Measuring a qubit

1. Mar 1, 2015

jimmycricket

I undertand that measuring a qubit's spin along a particular axis yields one of the eigenstates of the corresponding pauli matrices. I'm a little confused about what this type of measurement looks like mathematically. i.e how do you show the actual act of observing the qubits spin using the Pauli matrices?

2. Mar 1, 2015

matteo137

Let's say that your qubit is in the state $\vert\psi\rangle = c_0 \vert 0 \rangle + c_1 \vert 1 \rangle$, a simplistic description of a measurement along the $z$ direction is given by evaluating the expectation value of the Pauli $z$ operator, which is $\langle\psi\vert\sigma_z\vert\psi\rangle = \vert c_1\vert^2 - \vert c_0\vert^2$. (if $\sigma_z\vert 0\rangle =-\vert 0\rangle$).

Which means that if you measure your qubit $N$ times, you will obtain on average $\vert c_1\vert^2 N$ times $\vert 1\rangle$, and the rest $\left( 1- \vert c_1\vert^2\right) N = \vert c_0\vert^2 N$ times $\vert 0\rangle$.

3. Mar 1, 2015

Staff: Mentor

But you can't do that. It means that given an ensemble of $N$ identically prepared systems...

4. Mar 1, 2015

matteo137

Yes of course, I was too imprecise, sorry.

5. Mar 1, 2015

Staff: Mentor

There is no mathematical operation you can make, apart from saying that before a measurement of observable $\hat{A}$, with
$$\hat{A} | \phi_i \rangle = a_i | \phi_i \rangle$$
the system is in the state
$$| \psi \rangle = \sum c_i | \phi_i \rangle$$
and after having measured the value $a_j$, it is in the state
$$| \psi \rangle = | \phi_j \rangle$$

6. Mar 1, 2015

matteo137

Actually there is. You can say that you are making a measurement by introducing the Hamiltonian
$$H_{\text{meas}} = g A\otimes M$$
which couples the qubit to a meter ...

I can suggest you this good review: http://dml.riken.jp/pub/nori/pdf/PhysRep.520.pdf [Broken] (section 2.2.1)

Last edited by a moderator: May 7, 2017