#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(sqrt(1/6)+sqrt(1/3)*2+sqrt(1/3)*3)e^(-iE_nt)#

2 Answers
Mar 3, 2018

Well, I get #14/5E_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(0) = 0#.

And for context, we had let #phi_n(x) = sqrt(2/L) sin((npix)/L)#...

  • 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. #(d<< E >>)/(dt) = 0#...

So now...

#Psi_A(x,0) = 1/sqrt3 sqrt(2/L) sin((pix)/L) + 1/sqrt2 sqrt(2/L) sin((2pix)/L)#

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...

#<< E >> = (<< Psi | hatH | Psi >>)/(<< Psi | Psi >>) = 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#
#(delhatH)/(delt) = 0#

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

#color(blue)(<< E >>) = (1/3int_(0)^(L)Phi_1^"*"(x,t)hatHPhi_1(x,t)dx + 1/2int_(0)^(L)Phi_2^"*"(x,t)hatHPhi_2(x,t)dx)/(<< Psi | Psi >>)#

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 = (1/sqrt3)^2 + (1/sqrt2)^2 = 5/6#.

Therefore, #<< Psi | Psi >> = 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)]#

#= 6/5[1/3 E_1 + 1/2 4E_1]#

#= color(blue)(14/5 E_1)#

Mar 4, 2018

Answer:

# < E > = 1/6 E_0 +1/3E_1 + 1/2 E_2 = 6E_0#

Explanation:

Each stationary state corresponding to energy eigenvalue #E_n# picks up a phase factor #e^{-iE_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(x,0) = sqrt(1/6)phi_0(x) + sqrt(1/3)phi_1(x) + sqrt(1/2)phi_2(x)#

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**(x,t) hat{H} psi_A(x,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_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(x)# are energy eigenfunctions, so that #hat{H} phi_i(x) = E_i phi_i (x)#.

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(x) phi_j(x) dx = delta_{ij}#

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

# < E > = 1/6 E_0 +1/3E_1 + 1/2 E_2#

Using the standard result that #E_n = (n+1)^2 E_0#, we have #E_1 = 4E_0 # and #E_2 = 9E_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 > = (1/6 times 1 + 1/3 times 4 + 1/2 times 9) E_0 = 108/18 E_0 = 6E_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 # (sqrt{1/6})^2 + (sqrt{1/3})^2 + (sqrt{1/2})^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.