# 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?

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.

## 1 Answer

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.

• Good answer! Could you please also have a look at these two questions of mine: physics.stackexchange.com/questions/382397/… and math.stackexchange.com/questions/2622115/… ? – user182786 Jan 26 '18 at 14:31
• Sorry but I'm not familiar with those terms. In the first question I don't know what you mean by $\langle \psi \rangle$. I only understand the expectation value for an observable, not for a state. And I don't understand either what you mean by measurement operators. Maybe what you are trying to do is determining the state of one of the particles. But this is what I wrote in the answer. – MBolin Jan 26 '18 at 14:55
• And in the second question, as I said, I'm not familiar with writing 2-qubits states as matrices. I would write them as vectors $|00\rangle = [ 1, 0, 0, 0]^T$, $|01\rangle = [ 0,1,0,0]^T$, and so on. This way the projection onto say $|10\rangle$ would be $|10\rangle \langle 10| = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. But again maybe there is a way to do it writting the states as $2 \times 2$ matrices. – MBolin Jan 26 '18 at 15:02
• You mean there is nothing like expectation value of the whole system's quantum state (if it includes multiple particles)? Just expectation values for observables for single particle is valid? In your answer isn't "simultaneous state of A AND B", an observable in itself? – user182786 Jan 26 '18 at 15:16