# Simple Quantum Information Question

1. Mar 21, 2008

### Pbrunett

Hey folks, I have a pretty simple quantum information question that I was hoping somebody could answer.

let's say I have a pure state \ket{\psi} = \alpha \ket{10} + \beta \ket{11} + \gamma \ket{01} + \theta \ket{00}.

I then perform a measurement on only the first qubit and observe a value of 1. How do I now represent the state of my system? The temptation is to just renormalize the coefficients \alpha and \beta, but it's not clear to me whether this is correct or whether I have to use a density operator. Any advice would be awesome, this is a question that popped up while I was reading N and C for self-study. Thanks for your time!

2. Mar 22, 2008

### Ken G

It's a good question. I do believe it will be in the renormalized pure state you imagine, as long as the measurement on the first qubit is independent of the second qubit (perhaps the entangled particles are physically separated, for example). The way to think of this is to write the system as a linear combination of the two more obvious cases, the first having in effect \alpha=\theta and \beta=\gamma (so the second qubit is in its own independent pure state before and after the measurement on the first qubit), and the second having in effect \alpha=\beta and \gamma=\theta (so the renormalized result we are talking about is more obviously going to be correct). Rebuilding the combination will yield a pure state, and if you work it out, my money says it will be (\alpha \ket{10} + \beta \ket{11}) / (\alpha + \beta).

3. Mar 23, 2008

### funkstar

Remember that measurement is projection and renormalization.

Alright, so $$|\psi\rangle=a_{00}|00\rangle+a_{01}|01\rangle + a_{10}|10\rangle + a_{11}|11\rangle$$. A measurement of the first qubit as 1 means that $$|\psi\rangle$$ should be projected onto the subspace spanned by $$|10\rangle$$ and $$|11\rangle$$. Since $$|00\rangle$$ and $$|01\rangle$$ are orthogonal to this subspace, the projected state is simply $$|\psi\rangle_{\mathit{proj}}=a_{10}|10\rangle + a_{11}|11\rangle$$, which, however, is not generally a unit vector. So we renormalize in the standard way, and the state after measurement is $$|\psi_1\rangle= (a_{10}|10\rangle + a_{11}|11\rangle)/||\psi\rangle_{\mathit{proj}}|= (a_{10}|10\rangle + a_{11}|11\rangle)/\sqrt{|a_{10}|^2 + |a_{11}|^2}$$.

4. Mar 23, 2008

### Ken G

Oops, that's how I meant to renormalize it! Still working on the TeX...

5. Mar 24, 2008

### Pbrunett

Awesome, that's what I was hoping for. Thanks for the advice folks!

Last edited: Mar 24, 2008