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I'm a bit confused about some of the terminology being thrown around in my text. To start with:

A diagonal representation for an opertor $A$ on $V$ is a representation $A = \sum_i \lambda_i |i \rangle \langle i|$, where the vectors $|i \rangle$ form an orthogonal set of eigenvectors, with corresponding eigenvalues $\lambda_i$.

Now here is the statement of the Spectral Theorem:

Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthogonal basis for $V$.

What does it mean for an operator to be diagonal with respect to a basis? This was not part of the diagonal representation definition above. Do they mean that $M$ has a diagonal representation, as above, and, that using the specified basis, the matrix represenation of $M$ is a diagonal matrix?

By the Spectral theorem, we have $A = \sum_i \lambda_i |i \rangle \langle i|$, and so

$$A|0 \rangle = \lambda_0 |0 \rangle + 0|1\rangle + \cdots 0|n \rangle$$ $$A|1 \rangle = 0 |0 \rangle + \lambda_1|1\rangle + \cdots 0|n \rangle$$ ... $$A|n \rangle = + 0|0\rangle + 1|1\rangle + \cdots \lambda_n|n \rangle$$

So is the matrix representation of $A$ wrt the basis $\{|0 \rangle, \ldots, |n \rangle \}$ simply $diag \{\lambda_0, \ldots, \lambda_n \}$?

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  • $\begingroup$ $\uparrow$ Which text? Which page? $\endgroup$ – Qmechanic Oct 20 '16 at 19:14
  • $\begingroup$ Nielsen and Chuang's Quantum Computation and Information. $\endgroup$ – theQman Oct 20 '16 at 19:16
  • $\begingroup$ The basis' in the spectral theorem are exactly those (note that there could be many) which appear in the diagonal representation in the first definition. $\endgroup$ – pppqqq Oct 20 '16 at 19:32
  • $\begingroup$ That still doesn't really clarifiy things for me. I don't know what it means for an operator to be diagonal with respect to a basis. I only know that an operator is a diagonal if it can be expressed as $\sum \lambda_i |i \rangle \langle i|$, with eigenvalues and eigenvectors... $\endgroup$ – theQman Oct 20 '16 at 19:36
  • $\begingroup$ Given a basis $|i \rangle$, you can always write the operator as $\sum a_{ij} \langle j | i \rangle$. The operator is diagonal in that basis if only the $a_{ii}$ are nonzero. $\endgroup$ – knzhou Oct 20 '16 at 19:44
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Do they mean that M has a diagonal representation, as above, and, that using the specified basis, the matrix represenation of M is a diagonal matrix?

You've answered your own question exactly right. It's also implicit in the matrix definition: The diagonal matrix of eigenvalues is clearly the operator's matrix in the diagonalizing frame and the Hermitian conjugate of the normalized matrix of eigenvectors written as columns is the transformation that maps the beginning coordinates to the coordinates in the diagonalized frame.

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  • $\begingroup$ @J.Pak I can't see an answer to that right now. Will think on it. $\endgroup$ – WetSavannaAnimal Oct 21 '16 at 0:42
  • $\begingroup$ I'd appreciate it. (You have a great mathematical intuition, truly!) $\endgroup$ – user.3710634 Oct 21 '16 at 2:24
  • $\begingroup$ What do you mean by frame? $\endgroup$ – theQman Oct 21 '16 at 6:31
  • $\begingroup$ @theQman Co-ordinate system or basis, depending on context. The matrix of eigenvectors is the transformed "frame" in the sense that these are the basis vectors for the co-ordinate system you transform to to make the operator or matrix diagonal. The co-ordinates themselves transform contravariantly - i.e. mapped by the inverse of the column matrix of eigenvectors. You've actually got everything spot on in the quote of yours I reproduced, so I suggest - unless you're under a deadline to understand this - to let it stew subconsciously for a day or two, and you'll ..... $\endgroup$ – WetSavannaAnimal Oct 21 '16 at 6:43
  • $\begingroup$ @theQman ....likely find it's clear when you come back to it as the right notion is already in your head at some level. $\endgroup$ – WetSavannaAnimal Oct 21 '16 at 6:43
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Perhaps it would be helpful to remind ourself of some of the basic properties of orthogonal bases. Let's for this purpose assume the $\{|i\rangle\}$ represents an orthogonal basis of a vector space $V$. In addition to the orthogonality condition $$ \langle j|i\rangle=\delta_{ij} , $$ we also have the completeness condition $$ \sum_i |i\rangle\langle i| = {\cal I} , $$ which can serve as a way to resolve the identity operator ${\cal I}$.

