# Mix of States

Consider a free particle with a mass of $10^{-30} kg$ described by the wave function

$ \psi =\sqrt{1/7}*e^{ik_1x}+\sqrt{2/7}*e^{-ik_1x}+\sqrt{1/7}*e^{ik_2x}+\sqrt{3/7}e^{-ik_2x}$

where $k_1=2*10^7 m^{-1}$ and $k_2=3*10^7 m^{-1}$

a) if you measured the momentum of the particle, what possible values could you measure and what would be the probability of measuring each of those values?

b) For each of the possible values of momentum found in (a), what would the wavefunction of the particle be after measuring that value?

c) If you measured the energy of the particle, what possible values could you measure and what would be the probability of measuring each of those values?

A function $ψ(x)=A*\exp(ikx)$ is an eigenfunction of both the energy and of the momentum of the system because (1)

$Hψ=Eψ$

that is for a free particle $-[\hbar^2/2m] [d^2ψ/(dx^2)]=Eψ$ which means $E=(\hbar^2*k^2)/2m$

$pψ=p*ψ$

that is $(\hbar/i)*(d/dx) ψ=\hbar k*ψ$, which means $p=\hbar k$

If we have a superposition of states

$ψ(x)=A_1*exp(ik_1*x)+A_2*exp(ik_2*x)+⋯$

a)

The possible values of momentum after a measurement is made are exactly the values $\hbar k_1$,〖 $\hbar k_2$, $\hbar k_3$,… that are present in the initial state.

The probabilities of each measurement outcome are $P_i=|A_i |^2$

Therefore the outcome of the measurements of the momentum and their probabilities are:

$p_1=\hbar*k_1=1.055*10^{-34}*2*10^7=2.11*10^{-27} (kg*m/s),P_1=1/7$

$p_2=-\hbar*k_1=-2.11*10^{-27} (kg*m/s)$, $P_2=2/7$

$p_3=+\hbar*k_2=1.055*10^{-34}*3*10^7=3.165*10^{-27} (kg*m/s)$, $P_3=1/7$

$p_4=-\hbar*k_2=-3.165*10^{-27} (kg*m/s)$, $P_4=3/7$

b) After each measurement on the initial system the wavefunction will collapse to indicate exactly that measured state with a probability of $P=1$.

After measurement of p1 the wavefunction will be $ψ=1*exp(ik_1*x)$, after measurement of $p2$ the wavefunction will be $ψ=1*exp(-ik_1*x)$, after measurement of $p3$ the wavefunction will be $ψ=1*exp(ik_2*x)$ and after measurement of p4 the wavefunction will be $ψ=1*exp(-ik_2*x)$

c) Since

$E=p^2/2m=(\hbar^2 k^2)/2m$

There will be just two possible outcomes of the energy measurement (with probabilities):

$E_1=(p_1^2)/2m=(p_2^2)/2m=2.226*10^{-24} J$ with the probability $P_1=1/7+2/7=3/7$

$E_2=(p_3^2)/2m=(p_4^2)/2m=5.01*10^{-24} J$ with probability $P_2=1/7+3/7=4/7$