← Exercise library ← All exercises for this theme

Theme 3 — Postulates and Applications

Successive measurements on a three-level system

Keywords: postulates · Born rule · wave-function collapse · degenerate eigenvalue · expectation value · standard deviation · incompatible observables

Exercise 1 : Successive measurements on a three-level system

Consider a quantum system whose state space has dimension three, with an orthonormal basis $(\ket{u_1},\ket{u_2},\ket{u_3})$ . An observable $\hat A$ is defined by

\hat A\ket{u_1}=0,\qquad \hat A\ket{u_2}=a\ket{u_2},\qquad \hat A\ket{u_3}=a\ket{u_3},

where $a$ is a non-zero real number. The system is prepared in the state

\ket\psi=\frac{1}{\sqrt6}\Bigl(\ket{u_1}+2\ket{u_2}+i\ket{u_3}\Bigr).

Recall that, if $\hat P_n$ is the projector onto the eigenspace associated with the eigenvalue $a_n$ , the probability of obtaining $a_n$ is $\bra\psi\hat P_n\ket\psi$ , and the state immediately after the measurement is $\hat P_n\ket\psi/\sqrt{\bra\psi\hat P_n\ket\psi}$ .

Question 1

Verify that

\ket\psi

is normalised. What are the possible outcomes of a measurement of

\hat A

Solution

Since the basis is orthonormal, the squared norm of a vector is the sum of the squared moduli of its components:

\braket{\psi}{\psi}=\frac16\Bigl(|1|^2+|2|^2+|i|^2\Bigr)=\frac16(1+4+1)=1 .

The state is therefore normalised. According to the second postulate, the possible outcomes of a measurement of $\hat A$ are its eigenvalues. There are two: $0$ , associated with the single vector $\ket{u_1}$ , and $a$ , associated with the two-dimensional subspace spanned by $\ket{u_2}$ and $\ket{u_3}$ . The eigenvalue $a$ is thus doubly degenerate.

Question 2

Calculate the probabilities of the two outcomes, then the expectation value

\langle\hat A\rangle

and the standard deviation

\sigma_A

Solution

The projectors onto the two eigenspaces are

\hat P_0=\ket{u_1}\bra{u_1},\qquad \hat P_a=\ket{u_2}\bra{u_2}+\ket{u_3}\bra{u_3}.

For the non-degenerate eigenvalue $0$ , the Born rule gives

P(0)=|\braket{u_1}{\psi}|^2=\Bigl|\frac{1}{\sqrt6}\Bigr|^2=\frac16 .

For the degenerate eigenvalue $a$ , the squared moduli of the components in an orthonormal basis of the eigenspace must be summed:

P(a)=|\braket{u_2}{\psi}|^2+|\braket{u_3}{\psi}|^2=\frac46+\frac16=\frac56 .

We verify that $P(0)+P(a)=1$ . The expectation value is the probability-weighted average of the outcomes:

\langle\hat A\rangle=0\times\frac16+a\times\frac56=\frac{5a}{6}.

To find the standard deviation, first calculate $\langle\hat A^2\rangle$ . The operator $\hat A^2$ has the same eigenvectors as $\hat A$ , with eigenvalues $0$ and $a^2$ , so $\langle\hat A^2\rangle=a^2\times\frac56$ . Hence

\sigma_A^2=\langle\hat A^2\rangle-\langle\hat A\rangle^2=\frac{5a^2}{6}-\frac{25a^2}{36}=\frac{30a^2-25a^2}{36}=\frac{5a^2}{36}, \qquad \sigma_A=\frac{\sqrt5}{6}\,|a|\simeq0{,}37\,|a| .

Question 3

The measurement yielded the outcome

a

. What is the state

\ket{\psi'}

of the system immediately after the measurement? What would a second measurement of

\hat A

performed immediately afterwards yield?

Solution

According to the collapse postulate, the state is projected onto the eigenspace associated with the outcome obtained and then renormalised:

\hat P_a\ket\psi=\frac{1}{\sqrt6}\Bigl(2\ket{u_2}+i\ket{u_3}\Bigr), \qquad \norm{\hat P_a\ket\psi}=\sqrt{P(a)}=\sqrt{\frac56},

and therefore

\ket{\psi'}=\frac{\hat P_a\ket\psi}{\sqrt{P(a)}}=\frac{1}{\sqrt6}\sqrt{\frac65}\Bigl(2\ket{u_2}+i\ket{u_3}\Bigr)=\frac{1}{\sqrt5}\Bigl(2\ket{u_2}+i\ket{u_3}\Bigr).

