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Chapter 5: Q59E (page 191)

Determine the expectation value of the position of a harmonic oscillator in its ground state.

Short Answer

Expert verified

The expectation value of harmonic oscillator in ground state is 0.

Step by step solution

01

Expectation Value

Theprobabilistic value of an experiment in the quantum mechanics is called the expectation value. It is different from the most probable value of wave function.

02

Determination of expectation value of the position of a harmonic oscillator

The wave function of harmonic oscillator is given as:

ψx=mωπℏ1/2e-mω2ℏx2

The expectation value of harmonic oscillator in ground state is given as:

localid="1660044825093" x¯x=∫xψx2dx

Substitute all the values in the above equation.

localid="1660044830148" x¯x=∫xmωπℏ1/2e-mω2ℏx22dx

Substitute x = 0 for ground state

localid="1660044834568" x¯0=∫0mωπℏ1/2e-mω2ℏ022dxx¯0=0

Therefore, the expectation value of harmonic oscillator in ground state is 0.

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Most popular questions from this chapter

A study of classical waves tells us that a standing wave can be expressed as a sum of two travelling waves. Quantum-Mechanical travelling waves, discussed in Chapter 4, is of the form ψ(x,t)=Aei(kx=ωt). Show that the infinite well’s standing wave function can be expressed as a sum of two traveling waves.

Verify that Asin(kx) + Bcos(kx)is a solution of equation (5 - 12).

A classical particle confined to the positive x-axis experiences a force whose potential energy is-

U(x)=1x2-2x+1

a) By finding its minimum value and determining its behaviors at x=0and role="math" localid="1660119698069" x=∞, sketch this potential energy.

b) Suppose the particle has energy of 0.5J. Find any turning points. Would the particle be bound?

c) Suppose the particle has the energy of 2.0J. Find any turning points. Would the particle be bound?

Simple models are very useful. Consider the twin finite wells shown in the figure, at First with a tiny separation. Then with increasingly distant separations, In all case, the four lowest allowed wave functions are planned on axes proportional to their energies. We see that they pass through the classically forbidden region between the wells, and we also see a trend. When the wells are very close, the four functions and energies are what we might expect of a single finite well, but as they move apart, pairs of functions converge to intermediate energies.

(a) The energies of the second and fourth states decrease. Based on changing wavelength alone, argue that is reasonable.

(b) The energies of the first and third states increase. Why? (Hint: Study bow the behaviour required in the classically forbidden region affects these two relative to the others.)

(c) The distant wells case might represent two distant atoms. If each atom had one electron, what advantage is there in bringing the atoms closer to form a molecule? (Note: Two electrons can have the same wave function.)

Consider the delta well potential energy:

U(x)={0x≠0-∞x=0

Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E0.

(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by

ψ(x)=(2mE0h2)1/4e-(2mE0/ℏ)|x|

(b) Sketch ψ(x)and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?
See all solutions

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