- #1

Peter_Newman

- 155

- 11

I would like to start with an assumption. Suppose a system is in the state:

$$|\psi\rangle=\frac{1}{\sqrt{6}}|0\rangle+\sqrt{\frac{5}{6}}|1\rangle$$

The question is now: A measurement is made with respect to the observable Y. The expectation or average value is to calculate.

My first ideas are this:

$$Y=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$$

the eigenvalues are:

$$|u_1\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}, |u_2\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$$

The projection operator corresponding to a measurement of +1 is:

$$P_{u_1}=|u_1\rangle\langle u_1|=\left(\frac{|0\rangle+i|1\rangle}{\sqrt{2}}\right)\left(\frac{\langle0|-i\langle 1|}{\sqrt{2}}\right)$$

$$=\frac{1}{2}(|0\rangle\langle 0|-i|0\rangle\langle 1| +i|1\rangle\langle 0| +|1\rangle\langle 1| )$$

for ##P_{u_2}##

$$P_{u_2}=|u_2\rangle\langle u_2|=\frac{1}{2}(|0\rangle\langle 0|+i|0\rangle\langle 1| -i|1\rangle\langle 0| +|1\rangle\langle 1| )$$

accordingly

$$Pr(+1)=\langle \psi|P_{u_1}|\psi\rangle=\left(\frac{1}{\sqrt{6}}\langle 0|+\sqrt{\frac{5}{6}}\langle 1|\right)\left(\frac{1}{2}(|0\rangle\langle 0|-i|0\rangle\langle 1| +i|1\rangle\langle 0| +|1\rangle\langle 1| \right)\left(\frac{1}{\sqrt{6}}|0\rangle +\sqrt{\frac{5}{6}}|1\rangle\right)$$

I've calculated that and get to the following result:

$$Pr(+1)=0.5, Pr(-1)=0.5$$

The probabilities sum up to 1:

$$\langle \psi | P_{u_1}|\psi\rangle+\langle \psi | P_{u_2}|\psi\rangle=1$$

$$0.5+0.5=1$$

I would be interested to know if this is okay so far? My problem arises here:

The average value is: ##\langle X \rangle = (+1)Pr(+1)+(-1)Pr(-1)=0##

My question is, how can the average value be 0?

If something does not fit with the notation here, I would be very happy to be corrected.

So I would be very happy about answers and criticism. Thank you!