Problem 1
Sense-making

For part (a), we expect to produce a list of possible energy measurements, and their associated probabilities (a number between zero and 1). We also expect that the sum of all probabilities is exactly 1.

For part (b) the solution should be a graph of two variables, represented by a 3-D plot. The probability density graph should be symmetrical under interchange of x and y (i.e.  (ψ^2)(x,y)=(ψ^2)(y,x)  , because the state itself is
symmetrical under interchange of (Subscript[n, x]) and (Subscript[n, y]) and the side lengths of the well are equal. Since this is an infinite square well, the probability density should be zero at the edges

For part (c), the answer should have a response, and possibly additional content that looks like that of parts (a) and (b).

Part (d) requires only a written response, possibly with some math or graphs to support any claims.

(a)

Since this is a two-dimensional square well, there are energy degeneracies. Incidentally, the |3,1⟩ and the |1,  3⟩ states have the same energy, because for the infinite square well with equal side lengths,
					(Subscript[E, (Subscript[n, x]),(Subscript[n, y])])= (((π^2)(ℏ^2)(((Subscript[n, x])^2)+((Subscript[n, y])^2)))/(2(mL^2)))
	
Thus, 					(Subscript[E, 1,3]) =  (Subscript[E, 3,1])=  ((5(π^2)(ℏ^2))/(mL^2))
The other possible energy measurement is 
					(Subscript[E, 2,2]) = ((4(π^2)(ℏ^2))/(mL^2))
The probability of measuring the first energy value (((5(π^2)(ℏ^2))/(mL^2)))is, 
					
					P =  |⟨3,1|ψ  ⟩((|)^2) +|⟨1,3|ψ  ⟩((|)^2)= (((((    2    )/3)))^2)+(((((    2    )/3)))^2)= ((    8    )/9) 

The probability of measuring the second energy (((4(π^2)(ℏ^2))/(mL^2)))is, 

					P =  |⟨2,2|ψ  ⟩((|)^2)= (((((    1    )/3)))^2)= ((    1    )/9)

(b)

The probability density is the complex square of the state at time t=0: 

          ψ(x,y)=((2/3))((2/L))sin(((3π x)/L))sin(((π y)/L))-((1/3))((2/L))sin(((2π x)/L))sin(((2π y)/L))+((2/3))((2/L))sin(((3π x)/L))sin(((2π y)/L))

 A plot of this function is shown below (L=1):

Plot3D[(2/3*Sin[3Pi*x/2]Sin[Pi y/2]-1/3*Sin[Pi*x]Sin[Pi y]+2/3*Sin[Pi*x/2]Sin[3Pi y/2])^2,{x,0,2},{y,0,2}]

(c)

We would expect the probability of measuring particular energies to remain unchanged in time. One explanation for this is that, while the coefficients of the energy eigenstates are time-dependent, the complex-square of each coefficient is not. Thus, the answer to part (a) does not depend on time. In contrast, the answer to part (b) does depend on time. The complex squaring process will not eliminate all time dependence, because there will be
cross terms that do not negate each other. An animation of some of this time dependent probability density (with a conveniently chosen value for the particle mass) is shown below.

Animate[Plot3D[(2/3*Exp[I*5t]Sin[3Pi*x/2]Sin[Pi y/2]-1/3*Exp[I*4t]Sin[Pi*x]Sin[Pi y]+2/3*Exp[I*5t]Sin[Pi*x/2]Sin[3Pi y/2])(2/3*Exp[-I*5t]Sin[3Pi*x/2]Sin[Pi y/2]-1/3*Exp[-I*4t]Sin[Pi*x]Sin[Pi y]+2/3*Exp[-I*5t]Sin[Pi*x/2]Sin[3Pi y/2]),{x,0,2},{y,0,2}, PlotRange->{0, 1.8}], {t,0,10}]

(d)

With this new state, the biggest difference is that there is now only one possible energy measurement:  ((25(π^2)(ℏ^2))/(mL^2)). Thus, the state will in fact be ”stationary” in the sense that the probability density will be the same as time progresses, even though the state is a linear combination of energy eigenstates.