# What happens when the same observable is measured twice in a row?

I have the understanding that the outcome of measuring the same observable twice in a row on the same state is getting the square of the eigenvalue if and only if the state is an eigenstate of the observable. Is it true?

Otherwise $$\langle O \rangle \psi_n = n \psi_n$$ would not hold, right?

• True. The onservable should have an eigenvalue. – Aman pawar Oct 9 '18 at 18:18
• You do not get the square, but the same eigenvalue. – DanielC Oct 9 '18 at 18:22
• If you get a different value in the second measurement, you weren't paying attention to the experimental setup and something else happened between measurements, or your measurement tool is broken. – ohwilleke Oct 9 '18 at 18:27
• I suggest you take a look at the Stern-Gerlach experiment. en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment That page has a simple diagram showing the outcome of spin measurements. – R. Rankin Oct 10 '18 at 0:37

## 1 Answer

I have the understanding that the outcome of measuring the same observable twice in a row on the same state is getting the square of the eigenvalue

If you measure the observable $$O$$ and get a value $$o_1$$, then you immediately measure $$O$$ again, you'll get the value $$o_1$$ and not $$o^2_1$$. This is because the first measurement leaves the system in the state $$|o_1\rangle$$.

Note: measurement of an observable $$O$$ is not equivalent to acting on the state with $$O$$

This is most easily seen by acting on a superposition of eigenstates of $$O$$

$$O(c_1|o_1\rangle + c_2|o_2\rangle) = c_1o_1|o_1\rangle + c_2o_2|o_2\rangle$$

which is not an eigenstate of $$O$$. But, according to the measurement postulate, immediately after measuring $$O$$, the state will be an eigenstate of $$O$$.

• What do you mean by 'acting on the state with O'? – JD_PM Oct 9 '18 at 18:28
• @JD_PM, 'acting on the state with O' means the operator $O$ takes as input the state vector and produces as output another state vector, e.g., $|\psi_b\rangle = O|\psi_a\rangle$ – Alfred Centauri Oct 9 '18 at 18:36