Simple Quantum Information Question

Click For Summary
SUMMARY

The discussion centers on the measurement of a quantum state represented as \ket{\psi} = \alpha \ket{10} + \beta \ket{11} + \gamma \ket{01} + \theta \ket{00}. Upon measuring the first qubit and observing a value of 1, the state is projected onto the subspace spanned by |10⟩ and |11⟩. The resulting projected state is |\psi⟩_{\mathit{proj}}=a_{10}|10⟩ + a_{11}|11⟩, which requires renormalization to yield |\psi_1⟩= (a_{10}|10⟩ + a_{11}|11⟩)/\sqrt{|a_{10}|^2 + |a_{11}|^2}. This confirms that the measurement process involves projection and subsequent renormalization of the quantum state.

PREREQUISITES
  • Understanding of quantum states and notation, specifically Dirac notation.
  • Familiarity with quantum measurement theory and projection operators.
  • Knowledge of state renormalization in quantum mechanics.
  • Basic concepts of entanglement and independence of qubits.
NEXT STEPS
  • Study quantum measurement theory in detail, focusing on projection operators.
  • Learn about the implications of entanglement on measurement outcomes.
  • Explore the mathematical framework of quantum state renormalization.
  • Investigate the role of density operators in mixed states and measurements.
USEFUL FOR

Quantum physicists, students of quantum mechanics, and anyone interested in the principles of quantum measurement and state manipulation.

Pbrunett
Messages
5
Reaction score
0
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!
 
Physics news on Phys.org
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).
 
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}.
 
Oops, that's how I meant to renormalize it! Still working on the TeX...
 
Awesome, that's what I was hoping for. Thanks for the advice folks!
 
Last edited:

Similar threads

  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
8
Views
6K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K