Question

Sequential measurements. An operator $\stackrel{\mathbf{⏜}}{\mathbf{A}}$,representing observable A, has two normalized eigenstates ${\mathit{\psi }}_{\mathbf{1}}$and ${\mathit{\psi }}_{\mathbf{2}}$, with eigenvalues ${\mathit{a}}_{\mathbf{1}}$and ${\mathit{a}}_{\mathbf{2}}$, respectively. Operator $\stackrel{\mathbf{⏜}}{\mathbf{B}}$, representing observable $\stackrel{\mathbf{⏜}}{\mathbf{B}}$, has two normalized eigenstates ${\mathit{\varphi }}_{\mathbf{1}}$and ${\mathit{\varphi }}_{\mathbf{2}}$ , with eigenvalues ${\mathit{b}}_{\mathbf{1}}$and ${\mathit{b}}_{\mathbf{2}}$. The eigenstates are related by

${\mathit{\psi }}_{\mathbf{1}}\mathbf{=}\left(3{\varphi }_1+4{\varphi }_2\right)\mathbf{/}\mathbf{5}\mathbf{,}{\mathit{\psi }}_{\mathbf{2}}\mathbf{=}\left(4{\varphi }_1-3{\varphi }_2\right)\mathbf{/}\mathbf{5}$

(a) Observable Ais measured, and the value ${\mathit{a}}_{\mathbf{1}}$is obtained. What is the state of the system (immediately) after this measurement?

(b) If $\stackrel{\mathbf{⏜}}{\mathbf{B}}$is now measured, what are the possible results, and what are their probabilities?

(c) Right after the measurement of $\stackrel{\mathbf{⏜}}{\mathbf{B}}$,Ais measured again. What is the probability of getting ${\mathit{a}}_{\mathbf{1}}$? (Note that the answer would be quite different if I had told you the outcome of the $\stackrel{\mathbf{⏜}}{\mathbf{B}}$measurement.)

Short answer

a) The state is ${\psi }_1$.

b) The probabilities is 1 .

c) The probability is$\frac{337}{625}$ .

Step-by-step solution

Anti Hermitian operators.

Unitary transformations' infinitesimal generators are anti-Hermitian operators. As a result, each observable causes a state-space transition, and each observable causes a state-space transformation (or at least a Hermitian operator).

The state of the system (immediately) after the measurement.

a)

After measurement of A, the system can be found in only one of the two given eigenstates, namely ${\psi }_1$ or ${\psi }_2$. Since the value $a_1$has been obtained, one concludes that the system is (immediately) after the measurement in the state corresponding to the eigenvalue $a_1$; it is in the state ${\psi }_1$.

Their probabilities.

b)

The system is in the state ${\psi }_1$, but that state is not an eigenstate of the observable B, therefore expansion of the ${\psi }_1$state in the basis of the B observable. The relation between two basis has been given to us in the problem:

${\psi }_1=\frac{3}{5}{\varphi }_1+\frac{4}{5}{\varphi }_2$

Now, in whatever arbitrary state the system is in, the only two possible outcomes of the measurement are the states ${\varphi }_1$with eigenvalue $b_1$and ${\varphi }_2$with eigenvalue $b_2$. In this case, the probabilities are known, since one interprets the squared modules of the coefficients in the basis expansion as the probabilities of measuring the certain eigenvalue. Therefore, the probability to measure $b_1$ (probability to find the system in the state ${\varphi }_1$after the measurement) is 9/25 and the probability to measure $b_2$(probability to find the system in the state ${\varphi }_2$after the measurement) is 16/25 . It is important to notice that the vector is normalized, meaning the sum of the two probabilities is equal to 1 (meaning that the probability to find the system in any of the two given states is equal to one, as it should be, since these are all the possible outcomes).

Find the probability of getting a1 .

c)

First, let us rewrite the states ${\varphi }_1$and ${\varphi }_2$in the terms of the $\psi$basis. It can be trivially obtained by inverting the two expressions for ${\psi }_1$and ${\psi }_2$to obtain:

$$\begin{gathered} {\varphi }_1=\frac{\left(3{\psi }_1+4{\psi }_2\right)}{5} \\ {\varphi }_2=\frac{\left(4{\psi }_1+3{\psi }_2\right)}{5} \end{gathered}$$

The outcome of the B measurement is not known and therefore one cannot know in which state the system will be in after the B measurement. However, what one can do is sum up probabilities of all of the possible outcomes. See that if after the B measurement result obtained is $b_1$, the system is in the ${\varphi }_1$state and now if measured A , the probability to get $a_1$(find the system in the ${\psi }_1$state) is (given as the square modulus of the appropriate coefficient, as always)9/25 . The second scenario is if one obtain $b_2$after the B measurement, therefore the system is in the ${\varphi }_2$state, so, the probability to measure $a_1$in the A measurement in this scenario is16/25 . Multiply the mentioned probabilities with the probability to get the certain scenario after the B measurement, so the total probability, since the initial state is ${\psi }_1$is:

$$\begin{gathered} P=\frac{9}{25}.\frac{9}{25}+\frac{16}{25}.\frac{16}{25} \\ =\frac{337}{625} \end{gathered}$$

Thus, the sum of the probabilities is $\frac{337}{625}$.