# psi_A(x,0) = sqrt(1/6)phi_0(x) + sqrt(1/3)phi_1(x) + sqrt(1/2)phi_2(x) Calculate the expectation value <E> at any later time t=t_1, phi_n are energy eigenfunctions of the infinite potential well .Write the answer in terms of E_0?

## I get an answer of ${E}_{0} \left(\sqrt{\frac{1}{6}} + \sqrt{\frac{1}{3}} \cdot 2 + \sqrt{\frac{1}{3}} \cdot 3\right) {e}^{- i {E}_{n} t}$

Mar 3, 2018

Well, I get $\frac{14}{5} {E}_{1}$... and given your chosen system, it cannot be re-expressed in terms of ${E}_{0}$.

There are so many quantum mechanics rules broken in this question...

• The ${\phi}_{0}$, since we are using infinite potential well solutions, vanishes automatically... $n = 0$, so $\sin \left(0\right) = 0$.

And for context, we had let ${\phi}_{n} \left(x\right) = \sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right)$...

• It is impossible to write the answer in terms of ${E}_{0}$ because $n = 0$ does NOT exist for the infinite potential well. Unless you want the particle to vanish, I must write it in terms of ${E}_{n}$, $n = 1 , 2 , 3 , . . .$...

• The energy is a constant of the motion, i.e. $\frac{d \left\langleE\right\rangle}{\mathrm{dt}} = 0$...

So now...

${\Psi}_{A} \left(x , 0\right) = \frac{1}{\sqrt{3}} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) + \frac{1}{\sqrt{2}} \sqrt{\frac{2}{L}} \sin \left(\frac{2 \pi x}{L}\right)$

The expectation value is a constant of the motion, so we do not care what time ${t}_{1}$ we choose. Otherwise, this is not a conservative system...

$\left\langleE\right\rangle = \frac{\left\langle\Psi | \hat{H} | \Psi\right\rangle}{\left\langle\Psi | \Psi\right\rangle} = {E}_{n}$ for some $n = 1 , 2 , 3 , . . .$

In fact, we already know what it should be, since the Hamiltonian for the one-dimensional infinite potential well is time-INDEPENDENT...

hatH = -ℏ^2/(2m) (d^2)/(dx^2) + 0
$\frac{\partial \hat{H}}{\partial t} = 0$

and the (e^(-iE_nt_1//ℏ))^"*"(e^(-iE_nt_1//ℏ)) go to 1 in the integral:

$\textcolor{b l u e}{\left\langleE\right\rangle} = \frac{\frac{1}{3} {\int}_{0}^{L} {\Phi}_{1}^{\text{*"(x,t)hatHPhi_1(x,t)dx + 1/2int_(0)^(L)Phi_2^"*}} \left(x , t\right) \hat{H} {\Phi}_{2} \left(x , t\right) \mathrm{dx}}{\left\langle\Psi | \Psi\right\rangle}$

where we have let Phi_n(x,t) = phi_n(x,0)e^(-iE_nt_1//ℏ). Again, all the phase factors cancel out, and we note that the off-diagonal terms go to zero due to the orthogonality of the ${\phi}_{n}$.

The denominator is the norm of $\Psi$, which is

${\sum}_{i} | {c}_{i} {|}^{2} = {\left(\frac{1}{\sqrt{3}}\right)}^{2} + {\left(\frac{1}{\sqrt{2}}\right)}^{2} = \frac{5}{6}$.

Therefore, $\left\langle\Psi | \Psi\right\rangle = \frac{5}{6}$. That gives:

=> [(1/sqrt3)^2 (2/L) int_(0)^(L) sin((pix)/L)cancel(e^(iE_1t_1//ℏ)) [-ℏ^2/(2m) (d^2)/(dx^2)]sin((pix)/L)cancel(e^(-iE_1t_1//ℏ))dx + (1/sqrt2)^2 (2/L) int_(0)^(L) sin((2pix)/L)cancel(e^(iE_2t_1//ℏ)) [-ℏ^2/(2m) (d^2)/(dx^2)]sin((2pix)/L)cancel(e^(-iE_2t_1//ℏ))dx]/(5//6)

Apply the derivatives:

= 6/5[1/3 (2/L) int_(0)^(L) sin((pix)/L) [ℏ^2/(2m) cdot pi^2/L^2]sin((pix)/L)dx + 1/2 (2/L) int_(0)^(L) sin((2pix)/L) [ℏ^2/(2m) cdot (4pi^2)/L^2 ]sin((2pix)/L)dx]

Constants float out:

= 6/5[1/3 (ℏ^2pi^2)/(2mL^2) (2/L) int_(0)^(L) sin((pix)/L)sin((pix)/L)dx + 1/2 (4ℏ^2pi^2)/(2mL^2) (2/L) int_(0)^(L) sin((2pix)/L)sin((2pix)/L)dx]

And this integral is known for physical reasons to be halfway between $0$ and $L$, independent of $n$:

= 6/5[1/3 (ℏ^2pi^2)/(2mL^2) (2/L) L/2 + 1/2 (4ℏ^2pi^2)/(2mL^2) (2/L) L/2]

= 6/5[1/3 (ℏ^2pi^2)/(2mL^2) + 1/2 (4ℏ^2pi^2)/(2mL^2)]

$= \frac{6}{5} \left[\frac{1}{3} {E}_{1} + \frac{1}{2} 4 {E}_{1}\right]$

$= \textcolor{b l u e}{\frac{14}{5} {E}_{1}}$

Mar 4, 2018

$< E > = \frac{1}{6} {E}_{0} + \frac{1}{3} {E}_{1} + \frac{1}{2} {E}_{2} = 6 {E}_{0}$

#### Explanation:

Each stationary state corresponding to energy eigenvalue ${E}_{n}$ picks up a phase factor ${e}^{- i {E}_{n} t}$ on time evolution. The given state is not a stationary state - since it is the superposition of energy eigenstates belonging to different eigenvalues. As a result, it will evolve in time in a non-trivial manner. However, the Schroedinger equation that governs the time evolution of states is linear - so that each component energy eigenfunction evolves independently - picking up its own phase factor.

