# psi_A(x,0) = sqrt(1/6)phi_0(x) + sqrt(1/3)phi_1(x) + sqrt(1/2)phi_2(x)? More questions

## (d) What energy values will be observed as a result of a single measurement at t=0 and with what probabilities? How do these probabilities change with time? (c) What is the probability of measuring the energy to equal $\hat{E}$ as a result of a single measurement at t=0? At a later time t=t1? (e) The energy of the particle is found to be E2 as a result of a single measurement at t=t1. Write down the wave function ψA(x,t) which describes the state of the particle for t> t1. What energy values will be observed and with what probabilities at a time t2 > t1? (f) Construct another normalized wave function ψB(x, 0) which is linearly independent of ψA(x, 0), but yields the same value of $\hat{E}$ as well as the same set of measured energies with the same probabilities. (g) Construct another normalized wave function ψC(x, 0) which is linearly independent of ψA(x, 0), yields the same value of (Eˆ), but allows a different set of measured energies (which may include some but not all of E0, E1 and E2, plus others).

Mar 4, 2018

See below:

#### Explanation:

Disclaimer - I am assuming that ${\phi}_{0}$, ${\phi}_{1}$ and ${\phi}_{2}$ denote the ground, first excited and second excited states of the infinite well, respectively - the states conventionally denoted by $n = 1$, $n = 2$, and $n = 3$. So, ${E}_{1} = 4 {E}_{0}$ and ${E}_{2} = 9 {E}_{0}$.

(d) The possible results of energy measurements are ${E}_{0}$, ${E}_{1}$ and ${E}_{2}$ - with probabilities $\frac{1}{6}$, $\frac{1}{3}$ and $\frac{1}{2}$ respectively.

These probabilities are independent of time (as time evolves, each piece picks up a phase factor - the probability, which is given by the modulus squared of the coefficients - do not change as a result.
(c) The expectation value is $6 {E}_{0}$. The probability of an energy measurement yielding this as a result is 0. This is true for all times.
Indeed, $6 {E}_{0}$ is not an energy eigenvalue - so that an energy measurement will never give this value - no matter what the state.

(e)Immediately after the measurement that yields ${E}_{2}$, the state of the system is described by the wavefunction
${\psi}_{A} \left(x , {t}_{1}\right) = {\phi}_{2}$
At ${t}_{>} {t}_{1}$, the wavefunction is
 psi_A(x,t) = phi_2 e^{-iE_2/ℏ(t-t_1)}
The only possible value an energy measurement will yield on this state is ${E}_{2}$ - at all times ${t}_{2} > {t}_{1}$.

(f) The probabilities depend on the squared modulus of the coefficients - so
${\psi}_{B} \left(x , 0\right) = \sqrt{\frac{1}{6}} {\phi}_{0} - \sqrt{\frac{1}{3}} {\phi}_{1} + i \sqrt{\frac{1}{2}} {\phi}_{2}$
will work (there are infinitely many possible solutions). Note that since the probabilities have not changed, the energy expectation value will automatically be the same as ${\psi}_{A} \left(x , 0\right)$
(g) Since ${E}_{3} = 16 {E}_{0}$, we can get an expectation value of $6 {E}_{0}$ if we have ${E}_{1}$ and ${E}_{3}$ with probabilities $p$ and $1 - p$ if
$6 {E}_{0} = p {E}_{1} + \left(1 - p\right) {E}_{3} = 4 p {E}_{0} + 16 \left(1 - p\right) {E}_{0} \implies$
$16 - 12 p = 6 \implies p = \frac{5}{6}$
So a possible wavefunction (again, one of infinitely many possibilities) is

${\psi}_{C} \left(x , 0\right) = \sqrt{\frac{5}{6}} {\phi}_{1} + \sqrt{\frac{1}{6}} {\phi}_{3}$