(a) Prove the three-dimensional virial theorem

2T=rV

(for stationary states). Hint: Refer to problem 3.31,

(b) Apply the virial theorem to the case of hydrogen, and show that

T=-En;V=2En

(c) Apply the virial theorem to the three-dimensional harmonic oscillator and show that in this case

T=V=En/2

Short Answer

Expert verified

(a) In the case of stationary states ddtrp=0,sorv=2T.

(b) The given expression is verified.

(c) The given expression is verified.

Step by step solution

01

Define virial theorem

The virial theorem connects the gravitational potential energy, U, of a self-gravitating entity to its total kinetic energy, T, which results from the motions of its individual pieces.

The virial theorem connects a quantum system's expected kinetic energy to its potential. This is both theoretically interesting and crucial for computational methods such as "density functional theory."

02

(a) Verification of three-dimensional virial theorem

The three-dimensional virial theorem described as follows,

ddtrp=ihH,rp

It is known that role="math" localid="1658128835661" [H,rp]=i=13[Hj,rjpj]. Apply it to the above expression and then solve it.

ddtr.p=i=13[H,rj]pj+ri=H,pi=i=13p22m+V,rjpj+rjp22mV.pi=i=1312m+P2,ripi+riV.pi=i=1312mj=13pj,pi.ripi+riV,pi

Further evaluate the expression.

ddtr.p=i=1312mj=13(pj,pi.ripipi.ripjpi)+riVi,pi=i=1312mj=13-ihpjpiδij-ihpjpiδij+riVi,pi=i=13-1m-ihpjpi+riihVri=ih-P2m+r.V

Further simplify the expression.

ddtrp=H,r.pn=ih.ih-P2m+r.v=r.v+P2m=2T-r.v

Thus, in case of stationary states ddtrp=0,sorv=2T

03

Step 3: (b) Explanation for virial theorem and verification of the given expression

Write the expression in case of hydrogen atom.

Vr=-e4π0'0rV=e4π0'0r2r^r.V=e4π0'0r=-V

So, it can be observed the equation is true but 2T=-V.

It is known that T+V=En'. Substitute the values in this expression.

T+V=EnT-2T=En-T=EnT=-EnV=2En

Thus, the given expression is verified.

04

(c) Verification of the given expression

Write the expression for Harmonic oscillator.

V=122r2V=2rr^r.V=2r2=2V

Further solve the expression.

2T=2VT=V

It is known that as T+V=En . Substitute the values in this expression.

2T=EnT=V=En2

Thus, the given expression is verified.

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Most popular questions from this chapter

(a) Find the eigenvalues and eigenspinors of Sy .

(b) If you measured Syon a particle in the general state X(Equation 4.139), what values might you get, and what is the probability of each? Check that the probabilities add up to 1 . Note: a and b need not be real!

(c) If you measuredSy2 , what values might you get, and with what probabilities?

Work out the spin matrices for arbitrary spin , generalizing spin (Equations 4.145 and 4.147), spin 1 (Problem 4.31), and spin (Problem 4.52). Answer:

Sz=(s0000s-10000s-200000-s)Sx=2(0bs0000bs0bs-10000bs-10bs-20000bs-200000000b-s+10000b-s+10)Sy=2(0-ibs0000ibs0-ibs-10000-ibs-10-ibs-20000-ibs-200000000-ibs+10000-ibs+10)

where,bj(s+j)(s+1-j)

An electron is in the spin state

χ=A3i4

(a) Determine the normalization constant .

(b) Find the expectation values of Sx,Sy , and Sz.

(c) Find the "uncertainties" ,σSx , σSyandσSz . (Note: These sigmas are standard deviations, not Pauli matrices!)

(d) Confirm that your results are consistent with all three uncertainty principles (Equation 4.100 and its cyclic permutations - only with in place ofL, of course).

Coincident spectral lines. 43According to the Rydberg formula (Equation 4.93) the wavelength of a line in the hydrogen spectrum is determined by the principal quantum numbers of the initial and final states. Find two distinct pairs{ni,nf} that yield the same λ. For example,role="math" localid="1656311200820" {6851,6409} and{15283,11687}will do it, but you're not allowed to use those!

(a) Work out the Clebsch-Gordan coefficients for the case s1=1/2,s2=anything. Hint: You're looking for the coefficients A and Bin

|sm=A|1212|s2(m-12)+B|12(-12)|s2(m+12)

such that|sm is an eigenstate of . Use the method of Equations 4.179 through 4.182. If you can't figure out whatSx(2) (for instance) does to|s2m2 , refer back to Equation 4.136 and the line before Equation 4.147. Answer:

;role="math" localid="1658209512756" A=s2+12±m2s2+1;B=±s2+12±m2s2+1

where, the signs are determined bys=s2±1/2 .

(b) Check this general result against three or four entries in Table 4.8.

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