So, the starting wave-function

${\psi}_{A} \left(x , 0\right) = \sqrt{\frac{1}{6}} {\phi}_{0} \left(x\right) + \sqrt{\frac{1}{3}} {\phi}_{1} \left(x\right) + \sqrt{\frac{1}{2}} {\phi}_{2} \left(x\right)$

evolves in time $t$ to

psi_A(x,t) = sqrt(1/6)phi_0(x) e^{-iE_0 /ℏt} + sqrt(1/3)phi_1(x) e^{-iE_1/ℏ t} + sqrt(1/2)phi_2(x)e^{-iE_2/ℏ t}

Thus, the energy expectation value at time $t$ is given by

$< E > = {\int}_{-} {\infty}^{\infty} {\psi}_{A} \ast \left(x , t\right) \hat{H} {\psi}_{A} \left(x , t\right) \mathrm{dx}$
 = int_infty^infty (sqrt(1/6)phi_0(x) e^{iE_0/ℏ t} + sqrt(1/3)phi_1(x) e^{iE_1/ℏ t} + sqrt(1/2)phi_2(x)e^{iE_2ℏ t}) hat{H}(sqrt(1/6)phi_0(x) e^{-iE_0/ℏ t} + sqrt(1/3)phi_1(x) e^{-iE_1/ℏ t} + sqrt(1/2)phi_2(x)e^{-iE_2/ℏ t}) dx
 = int_-infty^infty (sqrt(1/6)phi_0(x) e^{iE_0/ℏ t} + sqrt(1/3)phi_1(x) e^{iE_1/ℏ t} + sqrt(1/2)phi_2(x)e^{iE_2/ℏ t}) times (sqrt(1/6)E_0phi_0(x) e^{-iE_0/ℏ t} + sqrt(1/3)E_1phi_1(x) e^{-iE_1/ℏ t} + sqrt(1/2)E_2phi_2(x)e^{-iE_2/ℏ t}) dx

where we have used the fact that the ${\phi}_{i} \left(x\right)$ are energy eigenfunctions, so that $\hat{H} {\phi}_{i} \left(x\right) = {E}_{i} {\phi}_{i} \left(x\right)$.

This still gives us nine terms. However, the final calculation is simplified a lot by the fact that the energy eigenfunctions are ortho-normalized, i.e. they obey

${\int}_{-} {\infty}^{\infty} {\phi}_{i} \left(x\right) {\phi}_{j} \left(x\right) \mathrm{dx} = {\delta}_{i j}$

This means that of the nine integrals, only three survive, and we get

$< E > = \frac{1}{6} {E}_{0} + \frac{1}{3} {E}_{1} + \frac{1}{2} {E}_{2}$

Using the standard result that ${E}_{n} = {\left(n + 1\right)}^{2} {E}_{0}$, we have ${E}_{1} = 4 {E}_{0}$ and ${E}_{2} = 9 {E}_{0}$ for an infinite potential well (you may be more used to an expression which says ${E}_{n} \propto {n}^{2}$ for an infinite well - but in these the ground state is labeled ${E}_{1}$ - here we are labeling it ${E}_{0}$ - hence the change). Thus

$< E > = \left(\frac{1}{6} \times 1 + \frac{1}{3} \times 4 + \frac{1}{2} \times 9\right) {E}_{0} = \frac{108}{18} {E}_{0} = 6 {E}_{0}$

Note :

1. While individual energy eigenfunctions evolve in time by picking up a phase factor, the overall wave function does not differ from the initial one by just a phase factor - this is why it is no longer a stationary state.
2. The integrals involved were like
int_-infty^infty psi_i(x) e^{+iE_i/ℏ t }E_j psi_j e^{-iE_j/ℏ t} dx = E_j e^{i(E_i-E_j)/ℏt} times int_-infty^infty psi_i(x) psi_j(x) dx
and these look like they are time dependent. However, the only integrals that survive are the ones for $i = j$ - and these are precisely the ones for which the time dependence cancels.
3. The last results fits with the fact that $\hat{H}$ is conserved - even though the state is not a stationary state - the energy expectation value is independent of time.
4. The original wave function is already normalized since ${\left(\sqrt{\frac{1}{6}}\right)}^{2} + {\left(\sqrt{\frac{1}{3}}\right)}^{2} + {\left(\sqrt{\frac{1}{2}}\right)}^{2} = 1$ and this normalization is preserved in time evolution.
5. We could have cut down a lot of work if we had made use of a standard quantum mechanical result - if a wave function is expanded in the form $\psi = {\sum}_{n} {c}_{n} {\phi}_{n}$ where the ${\phi}_{n}$ are eigenfunctions of a Hermitian operator $\hat{A}$, $\hat{A} {\phi}_{n} = {\lambda}_{n} {\phi}_{n}$, then $< \hat{A} > = {\sum}_{n} | {c}_{n} {|}^{2} {\lambda}_{n}$, provided, of course that the states are properly normalized.