Note that the measurement did not select $\ket{u_2}$ or $\ket{u_3}$ separately: because the eigenvalue is degenerate, the state after the measurement is the projection of the initial state onto the entire eigenspace, and it retains the relative proportions of the two components. The state $\ket{\psi'}$ belongs to the eigenspace with eigenvalue $a$ ; a new measurement of $\hat A$ therefore yields $a$ with probability $1$ .

Question 4

In the state

\ket{\psi'}

, we now measure a second observable

\hat B

defined by

\hat B\ket{u_1}=\ket{u_2},\qquad \hat B\ket{u_2}=\ket{u_1},\qquad \hat B\ket{u_3}=\ket{u_3}.

Determine the eigenvalues and eigenvectors of $\hat B$ , then the probabilities of the outcomes of this measurement. Do the observables $\hat A$ and $\hat B$ commute?

Hint

The subspace spanned by

\ket{u_1}

and

\ket{u_2}

is invariant under

\hat B

; in this subspace, the matrix of

\hat B

is a familiar Pauli matrix.

Solution

In the basis

(\ket{u_1},\ket{u_2},\ket{u_3})

, the matrix of

\hat B

\begin{aligned} B=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}. \end{aligned}

It is block diagonal. The $2\times2$ block acting on $\ket{u_1},\ket{u_2}$ is the matrix $\sigma_x$ , with eigenvalues $\pm1$ and eigenvectors $(\ket{u_1}\pm\ket{u_2})/\sqrt2$ . The vector $\ket{u_3}$ is an eigenvector with eigenvalue $1$ . Thus:

the eigenvalue $+1$ is doubly degenerate, with eigenspace spanned by $\ket{v_+}=(\ket{u_1}+\ket{u_2})/\sqrt2$ and $\ket{u_3}$ ;
the eigenvalue $-1$ is non-degenerate, with eigenvector $\ket{v_-}=(\ket{u_1}-\ket{u_2})/\sqrt2$ .

Let us calculate the components of $\ket{\psi'}=(2\ket{u_2}+i\ket{u_3})/\sqrt5$ along these vectors:

\braket{v_-}{\psi'}=\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt5}\bigl(0-2\bigr)=-\frac{2}{\sqrt{10}}, \qquad \braket{v_+}{\psi'}=\frac{2}{\sqrt{10}}, \qquad \braket{u_3}{\psi'}=\frac{i}{\sqrt5}.

The probabilities are therefore

P(B=-1)=\frac{4}{10}=\frac25, \qquad P(B=+1)=\frac{4}{10}+\frac15=\frac35 .

The two observables do not commute. Indeed, $\hat A\hat B\ket{u_1}=\hat A\ket{u_2}=a\ket{u_2}$ , whereas $\hat B\hat A\ket{u_1}=\hat B(0)=0$ : $[\hat A,\hat B]\ket{u_1}=a\ket{u_2}\neq0$ . Physically, $\hat B$ mixes the vectors $\ket{u_1}$ and $\ket{u_2}$ , which correspond to different eigenvalues of $\hat A$ .

Question 5

The measurement of

\hat B

yielded

-1

. We measure

\hat A

again. What are the probabilities of the outcomes? Comment on the result.

Solution

After the outcome

-1

, a non-degenerate eigenvalue, the state is

\ket{v_-}=(\ket{u_1}-\ket{u_2})/\sqrt2

, up to a global phase. A measurement of

\hat A

then gives

P(0)=|\braket{u_1}{v_-}|^2=\frac12,\qquad P(a)=|\braket{u_2}{v_-}|^2+|\braket{u_3}{v_-}|^2=\frac12 .

Although the first measurement of $\hat A$ yielded $a$ , and an immediately repeated measurement would have yielded $a$ again with certainty, the intermediate measurement of $\hat B$ has made the outcome $0$ possible once more. This is the same situation as the $z$ , $x$ , $z$ sequence in the Stern—Gerlach experiment: when two observables do not commute, measuring one changes the predictions concerning the other.