# How do you find the possible measurement values of an observable?

$$\newcommand{\ket}{\left|#1\right>}$$ Note: I considered posting this as an update to a prior question, but it seemed like it should be it's own post.

So this is a very basic question, but one I've had a lot of confusion about. Here is my attempt at a solution:

Let's consider the example:

Bob has a single qubit in a normalized state state $$\ket{\psi}$$ = $$\frac{1}{2}[\ket{0}-\ket{1}]$$. He wants to measure an observable $$\sigma_x$$, which is the x-Pauli matrix, $$\sigma_x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$. He wants to measure the result along the x-basis.

First, to determine the (theoretical) possible measurement values and the states they correspond to, he finds the eigenstates and eigenvalues of $$\sigma_x$$. They are:

• $$\lambda_1=-1$$; $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}-\ket{1}]=\ket{-}$$
• $$\lambda_1=1$$; $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}+\ket{1}]=\ket{+}$$

where the eigenvectors could be defined slightly differently, but we're using this as convention.

To find the probabilities of each possible outcome, he then applies the $$\sigma_x$$ operator to the state %\ket{\psi}} and writes the result in the x basis. Doing so, he gets:

$$\sigma_x\frac{1}{\sqrt(2)}[\ket{0}-\ket{1}]$$
$$=\frac{1}{\sqrt(2)}[\ket{1}-\ket{0}]$$
$$=0\ket{+}-(1)\ket{-}$$

indicating that the $$\ket{+}$$ result (measurement value +1) occurs with probability 0 and the $$\ket{-}$$ result occurs with probability $$|(-1)|^2$$ = 1.

Now, let's instead consider the probability if he wanted to measure the operator along the z-basis. In this case, he would considers the eigenvectors of the $$\sigma_x$$ operator in terms of the z-basis. As before,:

• $$\lambda_1=-1$$; $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}-\ket{1}]$$
• $$\lambda_1=1$$; $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}+\ket{1}]$$

$$=\frac{1}{\sqrt(2)}[\ket{1}-\ket{0}]$$. This indicates that the $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}+\ket{1}]=$$ state above occurs with probability 0 and the $$\vec{v_1}=\frac{1}{\sqrt(2)}[\ket{0}-\ket{1}]=$$ occurs with probability . . .

and this is where I start to get lost; the probabilities aren't adding to 1; so something's wrong.

Attempt at understanding where I went wrong: I know that the the possible measurement values of an observable are the eigenvalues of the corresponding Hermitian matrix, and that the state of the system corresponding to that eigenvalue is its eigenvector. I'm not sure I should have included the factor of $$\frac{1}{\sqrt(2)}$$ in front of the eigenvector; it's just that in class we always use $$\ket{\pm}$$ as the eigenvectors of $$\sigma_x$$.

• If you measure $\sigma_x$ it doesn't matter what basis you use to perform the calculation. In your case, the result for will always be $-1$ with probability 1, since your initial state $\psi$ is already an eigenstate of $\sigma_x$ with eigenvalue -1. I'm not sure what you think you are doing with your description of how you are calculating the probabilities... The probability result is always $P(-1) = |\langle -|\psi\rangle|^2$
– hft
Mar 15 at 3:21
• In general, the probability is always $P(v) = |\langle v|\psi\rangle|^2$, where $|v\rangle$ is the eigenvector of the Hermitian operator being measured, and which corresponds to the measured value $v$.
– hft
Mar 15 at 3:23
• $\newcommand{\ket}{\left|#1\right>}$ For a state $\ket{\psi}=\alpha\ket{0}+\beta\ket{1}$, measurement will yield 1(the $\ket{0}$ state) with probability $|\alpha|^2$ and -1 (the $\ket{1}$ state) with probability $|\beta|^2$ Mar 15 at 14:27
• Sure, that's consistent. Because $\alpha = \langle 0|\psi\rangle$ and $\beta = \langle 1 |\psi\rangle$,
– hft
Mar 15 at 15:05

Unfortunately, some parts of this question are unclear to me as currently written. I will do my best to try and answer, in the sense of addressing some aspects of your below statement

Attempt at understanding where I went wrong: I know that the the possible measurement values of an observable are the eigenvalues of the corresponding Hermitian matrix,

Yes. An "observable" operator corresponds to a Hermitian matrix, and potentially observable values are the eigenvalues of the matrix, which thankfully are always real.

and that the state of the system corresponding to that eigenvalue is its eigenvector.

Only after the measurement. That is probably what you meant.

If the measurement is a projective measurement, which is general enough for our purposes, then the state of the system after the measurement is: $$\frac{\hat P |\psi\rangle}{\sqrt{\langle \psi|\hat P^\dagger \hat P |\psi\rangle}}\;,$$ and the probability ($$p$$) that I got the result corresponding to the projector was (I'm using the past tense, since I can calculate this probability before making the measurement): $$p = \langle\psi|\hat P^\dagger \hat P|\psi\rangle\;,$$ where, if you would like, you can simplify the formula for projective measurements since, for those measurements, $$\hat P=\hat P^\dagger=\hat P^2 = \hat P^\dagger \hat P$$.

For example, for measuring "-" (i.e., the -1 eigenvalue of $$\sigma_x$$) the projective measurement operator of interest is $$\hat P_- = |-\rangle\langle-|$$

For measuring "+" (i.e., the +1 eigenvalue of $$\sigma_x$$) the projective measurement operator of interest is $$\hat P_+ = |+\rangle\langle+|$$

For measuring "0" (i.e., the +1 eigenvalue of $$\sigma_z$$) the projective measurement operator of interest is $$\hat P_0 = |0\rangle\langle 0|$$

For measuring "1" (i.e., the -1 eigenvalue of $$\sigma_z$$) the projective measurement operator of interest is $$\hat P_1 = |1\rangle\langle 1|$$

For example, if I measure $$\sigma_x$$ and I get the result "-" (i.e., the -1 eigenvalue), the post-measurement state collapses to: $$\frac{|-\rangle \langle-|\psi\rangle}{\sqrt{\langle \psi|-\rangle\langle- |\psi\rangle}}\;,$$ and the probability that I got this result was: $$p(-) = \langle \psi |-\rangle\langle - |\psi \rangle = |\langle - |\psi\rangle|^2$$

For example, if I measure $$\sigma_z$$ and I get the result "0" (i.e., the +1 eigenvalue), the post-measurement state collapses to: $$\frac{|0\rangle \langle0|\psi\rangle}{\sqrt{\langle \psi|0\rangle\langle0 |\psi\rangle}}\;,$$ and the probability that I got this result was: $$p(0) = \langle \psi |0\rangle\langle 0 |\psi \rangle = |\langle 0 |\psi\rangle|^2$$

I'm not sure I should have included the factor of $$\frac{1}{\sqrt(2)}$$ in front of the eigenvector;

It's an eigenvector regardless of its normalization.

it's just that in class we always use $$\ket{\pm}$$ as the eigenvectors of $$\sigma_x$$.

By definition $$|\pm\rangle \equiv \frac{1}{\sqrt{2}}\left(|0\rangle \pm |1\rangle\right)\;,$$ where the $$\frac{1}{\sqrt{2}}$$ is there for proper normalization (i.e., so that the vectors are unit vectors).