5
$\begingroup$

My knowledge of quantum mechanics is very limited, but I will try to ask a purely mathematical question here. If there is a text or resource that explains this, I would definitely appreciate any pointers! I have been unable to find any explanations online (most quantum mechanical resources seem too focused on the physics to ask such questions...).

I am coming at this from a computer science angle, so I will talk about qubits. To make things very simple and concrete, suppose I have two qubits in some quantum state: $$ a_{00} \left|00\right\rangle + a_{01} \left|01\right\rangle + a_{10} \left|10\right\rangle + a_{11} \left|11\right\rangle $$ where $a_{ij} \in \mathbb{C}$, $\sum_{ij} |a_{ij}|^2 = 1$.

Suppose I now measure the first bit. Given that I measure $0$, what is the quantum state of my system? Given that I measure $1$, what is the quantum state of my system?

A naive extension of Baye's rule leads me to conjecture that, for instance, if I measure zero for the first bit, then the quantum state is $$ \sqrt{\frac{|a_{00}|^2}{|a_{00}|^2 + |a_{01}|^2}} \left|00\right\rangle + \sqrt{\frac{|a_{01}|^2}{|a_{00}|^2 + |a_{01}|^2}} \left|01\right\rangle .$$

But is this really correct?

The general question would be that if I have a quantum state $\sum_i a_i \left|\phi_i\right\rangle$, and I "partially observe" $\phi_i$ (so I guess one would say I "partially" collapse the wave function?), how is the new quantum state mathematically determined?

$\endgroup$
3
$\begingroup$

Any observation in QM is tied to a hermitian operator, and the collapse of the wave function is always to the eigenstate of that operator corresponding to the measurement you got (at least as long as the possible outcomes are discrete). In your case we're dealing with a four-dimensional vector space over $\Bbb C$. Let's represent your quantum state as the column vector $$ \begin{pmatrix}a_{00}\\ a_{01} \\ a_{10}\\ a_{11}\end{pmatrix} $$ Then your "reading the first qbit"-observation is tied to the operator represented by the matrix $$ \begin{pmatrix}0 & 0 & 0 & 0\\ 0 &0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix} $$ When reading the first qbit as $1$, the state collapses to the eigenspace spanned by the eigenvectors with eigenvalue $1$, and while I do not know exactly how to handle degenerate states (i.e. that the eigenspace has more than one dimension), I would believe it collapses to $$ \begin{pmatrix}0\\ 0 \\ a_{10}\\ a_{11}\end{pmatrix} $$ which after a renormalization becomes $$ \begin{pmatrix} 0 \\ 0 \\ a_{10}\bigg/\sqrt{|a_{10}|^2 + |a_{11}|^2} \\ a_{11}\bigg/\sqrt{|a_{10}|^2 + |a_{11}|^2}\end{pmatrix} $$ Edit Wiki seems to confirm this, by the way. The vector $$ \begin{pmatrix}0\\ 0 \\ a_{10}\\ a_{11}\end{pmatrix} $$ is the projection of your original state onto the subspace of $C^4$ spanned by the eigenvectors of eigenvalue $1$ (that is, the last two coordinates).

$\endgroup$
  • $\begingroup$ The renormalization here is precisely conditional probability at work. Afterall you are dividing by the probability of the first bit being 1. $\endgroup$ – Alex R. Oct 17 '13 at 14:35
  • $\begingroup$ That's very clear, thank you! $\endgroup$ – usul Oct 17 '13 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.