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Coherent states of the harmonic oscillator. Among the stationary states of the harmonic oscillator (Equation 2.67) only n = 0 hits the uncertainty limit (σxσp=h/2); in general, σxσp=(2n+1)h/2, as you found in Problem 2.12. But certain linear combinations (known as coherent states) also minimize the uncertainty product. They are (as it turns out) Eigen functions of the lowering operator

ψn=1n!(a^+)nψ0(2.68).

a_|α>=α|a>(the Eigen value α can be any complex number).

(a)Calculate <x>,<x2>,<p>,<p2>in the state |α〉. Hint: Use the technique in Example 2.5, and remember that is the Hermitian conjugate of a-. Do not assume α is real.

(b) Find σx; show that σxσp=h/2.

(c) Like any other wave function, a coherent state can be expanded in terms of energy Eigen states: |α>=∑n=0∞Cn|n>.

Show that the expansion coefficients arecn=αnn!c0.

(d) Determine by normalizing |α〉. Answer: exp(-α2/2)

(e) Now put in the time dependence: |n>→e-iEntIh|n>,

and show that |αt|remains an Eigen state of a-, but the Eigen value evolves in time:α(t)=e-iӬt So a coherent state stays coherent, and continues to minimize the uncertainty product.

(f) Is the ground state (n=0>)itself a coherent state? If so, what is the Eigen value?

Short Answer

Expert verified

(a)<x>=<α\×α>=h2mӬ(α+α*).<x>=h2mӬ[1+α+α*2].<p>=<α\pα>=ihmӬ2<α\a+-a_α>=-ihmӬ2(α+α*).<p2>=hmӬ2[1-α+α*2].

(b)σxσp=h2mӬhmӬ2=h2

(c)localid="1655979681976" cn=<ψn\α>=1n!<a+nψ0|α>=1n!<ψ0|a-nα>=1n!αn<ψ0|α>=αnn!c0

(d)1=∑n=0∞|cn|2=|cn|2∑n=0∞|α|2nn!=|c0|2e|α|2⇒c0=e-|α|2/2.

(e)|α(t)>=∑n=0∞cne-iEntIh|n>=∑n=0∞αnn!e-|α|2/2e-i(n+12)Ӭt|n>=e-iӬt/2∑n=0∞(αe-iӬt)n!e-|α|2/2|n>

(f) Eigen value α=0

Step by step solution

01

(a) Calculate <x>,<x2>,<p>,<p2> in the state |α〉.

x=α\xα=h2mӬα|a++a-α=h2mӬa-α|α++α|α_α=h2mӬα+α*.x2=h2mӬa+2+a+a-+a-a+a-2x.Buta_a+=a_,a++a+a_=1+a+a_a?a+a^+a^-=1hӬH^-12a+a?(2.55).=h2mӬa+2+2a+a-+1+a-2.x2=h2mӬα|a+2+2a+a-+1+a-2α=h2mӬa-2α|α+2a_α|a_α+α|α+α|a-2α.=h2mӬα*2+2α*α+1+α2=h2mӬ1+α+α*2.p=α|pα=ihmӬ2α|a+-a-α=ihmӬ2a_α|α-α|a_α=ihmӬ2α-α*.p2=-hmӬ2a+2-a+a--a-a++a-2=-hmӬ2a+2-2a+a-+a-2=-hmӬ2α*2-2α*α-1+α2=hmӬ21-α-α*2

02

 Step2: (b) Finding σx and σp

σx2=x2-x2=h2mӬ1+α+α*2-α+α*2=h2mӬσp2=p2-p2=h2mӬ1-α-α*2+α+α*2=hmӬ2σxσp=h2mӬhmӬ2=h2

03

(c) Showing the expansion co-efficients.

Using Eq. 2.67 for ψn:

ψn=1n!a^+nψ0 (2.68).

cn=ψn|α=1n!a+nψ0|α=1n!ψ0|a+nα=1n!αnψ0|α=αnn!c0

04

(d) Determine by normalizing |α〉.

1=∑n=0∞cn2=c02∑n=0∞α2nn!=c02ea2⇒c0=e-a2/2.

05

 Step5:(e) showing that |αt|remains an Eigen state of a- 

|αt>=∑n=0∞cne-iEntIh|n>=∑n=0∞αnn!e-α2/2e-in+12Ӭt|n>=e-iӬt/2∑n=0∞αe-iӬtnn!e-α2/2|n>

Apart from the overall phase factor e-iӬt/2 (which doesn’t affect its status as an Eigen function of a_-a− ,

or its Eigen value), |αt>is the same as|α>, but with eigenvalueαt=e-iӬtα.

06

(f) Eigen value.

Equation 2.59 says a-|ψ0>=0, so yes, it is a coherent state, with Eigen value α = 0.

a^-ψ0=0 (2.59).

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Most popular questions from this chapter

Solve Equation 3.67 for Ψ(x). Note that ⟨x⟩and ⟨p⟩are constants.

(a) For a function f(x)that can be expanded in a Taylor series, show that f(x+x0)=eip^x0Ihf(x)

wherex_{0}

is any constant distance). For this reason, p^/his called the generator of translations in space. Note: The exponential of an operator is defined by the power series expansion: eQ^≡1+Q^+(1/2)Q^2+(1/3!)Q^3+...

(b) If ψ(x,t)satisfies the (time-dependent) Schrödinger equation, show that ψ(x,t+t0)=e-iH^t0/hψ(x,t)

where t_{0}is any constant time); -H^/his called the generator of translations in time.

(c) Show that the expectation value of a dynamical variableQ(x,p,t), at time , t+t0can be written34

<Q>t+t0=<ψx,t|eiH^t0/hQ^x^,p^,t+t0e-iH^t0/h|ψx,t>

Use this to recover Equation 3.71. Hint: Lett0=dt, and expand to first order in dt.

(a) Prove the following commutator identity:

[AB.C]=A[B.C]+[A.C]B

b) Show that

[xn,p]=ihnxn-1

(c) Show more generally that

[f(x),p]=ihdfdx

for any functionf(x).

Find the matrix elements <n|x|n'>²¹Â­Â­Â­Â­²Ô»å <n|p|n'>in the (orthonormal) basis of stationary states for the harmonic oscillator (Equation 2.67). You already calculated the "diagonal" elements (n=n') in Problem 2.12; use the same technique for the general case. Construct the corresponding (infinite) matrices, X and P . Show that(1/2m)P2+(mÓ¬2/2)X2=His diagonal, in this basis. Are its diagonal elements what you would expect? Partial answer:

<n|x|n'>=h2mӬ(n'δn,n'-1+nδn,n'-1)

The Hamiltonian for a certain three-level system is represented by the matrix

H=(a0b0c0b0a), where a, b, and c are real numbers.

(a) If the system starts out in the state |&(0)⟩=(010)what is |&(t) ?

(b) If the system starts out in the state|&(0)⟩=(001) what is|&(t) ?

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