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?

  • $\begingroup$ I'm not sure I completely understand the question: the matrix representation of an operator does depend on the basis in which it is expressed. If that basis is changed, the "components" of the operator change, just like with vectors: when you change the basis, the components of a vector change... $\endgroup$
    – Philip
    Jan 15, 2020 at 19:22
  • $\begingroup$ ..."intuitively at least". Can you explain a bit more? You are writing a weird unnamed operator defined by $\phi'$s in the orthogonal basis of $\phi$s, and you appear to marvel at it. What on earth is $\hat A$ and why exactly are you troubled by it? $\endgroup$ Jan 15, 2020 at 19:36
  • $\begingroup$ @Philip sure, but note that in equation (*) Zettili has not defined $\{ \left| \phi_n\right>\}$ at all. He has said $\{ \left| \phi_n\right>\}$ should be orthogonal and complete set. That's all. I defined $\{ \left| \phi_n\right>\}$ myself to show that for two different sets we would have different results. In other words, Even $\{ \left| \phi_n\right>\}$ which gives Zettili's result is just my guess, maybe Zettili had different set than mine. This equation was mentioned in second chapter. $\endgroup$
    – Paradoxy
    Jan 15, 2020 at 19:45
  • $\begingroup$ Part of your notational confusion is writing an operator, $\hat A$, as a matrix in a given basis, and then using the same symbol, improperly, to write it as a matrix in a different basis, without bothering to insert the similarity transformation relating these two matrices. $\endgroup$ Jan 15, 2020 at 19:52
  • $\begingroup$ @CosmasZachos $\hat{A}$ is an arbitrary operator. It can be anything, from hamiltonian of the system to anything else that is imaginable in quantum mechanics. Think of it as hamiltonian operator for the sake of simplicity. The weird thing about it is that "all operators" in this bases are diagonal at the same time. From momentum to hamiltonian and position and everything else. $\endgroup$
    – Paradoxy
    Jan 15, 2020 at 20:00

1 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


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?

  • $\begingroup$ Yes you are right. In fact this new representation would be a mess. I had trouble with it since it seemed to be diagonal for all operators. Thank you. $\endgroup$
    – Paradoxy
    Jan 15, 2020 at 20:09

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