How to write the probability of ending up in a state where qubit $A$ is in the state $|0\rangle_A$ when $A$ is measured, in the bra-ket notation?

Question

I need some help with the bra-ket notation. Suppose we have a normalized wavefunction for a two-qubit system (where $A$ and $B$ denote the two qubits respectively), like: $$|\psi\rangle_{A B} = a(|0\rangle_A \otimes |0\rangle_B) + b (|0\rangle_A \otimes |1\rangle_B) + c (|1\rangle_A \otimes |0\rangle_B) + d (|1\rangle_A \otimes |1\rangle_B)$$

Then what is the correct notation for writing the probability of wavefunction to collapse to a state where the state of qubit $A$ is $|0\rangle_A$ ? Is it $|_A\langle 0|\psi\rangle_{AB}|^2$ (which should be $|a|^2 + |b|^2$) ?

Also it would be very helpful if someone could link me a webpage or resource which discusses the properties of this operation i.e. multiplying a bra with the tensor product of two ket vectors. Does multiplying $\langle 0|_A$ with $|0\rangle_A \otimes |0\rangle_B$ or $|0\rangle_A \otimes |1\rangle_B$ equal $1$ ? And does multiplying $\langle 1|_A$ with $|1\rangle_A \otimes |0\rangle_B$ or $|1\rangle_A \otimes |1\rangle_B$ equal $0$ ? I'm not very sure about this.

Answer 1

I will first answer the second question. In a two particle system the basis is made up of tensor products of the form $| \psi\rangle_A \otimes | \phi \rangle_B$ that we can write $| \psi \phi \rangle$ remembering that position $1$ refers to particle $A$ and postion $2$ refers to particle $B$. The bra-ket multiplication of two such states is $\langle \psi_1 \phi_1 | \psi_2 \phi_2\rangle = \langle \psi_1 | \psi_2 \rangle \cdot \langle \phi_1 | \phi_2 \rangle $. In this sense it is meaningless to think about the multiplication of ${}_A\langle 0 |$ with anything, because this is not a possible state of the system.

And yes, that is the probability. One of the postulates of quantum mechanics is the following. The measurement of an observable $A$ on a state $| \psi \rangle $ can only give as an answer an eigenvalue $a_n$ of $A$, with probability

$P (a_n) = \sum\limits_{i=1}^{g_n} \big| \langle u_n^i | \psi \rangle \big|^2$

where $\{ | u_n^i \rangle, i= 1, ..., g_n \}$ is the subspace of eigenvectors of $A$ with eigenvalue $a_n$, and degeneracy $g_n$.

In your case you may think of the observable "state of A" $S_A$ with some eigenvalues $\{y, y, n, n\}$ in the basis $\{ |00\rangle, | 01 \rangle, |10\rangle , |11\rangle\}$. You want to know the probability of measuring the eigenvalue $y$ ("$A$ being in state $|0\rangle$"), with associate subspace $\{ |00\rangle, | 01 \rangle\}$. Therefore

$P(y) = \big|\langle 00 |\psi \rangle_{AB} \big|^2 + \big|\langle 01 |\psi \rangle_{AB} \big|^2 = \big| a \big|^2 + \big| b\big|^2 $

This is how I understand it, but maybe there is a simpler way to do it.