#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)#
I get an answer of
2 Answers
Well, I get
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
#<< 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 onedimensional infinite potential well is timeINDEPENDENT...
#hatH = ℏ^2/(2m) (d^2)/(dx^2) + 0#
#(delhatH)/(delt) = 0#
and the
#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 offdiagonal terms go to zero due to the orthogonality of the#phi_n# .
The denominator is the norm of
#sum_i c_i^2 = (1/sqrt3)^2 + (1/sqrt2)^2 = 5/6# .
Therefore,
#=> [(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
#= 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)#
Answer:
Explanation:
Each stationary state corresponding to energy eigenvalue
So, the starting wavefunction
evolves in time
Thus, the energy expectation value at time
where we have used the fact that the
This still gives us nine terms. However, the final calculation is simplified a lot by the fact that the energy eigenfunctions are orthonormalized, i.e. they obey
This means that of the nine integrals, only three survive, and we get
Using the standard result that
Note :
 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.
 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_iE_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.  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.  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.  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.