Quantum Measurements

## **Problem**

Let there be two operators

A
^

{\displaystyle \hat{A}}

 and

B
^

{\displaystyle \hat{B} }

 that represent their respective observables

A

{\displaystyle A}

 and

B

{\displaystyle B}

.

A
^

{\displaystyle \hat{A}}

 has two eigenvalues

a

1

{\displaystyle {a}\_{1} }

 and

a

2

{\displaystyle {a}\_{2}}

, each corresponding to respective normalized eigenstates:

ψ

1

=

1
5

(
3

ϕ

1

4

ϕ

2

)

{\displaystyle {\psi}\_{1} = \frac{1}{5}(3{\phi}\_{1} + 4{\phi}\_{2})}

ψ

2

=

1
5

(
4

ϕ

1

−
3

ϕ

2

)

{\displaystyle {\psi}\_{2} = \frac{1}{5}(4{\phi}\_{1} - 3{\phi}\_{2})}

B
^

{\displaystyle \hat{B} }

 also has two eigenvalues

b

1

{\displaystyle {b}\_{1} }

 and

b

2

{\displaystyle {b}\_{2} }

, which correspond respectively to normalized eigenstates

ϕ

1

{\displaystyle {\phi}\_{1}}

 and

ϕ

2

{\displaystyle {\phi}\_{2}}

.

You make an initial measurement of

A

{\displaystyle A}

, recording a value of

a
1

{\displaystyle a\_1}

. You then measure

B

{\displaystyle B}

, then

A

{\displaystyle A}

 again. What is the probability that you record

a
1

{\displaystyle a\_1}

 again?

## **Solution**

If

A

{\displaystyle A}

 is measured and

a

1

{\displaystyle {a}\_{1} }

​ is obtained, the state of the system after the measurement is

ψ

1

⟩

{\displaystyle |{\psi}\_{1}\rangle }

. Then,

B

{\displaystyle B}

 is now measured with outcomes

b

1

{\displaystyle {b}\_{1} }

​ of probability

⟨

ϕ

1

ψ

1

⟩

2

=

9
25

{\displaystyle {|\langle {\phi}\_{1}|{\psi}\_{1} \rangle|}^{2} = \frac{9}{25} }

 and

b

2

{\displaystyle {b}\_{2} }

​ of probability

⟨

ϕ

2

ψ

1

⟩

2

=

16
25

{\displaystyle {|\langle {\phi}\_{2}|{\psi}\_{1} \rangle|}^{2} = \frac{16}{25} }

.

Now

A

{\displaystyle A}

 is measured again, we can express

ϕ

1

⟩

{\displaystyle |{\phi}\_{1}\rangle}

 and

ϕ

2

⟩

{\displaystyle |{\phi}\_{2}\rangle}

 as:

ϕ

1

# ⟩

3
5

ψ

1

⟩

4
5

ψ

2

⟩

{\displaystyle |{\phi}\_{1}\rangle = \frac{3}{5} |{\psi}\_{1} \rangle + \frac{4}{5} |{\psi}\_{2}\rangle}

ϕ

2

# ⟩

4
5

ψ

1

⟩
−

3
5

ψ

2

⟩

{\displaystyle |{\phi}\_{2}\rangle = \frac{4}{5} |{\psi}\_{1} \rangle - \frac{3}{5} |{\psi}\_{2}\rangle}

The probability of obtaining

a

1

{\displaystyle {a}\_{1} }

​ if we knew the outcome of the measurement of

B

{\displaystyle B}

 as

b

1

{\displaystyle {b}\_{1} }

 and the system is in state

ϕ

1

⟩

{\displaystyle |{\phi}\_{1}\rangle}

 is​

⟨

ψ

1

ϕ

1

⟩

2

=

9
25

{\displaystyle {|\langle {\psi}\_{1}|{\phi}\_{1} \rangle|}^{2} = \frac{9}{25} }

. The probability of obtaining

a

1

{\displaystyle {a}\_{1} }

 if we knew the outcome of the measurement of

B

{\displaystyle B}

 as

b

2

{\displaystyle {b}\_{2} }

​ and the system is in state

ϕ

2

⟩

{\displaystyle |{\phi}\_{2}\rangle}

 is

⟨

ψ

1

ϕ

2

⟩

2

=

16
25

{\displaystyle {|\langle {\psi}\_{1}|{\phi}\_{2} \rangle|}^{2} = \frac{16}{25} }

​. The total probability of obtaining

a
1

{\displaystyle a\_1}

 in the series of measurement

A
→
B
→
A

{\displaystyle A \rightarrow B \rightarrow A}

 is the sum of the probability of obtaining

a

1

{\displaystyle {a}\_{1} }

 after obtaining

b

1

{\displaystyle {b}\_{1} }

 and the probability of obtaining

a

1

{\displaystyle {a}\_{1} }

 after obtaining

b

2

{\displaystyle {b}\_{2} }

.

P

=

⟨

ϕ

1

ψ

1

⟩

2

⋅

⟨
ψ

1

ϕ

1

⟩

2

⟨

ϕ

2

ψ

1

⟩

2

⋅

⟨

ψ

1

ϕ

2

⟩

2

=

9
25

⋅

9
25

16
25

⋅

16
25

=
0.5392

{\displaystyle \begin{align}
&= \frac{9}{25}\cdot\frac{9}{25} + \frac{16}{25} \cdot \frac{16}{25} \\[5 pt]
&= 0.5392
\end{align} }