Find x,p,x2,p2,T, for the nth stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied.

Short Answer

Expert verified

Answer

The values for nth stationary state of harmonic oscillator are

x=2mωψn*a-+a-ψndx

p=mdxdt=0

localid="1657818497766" x2=n+12mω

p2=n+12mω

localid="1657819069025" T=12n+12ω, Also, the uncertainty principal is satisfied.

Step by step solution

01

- A Classical and quantum oscillator


Beyond conventional turning points, the stationary states (states of definite energy) have nonzero values. A classical oscillator is least likely to be discovered in the ground state at the location of the minimum of the potential well, where a quantum oscillator is most likely to be found.

02

- The value of x and p

The value of xand ρwill be

x=2mωa++a-
p=imω2a1+a

So,

x=2mωψn*a-+a-ψndx

But, aψn=n+1ψn+1,aψn-nψn-1

x=2mωn+1ψn*ψn1dx+nψn*ψn-1dx

=0

And the value ofp=mdxdt=0

03

- The value of x2 and p2


According to the value of x, the value of

x2=2mωa++a22

=2mωa+2+a+a-+a-a++a-2

x2=2mωψn*a+2+a+a-+a-a++a-2

x2=2mω0+nψn2dx+(n+1)ψn2dx+0

=2mω(2n+1)

=n+12mω

Hence, the value is x2=n+12mω

Again, the value of,

p2=-mω2a--a-2

=-mω2a-2-a-a--a-a++a-2

So,

p2=-mω2[0-n-(n+1)+0]

=mω2(2n+1)

=n+12mω

Hence,

p2=n+12mω

04

- The value of T and satisfaction of uncertainty principal

The value of Twill be evaluated.

So,

T=p22m

=12n+12ω

Thus, the value ofT=12n+12ω

And to check the satisfaction of uncertainty principal,

σx=x2-x2

=n+12mω

σp=p2-p2

=n+12mω

σxσp=n+122

Thus, the uncertainty principal is satisfied.

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Most popular questions from this chapter

Consider the double delta-function potentialV(x)=-α[δx+a+δx-a]Whereand are positive constants

(a) Sketch this potential.

(b) How many bound states does it possess? Find the allowed energies, forα=ħ/maand forα=ħ2/4ma, and sketch the wave functions.

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  1. What is the most probable result? What is the probability of getting that result?
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