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Problems 1
Problems
1.1 What physical phenomenon requires us to work with probability amplitudes rather than just
with probabilities, as in other fields of endeavour?
Soln: Quantum interference.
1.2 Whatpropertiescausecompletesetsofamplitudestoconstitutetheelementsofavectorspace?
Soln: Meaning can be given to (a) adding two sets of amplitudes and (b) multiplying a set by an
arbitrary complex number.
1.3 V is the dual space of the vector space V. For a mathematician, what objects comprise V ?
′ ′
Soln: The linear functions on V.
1.4 Inquantummechanics,whatobjectsarethemembersofthevectorspaceV? Giveanexample
for the case of quantum mechanics of a member of the dual space V and explain how members of
′
V enable us to predict the outcomes of experiments.
′
Soln: Any complete set of amplitudes is a member of V. Since a complete set of amplitudes
uniquely specifies the physical state of the system, we can consider that physical states are the
members of V. V is inhabited by bras. A bra is a function that extracts from a ket the amplitude
′
for a specific event. For example E extracts from ψ the amplitude E ψ that a measurement
2 2
h | | i h | i
of energy determines that the system is in its second excited state.
1.5 Given that ψ =eiπ/5 a +eiπ/4 b , express ψ as a linear combination of a and b.
| i | i | i h | h | h |
Soln:
ψ =e iπ/5 a +e iπ/4 b
− −
h | h | h |
1.6 What properties characterise the bra a that is associated with the ket a ?
h | | i
Soln: If we choose any orthonormal basis that includes a , then a is the linear function that
| i h |
vanishes on every basis function except a .
| i
1.7 An electron canbe in one of two potential wells that are so close that it can‘tunnel’ fromone
to the other (see 5.2 for a description of quantum-mechanical tunnelling). Its state vector can be
§
written
ψ =aA +bB , (1.1)
| i | i | i
where A isthestateofbeinginthefirstwelland B isthestateofbeinginthesecondwellandall
| i | i
kets are correctly normalised. What is the probability of finding the particle in the first well given
that: (a) a=i/2; (b) b=eiπ; (c) b= 1 +i/√2?
3
Soln: (a) P = 1; (b) P =0; (c) P = 7 .
4 18
1.8 An electron can ‘tunnel’ between potential wells that form a chain, so its state vector can be
written
∞
ψ = a n , (1.2a)
n
| i | i
X−∞
where n is the state of being in the nth well, where n increases from left to right. Let
| i
n/2
1 i | |
a = − einπ. (1.2b)
n
√2 3
(cid:18) (cid:19)
a. What is the probability of finding the electron in the nth well?
b. What is the probability of finding the electron in well 0 or anywhere to the right of it?
Soln: (a) P = 213−|n|. (b) ∞n=13−n = 12 and P0 = 12 so P(n≥0)= 34.
P
2 Problems
Problems
2.1 How is a wavefunction ψ(x) written in Dirac’s notation? What’s the physical significance of
the complex number ψ(x) for given x?
Soln: ψ(x) xψ . ψ(x) is the amplitude to be found at x.
≡h | i
2.2 Let Q be an operator. Under what circumstances is the complex number aQb equal to the
h | | i
complex number ( bQa ) for any states a and b ?
∗
h | | i | i | i
Soln: When Q is Hermitian.
2.3 Let Q be the operator of an observable and let ψ be the state of our system.
| i
a. What are the physical interpretations of ψ Qψ and q ψ 2, where q is the nth eigenket
n n
h | | i |h | i| | i
of the observable Q and q is the corresponding eigenvalue?
n
b. What is the operator q q , where the sum is over all eigenkets of Q? What is the
n| nih n|
operator q q q ?
c. If u (x) is tnhenw|anviehfunn|cPtion of the state q , write down an integral that evaluates to q ψ .
n n n
Soln: (a) ψPQψ is the expectation value|ofiQ. q ψ 2 is the probability of obtainingh q| oin
n n
h | | i |h | i|
measuringthe observableQ.(b) q q is the identity operator. (c) q ψ = dxu (x)ψ(x).
n| nih n| h n| i ∗n
2.4 What does it mean to say that two operators commute? What is the significance of two
P R
observables having mutually commuting operators?