The important thing to realize is that this is not the only orthogonal basis that one can define for this vector space. In fact any unitary operation would convert this basis into a new basis, $$ U|i\rangle = |m\rangle $$ and it is a simple exercise to show the this new basis will also obey similar orthogonality and completeness conditions.

Let's now consider this case of an operator that is diagonal in our initial basis. This means we can write this operator as $$ A = \sum_i |i\rangle\lambda_i\langle i| . $$ What would happen if we convert this expression to the new basis $\{|m\rangle\}$?

To do this we operate on both sides of $A$ with the identity resolved in terms of the new basis (which we'll alternatively denote by either $|m\rangle$ or $|n\rangle$). See what happens $$\begin{align} {\cal I}A{\cal I} &= \sum_{mni} |m\rangle\langle m|i\rangle\lambda_i\langle i|n\rangle\langle n| \\ &= \sum_{mni} |m\rangle U_{mi} \lambda_i {U^{\dagger}}_{in}\langle n| \\ &= \sum_{mn} |m\rangle B_{mn} \langle n|\,.\end{align}$$ It is hopefully clear to see that the matrix $$ B_{mn} = \sum_i U_{mi} \lambda_i {U^{\dagger}}_{in} $$ would in general not be a diagonal matrix. In factor, the right-hand side $UDU^{\dagger}$ (where $D$ represent a diagonal matrix) is the spectral decomposition for some matrix.

This also implies that if one were to perform this process in reverse, one would be starting with a non-diagonal matrix, and then converting it to a diagonal matrix by an appropriate choice of basis. Let's see how that works. Let's assume I'm given a normal matrix $M$, expressed in some arbitrary basis $\{|a\rangle\}$. (I'm deliberately using different symbols here to avoid confusion with what we had before.) According to the spectral theorem, one can now express this as $$ M = U D U^{\dagger} , $$ where $U$ is a unitary matrix and $D$ is a diagonal matrix. Note that $M$ is still defined in terms of the basis $\{|a\rangle\}$ in which it is not diagonal. However we can remove the unitary matrices by operating on both sides as follows $$ U^{\dagger} M U = U^{\dagger} U D U^{\dagger} U = D . $$ Thus we end up with only the diagonal matrix. In the process we have redefined the basis in which the matrix is expressed. This redefinition comes about through the unitary matrix: $|a\rangle U = |i\rangle $ and $ U^{\dagger} \langle a| = \langle i|$. Therefore, the unitary matrix that is needed to diagonalize the matrix, also convert the basis to the special one in which the matrix becomes diagonal.

Let's look at the explicit questions:

"What does it mean for an operator to be diagonal with respect to a basis?"

It means that in this particular basis the operator (expressed as a matrix), one has non-zero elements on the diagonal only and these elements then represent the eigenvalues of the matrix. All other elements of the matrix are zero. The phrase "with respect to a basis" means that the rows (and columns) of the matrix are associated with particular element in that basis.

"Do they mean that $M$ has a diagonal representation, as above, and, that using the specified basis, the matrix representation of $M$ is a diagonal matrix?"

Yes indeed, provided that $M$ is a normal matrix, it always has a diagonal representation. (This is what the Spectral Theorem states.)

"So, is the matrix representation of $A$ wrt the basis $\{|0\rangle,...,|n\rangle\}$ simply diag$\{\lambda_0,...,\lambda_n\}$?"

Well, provided that this basis is the basis in which $A$ is diagonal, then yes, the diagonal matrix contains the eigenvalues on the diagonal.

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  • $\begingroup$ I'm reading through your response now. I think I am parsing through it without problem, but I'm unclear on what your conclusion is with regards to my question. $\endgroup$ – theQman Oct 24 '16 at 15:54
  • $\begingroup$ @theQman, I added some clarifications and addressed the specific questions. $\endgroup$ – flippiefanus Oct 25 '16 at 4:36

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