?The expectation value of the kinetic energy of a harmonic oscillator is most easily

Chapter 7, Problem P7E.11

(choose chapter or problem)

The expectation value of the kinetic energy of a harmonic oscillator is most easily found by using the virial theorem, but in this Problem you will find it directly by evaluating the expectation value of the kinetic energy operator with the aid of the properties of the Hermite polynomials given in Table 7E.1. (a) Write the kinetic energy operator \(\hat{T}\) in terms of x and show that it can be rewritten in terms of the variable y (introduced in eqn 7E.7) and the frequency \(\omega\) as

\(\hat{T}=-\frac{1}{2} \hbar \omega \frac{\mathrm{d}^{2}}{\mathrm{d} y^{2}}\)

The expectation value of this operator for an harmonic oscillator wavefunction with quantum number v is

\(\langle T\rangle_{v}=-\frac{1}{2} \hbar \omega \alpha N_{v}^{2} \int_{-\infty}^{\infty} H_{v} \mathrm{e}^{-y^{2} / 2} \frac{\mathrm{d}^{2}}{\mathrm{d} y^{2}} H_{v} \mathrm{e}^{-y^{2} / 2} \mathrm{d} y\)

where \(N_{v}\) is the normalization constant (eqn 7E.10) and α is defined in eqn 7E.7 (the term \(\alpha \text { arises from } \mathrm{d} x=\alpha \mathrm{d} y\)). (b) Evaluate the second derivative and then use the property \(H_{v}^{\prime \prime}-2 y H_{v}^{\prime}+2 v H_{v}=0\), where the prime indicates a derivative, to rewrite the derivatives in terms of the \(H_{v}\) (you should be able to eliminate all the derivatives). (c) Now proceed as in the text, in which terms of the form \(y H_{v}\) are rewritten by using the property \(H_{v+1}-2 y H_{v}+2 v H_{v-1}=0\); you will need to apply this twice. (d) Finally, evaluate the integral using the properties of the integrals of the Hermite polynomials given in Table 7E.1 and so obtain the result quoted in the text.

Text Transcription:

Tˆ

Ω

Tˆ=-1/2ωˆd^2dy^2

Tv=-y^1/2ωα N_v^2 ∫∞ − ∞ d^2 dy^2 H_ve^2-y^2/2y

N_v

α arises from dx = αdy

H_v′′ − 2yH_v ′ + 2vH_v = 0

H_v

yH_v

H_v+1 − 2yH_v + 2vH_v–1 = 0