Given that the commutator[P,Q]=0 for some observablesP and Q,does it follow that for all
6
ψ =0 we have [P,Q]ψ =0?
| i6 | i6
Soln: Commutation implies you always get the same result no matter in which order you use the
operators. If observables have commuting operators,there is a complete set of states in which both
observables have well-defined values.
No, [P,Q] = 0 implies only that there is some state φ such that PQφ = QP φ ; one may
6 | i | i 6 | i
readilyconstructanexampleofoperatorssuchthatPQi =QP i forbasisvectors i withi=2,
| i | i | i ∞
but PQ0 =QP 0 and PQ1 =QP 1 .
| i6 | i | i6 | i
2.5 Let ψ(x,t) be the correctly normalised wavefunction of a particle of mass m and potential
energyV(x). Write downexpressionsforthe expectationvalues of (a)x; (b) x2; (c) the momentum
p ; (d) p2; (e) the energy.
x x
What is the probability that the particle will be found in the interval (x ,x )?
1 2
Soln: (a) x = ∞ dxxψ 2; (b) x2 = ∞ dxx2 ψ 2;
h i | | h i | |
−∞ −∞
(c) p = i¯h ∞Rdxψ∗dψ; (d) p2 = ¯h2R ∞dxψ∗d2ψ; (e) E = ∞dxψ∗ −¯h2d2ψ +V(x)ψ .
h i − dx h i − dx2 h i 2m dx2
P(x ,x )= Zx2−d∞x ψ 2. Z−∞ Z−∞ (cid:18) (cid:19)
1 2 x1 | |
2.6 Write dRown the time-independent (tise) and the time-dependent (tdse) Schr¨odinger equa-
tions. Isitnecessaryforthewavefunctionofasystemtosatisfythetdse? Underwhatcircumstances
does the wavefunction of a system satisfy the tise?
Soln:
∂ ψ
tise: H E =E E ; tdse: i¯h | i =H ψ .
n n n
| i | i ∂t | i
The state of every system must satisfy the tdse, whereas the tise is satisfied only if the state has
well defined energy (which for real systems is never the case).
2.7 Why is the tdse first-order in time, rather than second-order like Newton’s equations of
motion?
Soln: Because ψ is a complete set of information, so for physics to be possible it must contain
completeinitialc|oniditionsforsolutionofthetime-evolutionequation. Ifthetdseweresecond-order
in t, this condition would be violated.
2.8 A particle is confined in a potential well such that its allowed energies are E = n2 , where
n
E
n=1,2,...isanintegerand apositiveconstant. Thecorrespondingenergyeigenstatesare 1 , 2 ,
E | i | i
..., n ,... At t=0 the particle is in the state
| i
ψ(0) =0.21 +0.32 +0.43 +0.8434 . (2.1)
| i | i | i | i | i
a. What is the probability, if the energy is measured at t = 0, of finding a number smaller than
6 ?
E
Problems 3
b. What is the meanvalue andwhat is the rms deviationof the energy ofthe particle in the state
ψ(0) ?
| i
c. Calculate the state vector ψ at time t. Do the results found in (a) and (b) for time t remain
| i
valid for arbitrary time t?
d. When the energy is measuredit turns out to be 16 . After the measurement, what is the state
E
of the system? What result is obtained if the energy is measured again?
