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Question: Consider the Stark effect (Problem 6.36) for the states of hydrogen. There are initially nine degenerate states, ψ3/m (neglecting spin, as before), and we turn on an electric field in the direction.

(a) Construct the matrix representing the perturbing Hamiltonian. Partial answer: <300|z|310>=-36a,<310|z|320>=-33a,<31±1|z|32±1>=-(9/2)a,,

(b) Find the eigenvalues, and their degeneracies.

Short Answer

Expert verified

Answer

(b)The value Eigenvalues and corresponding degeneracies are

Step by step solution

01

Definition of the Stark effect.

Due to the existence of an external electric field, the Stark effect causes the spectral lines of atoms and molecules to move and split.

It's the electric-field equivalent of the Zeeman phenomenon, in which the presence of a magnetic field splits a spectral line into many components

02

Step 2: (a) Creation of the matrix

eEext383a6·3a27·8150x76-32xe-xdx=eaEext36036·6!-32·7!310H'320=-33eaEext

Calculate perturbations in energy for . In order to do that, the wave functions of hydrogen must be known.

ψ300=1813πa327-18ra+2r2a2e-r/3aψ31±1=181πa36-rarae-r/3asinθe±iφψ310=281πa36-rarae-r/3acosθψ32±2=1162πa3r2a2e-r/3asin2θe±2iφψ32±1=181πa3r2a2e-r/(3a)sinθcosθe±iφψ320=1816πa3r2a2e-r/(3a)3cos3θ-1

Perturbation matrix has 81 elements and it is a 9x9 matrix. The value of maximum number of elements is zero, except for 8 elements. So, determine the value of 3 of them.

Matrix elements which have 02πe±miφdφ vanish. Those elements are as follows,

300H'31±1,300H'32±1,300H'32±2310H'31±1,310H'32±1,310H'32±2320H'31±1,320H'32±1,320H'32±231±1H'32±2,31±1H'32±1,32±1H'32±2

Even the diagonal elements also vanish, as integral with vanish. As 0πcos2k+1ϑsinϑdϑ=0for k = 0, 1, 2 , below matrix elements are zero:

311H'31-1,311H'32-1,300H'322,300H'320,322H'32-2

Thus, that leaves us with the matrix elements as follows,

300H'310,31±1H'32±1,310H'320300H'310=eEext813πa3281πa327-18ra+2r2a2e-r/3arcosθ6-rarae-r/3acosθr2drsinθdθdφ

Simplify the above expression.

eEext22383a4027-18ra+2r2a26-rae-2r/3ar4dr0πcos2θsinθdθ

Find the value of dr .

localid="1659008811088" x=2r3ar=3a2xdr=3a2dx

Substitute the above value in the expression.

eEext22383a4·233a250x427-27x+92x26-32xe-xdx=eaEext32460x4162-4052x+1352x2-274x3e-xdx=eaEext3246162·4!-4052·5!+1352·6!-274·7!=eaEext3246·-5832300H'310=-36eaEext

Calculate for the next function.

310H'320=eEext816πa3281πa3r2a23cos2θ-1e-r/(3a)rcosθ6-rarae-r/3acosθr2drsinθdθdφ=eEext383a606-rar6e-2r/3adr0π3cos4θsinθ-cos2θsinθdθSubstitute the value of in the above expression.

eEext383a6·3a27·8150x76-32xe-xdx=eaEext36036·6!-32·7!310H'320=-33eaEext

Evaluate the next function.

31±1H'32±1=eEext81πa3181πa36-rarae-r/(3a)e±iφsinθrcosθr2a2e-r/(3a)eiφcosθsinθr2drsinθdθdφ=2eEext38a606-rar6e-2r/3adr0πcos2θsin3θdθ

Substitute the value of in the above expression.

2eEext38a6·3a27·4150x66-32xe-xdx=eaEext720·6·6!-32·7!31±1H'32±1=-9eaEext2

Write the perturbation matrix.

Thus, reversed the rows and columns to create attractive blocks that may be diagonalized separately. Rows and columns can be rearranged, but carefully: W must be Hermitian in the final matrix.: Wij=Wji*.

03

(b) Determination of the value of eigenvalues and degenderizes  

There are one 3X3 block, two 2X2 blocks and one 1X1 block.

Find the determinant.

-E36λ036λ-E33λ033λ-E=0

Simplify the above.

-EE2-27λ2-36λ(-E·36λ)=0E54λ2-E2+27λ2=0E2=81λ2E=9eaEext

Solve further.

-E9λ/29λ/2-E=0E2-81λ24=0-E00-E=0E2=0E=0

Write the perturbation energy of E3 .

.

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Most popular questions from this chapter

Question: Use the virial theorem (Problem 4.40) to prove Equation 6.55.

Starting with Equation 6.80, and using Equations 6.57, 6.61, 6.64, and 6.81, derive Equation 6.82.

Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as inProblem 2.46).

(a) Show that the stationary states can be written in the formψn(x)=1Le2πinx/L,(-L/2<x<L/2),

wheren=0,±1,±2,....and the allowed energies areEn=2mnπL2.Notice that with the exception of the ground state (n = 0 ) – are all doubly degenerate.

(b) Now suppose we introduce the perturbation,H'=-V0e-x2/a2where aLa. (This puts a little “dimple” in the potential at x = 0, as though we bent the wire slightly to make a “trap”.) Find the first-order correction to En, using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that aLato extend the limits from ±L/2to±after all, H′ is essentially zero outside -a<x<a.

E±1=12Waa+Wbb±Waa-Wbb2+4Wab2(6.27).

(c) What are the “good” linear combinations ofψnandψ-n, for this problem? Show that with these states you get the first-order correction using Equation 6.9.

En'=ψn0H'ψn0(6.9).

(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates ofH0and A are precisely the ones you used in (c).

For the harmonic oscillator[Vx=1/2kx2], the allowed energies areEN=(n+1/2)ħω,(n=0.1.2,..),whererole="math" localid="1656044150836" ω=k/mis the classical frequency. Now suppose the spring constant increases slightly:k(1+ο')k(Perhaps we cool the spring, so it becomes less flexible.)

(a) Find the exact new energies (trivial, in this case). Expand your formula as a power series inο,, up to second order.

(b) Now calculate the first-order perturbation in the energy, using Equation 6.9. What ishere? Compare your result with part (a).

Hint: It is not necessary - in fact, it is not permitted - to calculate a single integral in doing this problem.

Show thatP2is Hermitian, butP4is not, for hydrogen states withl=0. Hint: For such statesψis independent ofθandϕ, so

localid="1656070791118" p2=-2r2ddr(r2ddr)

(Equation 4.13). Using integration by parts, show that

localid="1656069411605" <fp2g>=-4πh2(r2fdgdr-r2gdfdr)0+<p2fg>

Check that the boundary term vanishes forψn00, which goes like

ψn00~1π(na)3/2exp(-r/na)

near the origin. Now do the same forp4, and show that the boundary terms do not vanish. In fact:

<ψn00p4ψm00>=84a4(n-m)(nm)5/2+<p4ψn00ψm00>

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