# I Measurement with respect to the observable Y

#### Peter_Newman

Hello,

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!

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#### jambaugh

Gold Member
You may calculate the expectation value more directly without resorting to eigen-value decomposition:
$$\langle Y\rangle =\psi^\dagger Y \psi = \left(\begin{array}{cc}\sqrt{1/6} & \sqrt{5/6}\end{array}\right) \left(\begin{array}{rr} 0 & i \\ -i & 0\end{array}\right)\left(\begin{array}{r}\sqrt{1/6}\\ \sqrt{5/6}\end{array}\right)=$$
$$= \left(\begin{array}{cc}\sqrt{1/6} & \sqrt{5/6}\end{array}\right) \left(\begin{array}{r}i\sqrt{5/6}\\ -i\sqrt{1/6}\end{array}\right)= i\frac{\sqrt{5}}{6} - i\frac{\sqrt{5}}{6} =0$$

So you see I am getting 0 as well. That's not impossible. You gave the two eigen-vectors for Y and their respective eigen-values were +1 and -1. You also noted that their separate probabilities, given your initial system "state" were each 1/2 as you verified the total probability was 1. It is then quite clear that if you have outcome +1 50% of the time and outcome -1 50% of the time you'll have an average outcome of 0. Quite literally "you win some, you lose some".

#### Peter_Newman

"Measurement with respect to the observable Y"

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