Soln: (a)0.13;(b)theprobabilitiesareP =0.04,P =0.09,P =0.16,P =0.71so E =13.2 ,
1 2 3 4
E2 = 196 2, and σ = 4.67 . (c) ψ(t) = 0.2e i t/h¯ 1 + 0.3e i4 t/h¯ 2 + 0.4eh ii9 t/h¯ 3 E+
E −E − E − E
0h.84i3e i16 t/Eh¯ 4 . (d) ThereafterEψ = 4| andi E =16 with|ciertainty. | i | i
− E
| i | i | i E
2.9 A system has a time-independent Hamiltonian that has spectrum E . Provethat the prob-
n
{ }
ability P that a measurement of energy will yield the value E is is time-independent. Hint: you
k k
can do this either from Ehrenfest’s theorem, or by differentiating E ,tψ w.r.t. t and using the
k
tdse. h | i
Soln: This follows from Ehrenfest’s theorem: the expectation value of any function of H is con-
stant, so the probability distribution defined by the P is time-independent. For a more direct
k
answer, consider the amplitude a (t) to get E : a = E ,tψ , so using the tdse and its adjoint
k k k k
h | i
∂a ∂ E ,t ∂ ψ 1
k k
= h | ψ + E ,t | i = ( E ,tH ψ + E ,tH ψ )=0.
k k k
∂t ∂t | i h | ∂t i¯h −h | | i h | | i
2.10 Let ψ(x) be a properly normalised wavefunction and Q an operator on wavefunctions. Let
q be the spectrum of Q and u (x) be the corresponding correctly normalised eigenfunctions.
r r
{ } { }
Write downanexpressionforthe probabilitythatameasurementofQwillyieldthe value q . Show
r
that rP(qr|ψ)=1. Show further that the expectation of Q is hQi≡ ∞ ψ∗Qˆψdx.1
Soln: −∞
P 2 R
Pi = dxu∗i(x)ψ(x) where Qui(x)=qiui(x).
(cid:12)Z (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
1= dx ψ 2 = dx (cid:12) a u (cid:12)a u = a a dxu u = a 2 = P
| | ∗i ∗i! j j ∗i j ∗i j | i| i
Z Z i j ij Z i i
X X X X X
Similarly
dxψ Qψ = dx a u Q a u = a a dxu Qu
∗ ∗i ∗i! j j ∗i j ∗i j
Z Z i j ij Z
X X X
= a a q dxu u = a 2q = P q
∗i j j ∗i j | i| i i i
ij Z i i
X X X
which is by definition the expectation value of Q.
2.11 Find the energy of neutron, electron and electromagnetic waves of wavelength 0.1nm.
Soln:
p2 (h/λ)2 m
E = = =1.32 10 20J=0.0821eV; E = n =1837E =150.8eV;
n − e n
2m 2m × m E
e n
E =hν =hc/λ=1.988 10 15J=12.407keV.
γ −
×
2.12 Neutrons are emitted from an atomic pile with a Maxwellian distribution of velocities for
temperature 400K. Find the most probable de Broglie wavelength in the beam.
Soln: Therateatwhichaneutronhitsthewallofthepileisproportionaltoitsmomentump,sothe
probability density of neutrons hitting the wallis proportionalto p3e p2/2mkT =(2mkT)3/2x3e x2,
− −
where p = √2mkTx. The probability density is extremised when (3x2 2x4)e x2 = 0, i.e., when
−
−
x = 3/2 and p = √3mkT. Thus the most probable de Broglie wavelength is λ = h/√3mkT =
0.126nm.
p
1In an elegant formulation of quantum mechanics, this last result is the basic postulate of the theory, and one
derives other rules for the physical interpretation of the qn, an, etc., from it – see J. von Neumann, Mathematical
Foundations of Quantum Mechanics,PrincetonUniversityPress.
4 Problems
2.13 A beam of neutrons with energy E runs horizontally into a crystal. The crystal transmits
half the neutrons and deflects the other half vertically upwards. After climbing to height H these
neutronsaredeflectedthrough90 ontoahorizontalpathparalleltotheoriginallytransmittedbeam.
◦
The two horizontal beams now move a distance L down the laboratory, one distance H above the
other. After goingdistanceL, the lowerbeam is deflectedverticallyupwardsandis finally deflected
into the path of the upper beam such that the two beams are co-spatial as they enter the detector.
Given that particles in both the lower and upper beams are in states of well-defined momentum,
show that the wavenumbers k,k of the lower and upper beams are related by
′
m gH
n
k k 1 . (2.2)
′
≃ − 2E
(cid:18) (cid:19)
In an actual experiment (R. Colella et al., Phys. Rev. Let., 34, 1472, 1975) E = 0.042eV and
LH 10 3m2 (theactualgeometrywasslightlydifferent). Determinethephasedifferencebetween
−
∼
the two beams at the detector. Sketch the intensity in the detector as a function of H.
