I'm working through Quantum Computing for Computer Scientists (Yanofsky & Mannucci, 2008), and am getting a little confused about Observables. From what I understand an observable is a question represented by a hermitian matrix. But that's as far as it goes. When we use an observable to make a measurement we obtain a real result, which can change the state space $|\Psi\rangle$. There is mention in the book that "after an observation" (presumably after a measurement is taken) then the result should be an eigenvalue, and the state of the system should collapse into the state which is the eigenvector corresponding to that eigenvalue.

Then, in example 4.3.1 on p.126 the authors use the observable $$\Omega=\begin{bmatrix}-1&-i\\i&1\end{bmatrix},$$ which they state has eigenvalues $\lambda_1=-\sqrt{2}$ and $\lambda_2=\sqrt{2}$, with corresponding eigenvectors $|e_1\rangle=[-0.923i,-0.382]^T$ and $|e_2\rangle=[-0.382,0.923i]^T$.

It goes on to say "now, let us suppose that afer an observation of $\Omega$ on $|\Psi\rangle=\frac{1}{2}[1,1]^T$, the actual value observed is $\lambda_1$. The system has "collapsed" from $|\Psi\rangle$ to $|e_1\rangle$.

I'm finding it difficult to understand this. Do the authors mean to perform a measurement, i.e.

$$\Omega|\Psi\rangle=\begin{bmatrix}-1&-i\\i&1\end{bmatrix}\frac{1}{2}\begin{bmatrix}1\\1\end{bmatrix}= \begin{bmatrix} -\frac{i}{2}-\frac{1}{2} \\ \frac{i}{2}+\frac{1}{2} \\ \end{bmatrix} $$

But then how have we observed $\lambda_1=\sqrt{2}$ ?

I think I've got the wrong end of the stick because the authors say "now, let us suppose that after an observation...," so maybe there is no calculation to be made, but it's very confusing.

Can anybody help me understand this?


1 Answer 1


When you give an observable, such as $\Omega$, that is used to define the measurement basis. It is not something that you would usually use to directly perform calculations (you can, and for $2\times 2$ matrices, we often do, as I'll detail below).

Normally, you want to take your observable, $\Omega$, and find the eigenvalues and eigenvectors. More specifically, you want projectors onto the different eigenspaces (this distinction is important if your matrix has degeneracy). So, we take $$ P_1=|e_1\rangle\langle e_1|,\qquad P_2=|e_2\rangle\langle e_2|. $$ In a sense, the eigenvalues are irrelevant, and hence so is $\Omega$. Our outcome is a state $$ |\Phi_i\rangle=P_i|\Psi\rangle, $$ up to normalisation (for rank 1 projectors, as here, the renormalised version is just $|e_i\rangle$), and the probability of getting outcome $i$ is $\langle\Phi_i|\Phi_i\rangle$. Note that you have to explicitly describe the branching outcomes (here, two different possibilities, with different probabilities). One simple calculation such as a matrix multiplication can't give you that.

Now, it turns out that you can use the matrix $\Omega$ directly if you want. This is because, by completeness, $$ \sum_iP_i=I, $$ and we know that $\Omega=\sqrt{2}(-P_1+P_2)$. Hence, we can rearrange for $P_1$: $$ P_1=\frac{1}{2}(I-\Omega/\sqrt{2}), $$ and we can perform the same calculation as previously without having directly calculated the eigenvectors first. There's always a trick like this for $2\times 2$ matrices. For $n\times n$ matrices, you'd probably have to describe each of your projectors as a polynomial in $\Omega$ up to a power $n-1$, which is why we don't usually do this for anything other than $2\times 2$ matrices - it just gets more messy.

  • $\begingroup$ Thank you. So as I now understand it, our observable $\Omega$ is simply a matrix representation of our question, which we don't really compute with at all. What matters is $\Omega$'s eigenvectors $|e_i\rangle$, which become our basis for measurement of our quantum system. To actually perform a calculation (measure) we work out $P_i=|e_i\rangle\langle e_i|\in M_{n,n}(\mathbb{C})$, and then then compute the outcomes as $|\Phi_i\rangle=P_i|\Psi\rangle$, the probability of getting each of which is $\langle\Phi_i|\Phi_i\rangle$. $\endgroup$
    – user8328
    Sep 11, 2019 at 12:17
  • $\begingroup$ @Pixel Yes, exactly. $\endgroup$
    – DaftWullie
    Sep 11, 2019 at 13:41
  • $\begingroup$ @DaftWullie "In a sense, the eigenvalues are irrelevant," ah! I feel like my rumblings in this question have been vindicated =) $\endgroup$
    – glS
    Sep 11, 2019 at 14:28

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