Measure the state of the second qubit

[Kara386]

Homework Statement

For the state
$(4|00\rangle+3i|11\rangle)\otimes (|0\rangle+i|1\rangle) + (2|01\rangle -i|10\rangle)\otimes(|0\rangle-|1\rangle)$
What's the probability of zero being the outcome of measuring the second bit and what is the state of the other two qubits after measurement?

Homework Equations

The Attempt at a Solution

Expanding the tensor product, for half the states the second qubit is zero so the probability is half, but I don't know what I can say about the state of the other two. Because in states where the second qubit is zero, the other qubits are sometimes one and sometimes zero I think? Unless I've calculated the states wrong? I had the final state after measurement as this:

$\frac{1}{\sqrt{34}} (4|000\rangle + 4i|001\rangle -i|100\rangle +i|101\rangle)$
Which tells me about all three qubits after measurement not just two of them.
Thanks for any help, I really appreciate it!

 

[DrClaude]

Expanding the tensor product, for half the states the second qubit is zero so the probability is half

The part in bold is not correct.

$\frac{1}{\sqrt{34}} (4|000\rangle + 4i|001\rangle -i|100\rangle +i|101\rangle)$
Which tells me about all three qubits after measurement not just two of them.

That's just a question of notation. You can factor out the state of qubit 2, and write the remaining kets for qubits 1 and 3. (To avoid confusion, I would write something like $| 00 \rangle_{1,3}$ for state $| 000 \rangle$, etc.)

 

[Kara386]

The part in bold is not correct.That's just a question of notation. You can factor out the state of qubit 2, and write the remaining kets for qubits 1 and 3. (To avoid confusion, I would write something like $| 00 \rangle_{1,3}$ for state $| 000 \rangle$, etc.)

Ah, if I actually normalise the wavefunction and use the coefficients then I get a probability of 17/30. And I'll adopt that notation for my answer, thank you, I really appreciate your help!

 

[DrClaude]

Ah, if I actually normalise the wavefunction and use the coefficients then I get a probability of 17/30.

That's better!