Soln: We take the zero of gravitationalpotential to be at the level of the lower beam. Then
p2 p2 2m2gH mgH
E = =mgH + ′ p′2 =p2 1 =p2 1 .
2m 2m ⇒ − p2 − E
(cid:18) (cid:19) (cid:18) (cid:19)
Square rooting each side and simplifying with the binomial theorem, we find
mgH mgHLk m2gHL
k k 1 and ∆φ=(k k )L = 56.4.
′ ≃ − 2E − ′ ≃ 2E ¯h2k ≃
(cid:18) (cid:19)
The intensity in the detector will be proportional to 1+ei∆φ 2 = 4cos2(∆φ/2), so the plot is a
sinusoid in H with wavenumber m2gL/¯h2k. | |
2.14 AparticlemovesinthepotentialV(x)andisknowntohaveenergyE . (a)Canithavewell-
n
definedmomentumforsomeparticularV(x)? (b)Canthe particlesimultaneouslyhavewell-defined
energy and position?
Soln: (a) A state of well defined momentum is eipx/h¯. This is an eigenfunction of H only if
·
V =constant. (b) A state of well defined position is δ(x), which is not a plane wave so not a state
of well defined momentum.
2.15 The states 1 , 2 form a complete orthonormalset of states for a two-state system. With
{| i | i}
respect to these basis states the operator σ has matrix
y
0 i
σy = i −0 . (2.3)
(cid:18) (cid:19)
Could σ be an observable? What are its eigenvalues and eigenvectors in the 1 , 2 basis? Deter-
{| i | i}
mine the result of operating with σ on the state
y
1
ψ = (1 2 ). (2.4)
| i √2 | i−| i
Soln: It is Hermitian so could be the operator of an observable. Its eigenvalues are 1 and 1 and
the associated eigenkets are 2 1/2(i, 1) and 2 1/2(i,1), respectively. −
− −
−
0 i 1 1 i
σy|ψi=2−1/2 i −0 1 = √2 i .
(cid:18) (cid:19)(cid:18)− (cid:19) (cid:18) (cid:19)
2.16 A three-state system has a complete orthonormal set of states 1 , 2 , 3 . With respect to
| i | i | i
this basis the operators H and B have matrices
1 0 0 1 0 0
H =¯hω 0 1 0 B =b 0 0 1 , (2.5)
−
0 0 1 0 1 0
−
Problems 5
where ω and b are real constants.
a. Are H and B Hermitian?
b. Write down the eigenvalues of H and find the eigenvalues of B. Solve for the eigenvectors of
both H and B. Explain why neither matrix uniquely specifies its eigenvectors.
c. Show that H and B commute. Give a basis of eigenvectors common to H and B.
Soln: (a) They are both real symmetric matrices, therefore Hermitian. (b) The e-values of H are
¯hω and ¯hω (twice) while those of B are b twice and b. Since both matrices have degenerate
− −
eigenvectors neither defines a unique set of basis eigenvectors. (c) Show commutation by brute
algebra. The eigenvectorsof H are (1,0,0)andany linear combinationof (0,1,0)and (0,0,1),while
those of B are (0,1, 1) and (x,y,y), where x,y are arbitrary. To keep H happy, in any set of
−
mutual eigenvectors we must have (1,0,0), which is the eigenket of B with x = 1, y = 0. To keep
B happy we must also have (0,1, 1), which is an eigenket of H. The third must be of the form
−
(x,y,y) and orthogonal to (1,0,0) so it can only be (0,1,1).
2.17 Given that A and B are Hermitian operators, show that i[A,B] is a Hermitian operator.
Soln: Using the rule that to dagger a product we reversethe order of the factors and dagger each
factor individually, we have
(i[A,B])† =[B†,A†]( i)=i[A,B]
−
as required.
