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In an interesting version of the energy-time uncertainty principle31, ∆t=τ/πwhere τis the time it takesΨ(x,t)to evolve into a state orthogonal toΨ(x,0) . Test this out, using a wave function that is an equal admixture of two (orthonormal) stationary states of some (arbitrary) potential:Ψ(x,0)=(1/2)[Ψ1x+Ψ2(x)]

Short Answer

Expert verified

This is consistent with the original time-energy uncertainty principle:

∆E.∆t=h2.

Step by step solution

01

Wave function

In the energy-time uncertainty principle ∆t=τ/π, whereτis the time taken by Ψ(x,t)to grow into a state orthogonal to Ψ(x,0), testing this is required. Start with a wave function that is a combination of two stationary states, that is:

Ψ(x,0)=12(ψ1(x)+ψ2(x))........(1)

The wave function at the later time is, therefore:
Ψx,τ=12ψ1xe-iE1τ/h+ψ2xe-iE2τ/h ........(2)

Multiply each term by the appropriate wiggle factor to get the position-space wave function at any future time t.

02

Substitute values from (1) & (2)

For the two functions in (1) and (2) to be orthogonal,

Ψ(x,τ)|x,0=0

Substitute from (1) and (2) to get :

Ψ(x,τ)|x,0=∫∞∞Ψ(x,τ)Ψ(x,0)dxΨ(x,τ)|x,0=∫∞∞12Ψ1*(x)eiE1τ/h+12Ψ2*(x)eiE1τ/h12ψ1x+12ψ2xdxΨ(x,τ)|x,0=∫∞∞12ψ1*xψ1*eiE1τ/h+12ψ1*xψ2(x)e*iE1τ/h+12ψ2*xψ1xeiE2τ/h+12ψ2*xψ2xeiE2τ/hdxΨ(x,τ)|x,0=12eiE1τ/h∫∞∞ψ1*xψ1xdx+12eiE1τ/h∫∞∞ψ1*xψ2xdx12eiE2τ/hψ2*xψ1xdx+12eiE2τ/hψ2*xψ2xdx12eiE1t/hψ1|ψ1+eiE1t/hψ1|ψ2+eiE2t/hψ2|ψ1+eiE2t/hψ2|ψ2=0

but:

ψ1|ψ1=1ψ2|ψ2=1ψ1|ψ2=0ψ2|ψ1=0

The exponential function is never zero, thus:

12eiE1t/h+eiE1t/h=0.eiE2t/h=-eiE1t/heiE2-E1t/h=-1

buteiπ+1=0,so:eiE2-E1t/h=-1eiE2-E1t/h=eiπ

Thus by the definition of ∆t:

E2-E1h=Ï€

Taking τ/π=∆tand E2-E1=∆Egiving the standard deviation in energy as:

∆E∆t=h>h2

This confirms that the energy-time uncertainty principle is obeyed. This is consistent with the original time-energy uncertainty principle.

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Most popular questions from this chapter

(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).

Test the energy-time uncertainty principle for the wave function in Problem2.5and the observable x, by calculatingσHσXandd<x>/dtexactly.

Show that projection operators are idempotent: P^2=P^. Determine the eigenvalues of P^ , and characterize its eigenvectors.

Consider the wave functionΨ(x,0)={12±¹Î»ei2Ï€³æ/λ,-²Ôλ<x<²Ôλ0,

wherenis some positive integer. This function is purely sinusoidal (with wavelengthλ)on the interval-²Ôλ<x<²Ôλ, but it still carries a range of momenta, because the oscillations do not continue out to infinity. Find the momentum space wave functionΦ(p,0). Sketch the graphs of|Ψ(x,0)|2and|Φ(p,0)|2, and determine their widths,wxandwp(the distance between zeros on either side of the main peak). Note what happens to each width asn→∞. Usingwxandwpas estimates of∆xand∆p, check that the uncertainty principle is satisfied. Warning: If you try calculatingσp, you're in for a rude surprise. Can you diagnose the problem?

(a) Write down the time-dependent "Schrödinger equation" in momentum space, for a free particle, and solve it. Answer: exp(-ip2t/2mh)Φ(p,0).

(b) Find role="math" localid="1656051039815" Φ(p,0)for the traveling gaussian wave packet (Problem 2.43), and construct Φ(p,t)for this case. Also construct |Φ(p,t)|2, and note that it is independent of time.

(c) Calculaterole="math" localid="1656051188971" pandrole="math" localid="1656051181044" p2by evaluating the appropriate integrals involvingΦ, and compare your answers to Problem 2.43.

(d) Show thatrole="math" localid="1656051421703" <H>=<p>2/2m+<H>0(where the subscript denotes the stationary gaussian), and comment on this result.

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