# Perturbation Theory (Physics 325)

Consider the time dependent system from the tutorial represented below (the sections in the figure refer to the sections of the tutorial during which you consider that potential). Let the energy eigenfunctions and eigenvalues of the unperturbed system be given by $φ_m^0 (x)$ and $E_m^0$, respectively.

a) Suppose that the system is initially in the tenth excited state of the unperturbed potential.

i) What are the possible results of an energy measurement of this system at $t=T/2$ ?

a i)

The initial STATIC state can be written as s superposition of 10 fundamental states, having nine coefficients =0 and one coefficient =1

$Ψ(x,0)=∑_{(n=1)}^9 c_n*φ_n^0 (x)+c_10*φ_10^0 (x)$

with $c_n=0$ for $n=1-9$ and $c_10=1$

Of course

$∑_{(i=1)}^{10} |c_i |^2=1$

In case there is no time dependent perturbation the time-dependent initial state (for $t<0$)

$Ψ(x,t)=c_1 φ_1^0 (x)*e^{-(i E_1^0)/hbar t}+⋯+c_{10} φ_{10}^0 (x)*e^{-(i E_{10}^0)/ℏ t}$

With the same STATIC coefficients as above (9 coefficients =0 and $c10=1$)

i)

At $t=0$ the small perturbation  $H’$ is applied

$H’ (t)=$

$=V_0’$       if $0<t<T$
$=0$             if $t≥T$
All coefficients will begin to vary with time and the state for $t=T/2$ is

$Ψ(x,t)=c_1 (t)*φ_1^0 (x)*e^{-(i E_1^0)/hbar t}+⋯+c_{10} (t)*φ_{10}^0 (x)*e^{-(i E_{10}^0)/ℏ t}$

Therefore the possible values of the measured energies will still be the same, that is $E_1^0,E_2^0,…,E_{10}^0$

but this time the probability of measuring an energy $E_1^0,…,E_9^0$ will be greater than zero.

$P(E_n^0 )=|c_n (t) |^2$

Therefore the probabilities will depend on the time in the interval $0<t<T$.

To find the coefficients c_n (t) we apply an equation similar to [equation 9.13] only that now there are possible 10 states in the superposition

$(dc_{10} (t))/d t=-i/ℏ {c_{10}*H_{10,10}’+∑_(n=1)^9 H_{10,n}’*exp⁡(-i(E_n-E_10)/ℏ t)*c_n (t)}$

$(dc_n(t) )/d t=-i/ℏ {c_n*H_{n,n}’+∑_(n=1)^9 H_{n,10}’ *exp⁡(i*(E_n-E_10)/ℏ t)*c_{10} (t) }$

In the first order approximation

$c_{10}(t) =1$ and

$c_n (t)=-i/ℏ*∑_(n=1)^9 ∫_0^(t=T/2) H_{n,10}’ (t)*exp⁡(i*(E_n-E_10)/ℏ t)*d t$

$c_{10}(t) =1$ and

$c_n (t)=-i/ℏ*∑_{(n=1)}^9 ∫_0^(t=T/2) H_{n,10}’ (t)*exp⁡(i*(E_n-E_10)/ℏ t)*d t$

ii) What are the possible results of an energy measurement of this system at =3T/2 ? Explain how you would find the probability of each result. Indicate whether or not your probabilities should depend on time.

At $t=3T/2$ the superposition of states (1,..,10) will continue to exist. The wave function will be again a superposition as above

$Ψ(x,t)=c_1 (t)*φ_1^0 (x)*e^{-(i E_1^0)/hbar t}+⋯+c_{10} (t)*φ_{10}^0 (x)*e^{-(i E_{10}^0)/ℏt}$

but now in the red equations above all $H_{n,m}’=0$ since the perturbation is turned again off. Therefore all coefficients will stop varying with time and will continue to have the same value as they had when $t=T$. The possible results of measuring the energy will be again $E_1^0,E_2^0,…,E_{10}^0$ with the nonzero probabilities

$P(E_n^0 )=|c_n (t=T) |^2$

Therefore for $t>T$ the probabilities of measuring each energy will not depend on time.

iii) What are the possible results of an energy measurement of this system as t approaches infinity? Explain how you would calculate the probability of each result.As t approaches infinity and the perturbation stays off, the case ii) above will remain unchanged. The same energies $E_1^0,E_2^0,…,E_{10}^0$ could be measured with the same probabilities as they were measured at $t=T$.

iv) Does the probability density for the system change with time during the interval from =0 to T ? Does it change with time when $t>T$ ? Explain.

For $0 < t < T$ the probability of each energy level changes with time (since the coefficients $c_n=c_n (t)$) are variable. Therefore the probability density $P(x,t)=Ψ*(x,t)Ψ(x,t)$  will vary with time. When the perturbation is turned off at $t=T$ since the coefficients will have the same value (nonzero) as they had at $t=T$ the probability density will stay also the same with time $t>T$.