2.18 Given a ordinary function f(x) and an operator R, the operator f(R) is defined to be
f(R)= f(r )r r , (2.6)
i i i
| ih |
i
X
where r are the eigenvalues of R and r are the associated eigenkets. Show that when f(x) =x2
i i
| i
this definition implies that f(R)=RR, that is, that operating with f(R) is equivalent to applying
the operator R twice. What bearing does this result have in the meaning of eR?
Soln: We always have R= r r r , so
i i| iih i|
RR= rj|rjiPhrj| ri|riihri|= rjri|rjihrj|riihri|= ri2|riihri|,
j i ij i
X X X X
which this is precisely what we get if we put f(r )=r2.
i i
2.19 Showthatifthere is acomplete setofmutualeigenketsofthe HermitianoperatorsAandB,
then [A,B]=0. Explain the physical significance of this result.
Soln: Let i be the complete set of mutual eigenkets (Ai = a i and B i = b i ) and let
i i
| i | i | i | i | i
ψ = ψ i be an arbitrary ket. We show that [A,B]ψ =0:
| i i i| i | i
P[A,B]ψ =(AB BA) ψi i =A ψibi i B ψiai i = ψi(biai aibi)i .
| i − | i | i− | i − | i
i i i i
X X X X
2.20 Given that for any two operators (AB) =B A , show that
† † †
(ABCD) =D C B A . (2.7)
† † † † †
Soln: We group the four factors into two pairs and apply what’s given twice:
(ABCD) =(CD) (AB) =D C B A .
† † † † † † †
2.21 Prove for any four operators A,B,C,D that
[ABC,D]=AB[C,D]+A[B,D]C+[A,D]BC. (2.8)
Explain the similarity with the rule for differentiating a product.
Soln: First we prove [AB,C]=A[B,C]+[A,C]B:
[AB,C]=ABC CAB =ABC ACB+ACB CAB =A(BC CB)+(AC CA)B.
− − − − −
6 Problems
Now we apply this rule recursively:
[ABC,D]=[(AB)C,D]=AB[C,D]+[(AB),D]C =AB[C,D]+A[B,D]C+[A,D]BC.
Analogously
d df dg dh
(fgh)= gh+f h+fg .
dx dx dx dx
2.22 Show that for any three operators A, B and C, the Jacobi identity holds:
[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0. (2.9)
Soln: We expand the inner commutator and then use the rule for expanding the commutator of a
product:
[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=[A,BC] [A,CB]+[B,CA] [B,AC]+[C,AB] [C,BA]
− − −
=[A,B]C+B[A,C] [A,C]B C[A,B]+[B,C]A+C[B,A] [B,A]C
− − −
A[B,C]+[C,A]B+A[C,B] [C,B]A B[C,A]
− − −
=2([A,B]C C[A,B] [A,C]B+B[A,C]+[B,C]A A[B,C])
− − −
=2 ABC BAC CAB+CBA ACB+CAB+BAC BCA+BCA
− − − −
(cid:0) CBA ABC+ACB =0
− −
(cid:1)
2.23 Show that a classical harmonic oscillator satisfies the virial equation 2 KE = α PE and
h i h i
determine the relevant value of α.
Soln: Classicalharmonicmotionisthesolutionx=Acosωttomx¨= mω2x,wheretheamplitude
Aisarbitrary. Thetime-averagedkineticenergyis 1m x˙2 = 1mω2A2.−Thetime-averagedpotential
2 h i 4
energy is 1mω2 x2 = 1mω2A2, so the virial theorem is satisfied for α = 2, as it should be for a
2 h i 4
quadratic potential.
2.24 Given that the wavefunction is ψ =Aei(kz ωt)+Be i(kz+ωt), where A and B are constants,
− −
show that the probability current density is
J=v A2 B 2 ˆz, (2.10)
| | −| |
(cid:0) (cid:1)
where v =¯hk/m. Interpret the result physically.
Soln: We have ∇ψ =ikˆz(Aeikz Be ikz), so
−
−
¯hkˆz
J= (Aeikz +Be−ikz)( A∗e−ikz +B∗eikz) (A∗e−ikz +B∗eikz)(Aeikz Be−ikz) .
