Solving Quantum Computation: Diagonalizing Y & Z Operators

[James Jackson]

I'm just looking at another quantum computation question. It is stated like so:

The operators Y and Z on $$C^2$$ are defined by:

$$Y|0\rangle =i|1\rangle ; Y|1\rangle = -i|0\rangle$$
$$Z|0\rangle = |0\rangle ; Z|1\rangle = -|1\rangle$$

Write Z in diagonal form
Write Y in Dirac form with respect to the basis $$\{ 0\rangle , |1\rangle\}$$

Now, I'm confusing myself something silly. I'm under the impression that the diagonal form of an operator is given by:

$$A=\sum \lambda_{n}|n\rangle\langle n|$$

where $$|n\rangle$$ are the eigenvectors and $$\lambda_n$$ are the eigenvalues of A.

But I would also take this to be the Dirac form, so I'm clearly missing something.

The eigenvalues of Z are clearly $$\{1,-1\}$$ with eigenvectors $$\{ |0\rangle ,|1\rangle\}$$, so the diagonal form is $$Z=|0\rangle\langle 0|-|1\rangle\langle 1|$$.

I suppose my question breaks down to 'What is meant by the Dirac form of an operator?'

Any hints?

Edited to remove me being stupid and working out eigenvectors incorrectly.

Edit: Or, by Dirac form of an operator, do they mean the matrix representation which, for Y, is given by:

$$Y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)$$

 

[PBRMEASAP]

My guess is that by "diagonal form" they mean write Z as a diagonal matrix. And by "Dirac form" they mean write Y as a sum of projection operators as you did for Z. But I agree with you, in that I would say the "Dirac Form" version of Z written in terms of eigenvectors is also in "diagonal form".

 

[James Jackson]

To anyone who's remotely interested, it transpires that given an operator, A, in matrix respresentation, then the Dirac form of the operator is:

$$\hat A=\sum_{ij} A_{ij} |i\rangle\langle j|$$

and given the eigenvalues $\lambda_i$ and eigenvectors $|\lambda_i\rangle$ of an operator, the diagonal representation is (of course):

$$\hat A = \sum_i \lambda_i |\lambda_i\rangle\langle\lambda_i |$$

As a 'learning point', note that the Dirac form of an operator is identical to the diagonal form for a diagonal matrix. An obvious result, but perhaps it will be useful for others.