end{bmatrix} + \frac{1}{2}\begin{bmatrix}1 \\-1 \end{bmatrix} \begin{bmatrix}1 & 1 \end{bmatrix} +\frac{1}{2}\begin{bmatrix}1 \\-1 \end{bmatrix} \begin{bmatrix}1 & -1 \end{bmatrix} = \frac{1}{2}\begin{bmatrix}1 & 1 \\1 & 1\end{bmatrix} + \frac{1}{2}\begin{bmatrix}1 & -1 \\1 & -1\end{bmatrix} + \frac{1}{2}\begin{bmatrix}1 & 1 \\-1 & -1\end{bmatrix} +\frac{1}{2}\begin{bmatrix}1 & -1 \\-1 & 1\end{bmatrix} = \frac{1}{2}\begin{bmatrix}4 & 0 \\0 & 0\end{bmatrix} $$ But it's not good! Since in this case I have:

$$\hat{A}=\begin{bmatrix}2A_{11} & 0 \\0 & 0\end{bmatrix}$$

Because $\hat{A}$ is completely arbitrary, that would mean any operator in $\{ \left| \phi_n' \right>\text{}\}$ will take the form of $\hat{A}$ which doesn't seem to be correct intuitively at least. In Quantum Mechanics from Zettili, author has written equation (*) without mentioning $\{ \left| \phi_n\right>\}$ at all, which makes me think that $\{ \left| \phi_n \right>\text{}\}$ is indeed arbitrary, as we can see from (*) itself. But from above calculation, it seems as if matrix representation of operator $\hat{A}$ depends on $\{ \left| \phi_n\right>\}$. What am I missing here?

Answer 1

I may have misunderstood your question, but bear with me. I don't think your calculations are right at all. I agree with you up to $$\hat{A} = \sum_n \sum_m \left<\phi_n\right|\hat{A}\left|\phi_m\right> \left| \phi_n\right> \left<\phi_m\right|$$

Now, using your initial basis, $\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\begin{pmatrix}0 \\ 1\end{pmatrix}$, you can write the above equation as

$$\left<\phi_1\right|\hat{A}\left|\phi_1\right> \begin{pmatrix}1 & 0\\ 0 &0\end{pmatrix} + \left<\phi_1\right|\hat{A}\left|\phi_2\right> \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+ \left<\phi_2\right|\hat{A}\left|\phi_1\right> \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} + \left<\phi_2\right|\hat{A}\left|\phi_2\right> \begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.$$

Of course, if you now label $\left<\phi_n\right|\hat{A}\left|\phi_m\right>=A_{nm}$, you get

$$\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}.$$

The mistake you seem to make is that you expand the sum without the $\left<\phi_n\right|\hat{A}\left|\phi_m\right>$ terms (which implicitly depend on the summing variables) and then add them on ''by hand'' after summing the matrices. You can't do that!

You can do exactly what I've done above with the other basis, and you'll find that

$$\hat{A} = \left<\phi^\prime_1\right|\hat{A}\left|\phi^\prime_1\right> \frac{1}{2}\begin{pmatrix}1 & 1\\ 1 &1\end{pmatrix} + \left<\phi^\prime_1\right|\hat{A}\left|\phi^\prime_2\right> \frac{1}{2}\begin{pmatrix}1 & -1\\ 1 & -1\end{pmatrix}+ \left<\phi^\prime_2\right|\hat{A}\left|\phi^\prime_1\right> \frac{1}{2}\begin{pmatrix}1 & 1\\ -1 & -1\end{pmatrix} + \left<\phi^\prime_2\right|\hat{A}\left|\phi^\prime_2\right> \frac{1}{2}\begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix}.$$

If you wish, you could now call $A_{nm}^\prime = \left<\phi^\prime_n\right|\hat{A}\left|\phi^\prime_m\right>$, but remember that $A^\prime_{nm}\neq A_{nm}$, as is easy to see by expanding your ''new'' basis in terms of the old one. For example, take the following component

$$A^\prime_{11} = \left<\phi^\prime_1\right|\hat{A}\left|\phi^\prime_1\right> = \frac{1}{2}\left( \left<\phi_1\right| + \left<\phi_2\right| \right) \hat{A} \,\left( \left|\phi_1\right> + \left|\phi_2\right> \right) = \frac{A_{11} + A_{12} + A_{21} + A_{22}}{2} \neq A_{11}.$$

Can you use this idea to actually calculate the matrix for $\hat{A}$ "with" your new basis now?