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The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change whengradually increases from1to2. If the increase occurs at a constant rate, (诲伪/dt=c)what is the dynamic phase change for this process?

Short Answer

Expert verified

The geometric phase is Yn(t)=0.The dynamic phase change for this process is

(t)=m6h2c(23-13).

Step by step solution

01

Define Dynamic phase.

Geometric phase is a phase difference gained during the course of a cycle when a system is subjected to cyclic adiabatic processes, which derives from the geometrical features of the Hamiltonian's parameter space in both classical and quantum mechanics.Dynamic phases take inspiration from dynamic timers and static phases to generate performance metrics for all functions performed in a single phase invocation.

02

Obtain the geometric phase.

Let the time independent wave function for the bound state be,=尘伪he-尘伪h.....

(1)

The equation 10.42 is given by,localid="1656050634693" role="math" Yn(t)=iRlRlnl饾湑n饾湑RdR......(2)Inthiscase,R=.So,饾湑饾湑R=mh121e-尘伪xh2+尘伪h-mxh2e-尘伪xh2

Thus,

饾湑饾湑R=mh12hm-mmh3xe-2mxh2=m2h2-m2h4xe-2mxh2|饾湑饾湑R=2m2h20e-2mxh2dx-m2h40xe-2mxh2dx=mh2h22m-2m2h4h22m2=12-12=0Substitutethevalueintheequation(2)toget:Yn(t)=0

03

Obtain the dynamic phase.

The dynamic phase change for this process is given by, n(t)=1h0tEn(t')dt'.....(3)

But the energy in this case is E=-尘伪22h2.Thus,

(t)=-1h0T-m22h2dt'=m2h3122dt'dd=m2h3C122d=m6h2c(23-13)(t)=m6h2c(23-13)

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Most popular questions from this chapter

(a) Derive the equation 10.67 from Equation 10.65.

(b) Derive Equation 10.79, starting with Equation 10.78.

Show that if n is real, the geometric phase vanishes. (Problems 10.3 and 10.4 are examples of this.) You might try to beat the rap by tacking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: n(t)e颈蠁苍n(t), wheren(R) is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions.

Check the Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33, and show that the sum of the squares of the coefficients is 1, as required for normalization.

The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficientsCm(t)in Equation. Suppose the system starts out in theth state; in the adiabatic approximation, it remains in theth state, picking up only a time-dependent geometric phase factor (Equation):

Cm(t)=mneiyn(t)

(a) Substitute this into the right side of Equationto obtain the "first correction" to adiabaticity:

Cm(t)=cm(0)-0t<mt'lt'n(t')>eiyn(t')ei(n(t')-t!(t'))dt'.[10.95]

This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equationon the right side of Equation, and so on.


(b) As an example, apply Equationto the driven oscillator (Problem). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which

Cn+1(t)=-尘蝇2hn+10tf(t')e颈蝇迟dt'Cn+1(t)=-尘蝇2hn0tf(t')e颈蝇迟dt'

Work out to analog to Equation 10.62 for a particle of spin I.

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