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.