# Well-Mix of States

Consider a particle of mass $m$ in the infinite square well of width $L$. Its initial wave function (at time $t=0$) is a coherent mixture of the second and third energy eigenstates $\psi(x,0)=A[\phi_2(x)+2i\phi_3(x)]$

a) Normalize $\psi(x,0)$. That is, find $A$.

b) Determine the expression for $\psi(x,t)$

c) Write the integral expression for $<x>$ for this particle.

d) Does $<x>$ chnge in time? Explain either with qualitative or quantitative arguments.

e) Determine the expectation value of the energy at time $t=0$

f) Does $<E>$ change with time? Explain using qualitative arguments. If you measure the energy at time $t>0$ what values would you get and with what probabilities?

A mix of elementary states $|\phi_n>$ is described by the wavefunction

$\psi=\sum_n C_n\phi_n$ where $C_n$ are complex coefficients.

Since all elementary states $\phi_n$ are orthogonal (and normate) one to each other $<\phi_i|\phi_j>=\delta_{ij}$ then

$1=<\psi|\psi>=\sum_n |C_n|^2$

With the wave function given in text we have

$1=A^2(1+4)$ so that $A=1/\sqrt{5}$

For a square well the elementary states for energy levels 2 and 3 are

$\phi_2=\sin(2\pi x/L)$ and $\phi_3 =\sin(3\pi x/L)$ so that

$\psi(x,0)=\frac{1}{\sqrt{5}}[\sin(2\pi x/L) +2i\sin(3\pi x/L)]$

When time changes $(t>0)$ then the time dependent Schroedinger equation applies for each elementary state:

$H|\phi_n(t)>=i\hbar \frac{d}{dt}|\phi_n(t)>$

so that the formal solution is

$\phi_n(t)=\phi_n(0)*\exp(-iHt/\hbar)$

and he actual exact solution is

$|\phi_n(t)>=\phi_n(0)\exp(-iE_nt/\hbar)$

which means that

$\psi(x,t)=\frac{1}{\sqrt{5}}\left[\phi_2(0)\exp(-iE_2t/\hbar)+2i\phi_3(0)\exp(-iE_3t/\hbar)\right]$

Since all $\phi_n$ are orthonormate then $<x>=<\psi|x|\psi>=|C_2|^2<\phi_2|x|\phi_2>+|C_3|^2<\phi_3|x|\phi_3>$

If one considers $C_n(t)=\exp(-iE_n t/\hbar)*C_n(0)$

then $|C_n(t)|^2=|C_n(0)|^2$

so that

$<x>=1*(1/5)\int_0^L x\sin^2(2\pi x/L)dx+4*(1/5)\int_0^L x\sin^2(3\pi x/L)dx$

and thus $<x>$ does not change with time $t$.

For the energy one has

$<E>=|C_2|^2*E_2+|C_3|^2*E_3=(1/5)E_2+(4/5)E_3=$

$=(1/5)*2^2*\frac{\hbar^2\pi^2}{2mL^2}+(4/5)*3^2*\frac{\hbar^2\pi^2}{2mL^2}=8*\frac{\hbar^2\pi^2}{2mL^2}$

and since $|C_n(t)|^2=|C_n(0)|^2$ then at a latter time one will get the same average energy.

Moreover since elementary states are orthonormate and

$<\psi(t)|H|\psi(t)>=|C_2|^2*E_2<\phi_2|\phi_2>+|C_3|^2*E_3<\phi_3|\phi_3>$

at a latter time $t>0$ you can measure the same energies $E_2$ and $E_3$ with the same probabilities $|C_2|^2$ and $|C_3|^2$ as initially.

Basically until a measurement is made on the system its state will remain described by the same mix of states. Therefore the average energy will not vary with time. Just after the measurement is made the state of the system will change (the wave function will collapse to indicate the exact state whose energy was just measured) so that the probability of an energy will be different only if a measurement was made on the system.