−2m − − −
(cid:2) (cid:3)
When we multiply out the brackets on the right, the terms involving both A and B cancel and
the other terms add up to give the required expression. Each wave represents a stream of particles
movingatspeedv,thatwithcoefficientAmovingtoincreasingz andthatwithcoefficientB moving
in the opposite direction.
Problems 7
Problems
3.1 After choosing units in which everything, including ¯h = 1, the Hamiltonian of a harmonic
oscillator may be written H = 1(p2+x2), where [x,p] = i. Show that if ψ is a ket that satisfies
2 | i
H ψ =E ψ , then
| i | i 1(p2+x2)(x ip)ψ =(E 1)(x ip)ψ . (3.1)
2 ∓ | i ± ∓ | i
Explain how this algebra enables one to determine the energy eigenvalues of a harmonic oscillator.
Soln: Let A x+ip so
≡
A A=(x ip)(x+ip)=p2+x2+i[x,p]=p2+x2 1=2H 1.
†
− − −
and
[A†,A]=[x ip,x+ip]=2i[x,p]= 2.
− −
H ψ =E ψ EA ψ =A H ψ =HA ψ +[A ,H]ψ (3.2)
† † † †
| i | i ⇒ | i | i | i | i
Now [A ,H]= 1[A ,A A]= 1A [A ,A]= A . Substituting this into (3.2)
† 2 † † 2 † † − †
HA ψ =(E+1)A ψ .
† †
| i | i
Thus by multiplying aneigenstate ofenergy E we geta state of energyE+1. Trivialmodifications
of the above yield
HAψ =(E 1)Aψ .
| i − | i
sowithAwecanmakestatesoflowerenergy. But ψ H ψ = 1(pψ 2+ xψ 2)>0,sosomething
h | | i 2 | | i| | | i|
must terminate the sequence of lower-energy states. It can only be stopped by A annihilating a
state of lowest energy
0= A0 2 = 0A A0 = 0(2H 1)0 =2E 1.
† 0
| | i| h | | i h | − | i −
Thus we find that E = 1 and thus that E =n+ 1.
0 2 n 2
3.2 Given that AE = αE and E = (n+ 1)h¯ω, where the annihilation operator of the
harmonic oscillator|isni | n−1i n 2
mωx+ip
A , (3.3)
≡ √2m¯hω
show that α=√n. Hint: consider AE 2.
n
| | i|
Soln: We have
1
α2 = E A AE = E (mωx ip)(mωx+ip)E
n † n n n
| | h | | i 2m¯hωh | − | i
1 1 E
= E m2ω2x2+p2+imω[x,p] E = (2mE mω¯h)= n 1 =n.
2m¯hωh n| | ni 2m¯hω n− ¯hω − 2
(cid:18) (cid:19)
Since the phase of E(cid:0) 1 is arbitrary, so is t(cid:1)he phase of α. We choose to make it real.
n
| − i
3.3 The pendulum of a grandfather clock has a period of 1s and makes excursions of 3cm either
side of dead centre. Given that the bob weighs 0.2kg, around what value of n would you expect its
non-negligible quantum amplitudes to cluster?
Soln:
2π
ω = x=0.03cosωt v =0.03 2π E = 1mv2 =0.1(6π/100)2J
1s ⇒ max × ⇒ 2 max
So
E (6π)2 10 5
n = × − 5 1030
≃ ¯hω 1.05 10 34 2π ∼ ×
× − ×
3.4 Showthatthe minimumvalue ofE(p,x) p2/2m+1mω2x2 withrespecttothe realnumbers
≡ 2
p,x when they are constrained to satisfy xp = 1¯h, is 1¯hω. Explain the physical significance of this
2 2
result.
8 Problems
Soln: Use a Lagrange multiplier λ:
∂
0= (E λxp) p
∂p − 0= λx (1)
m −
∂ ⇒ 0=mω2x λp (2)
0= ∂x(E−λxp) −
p (1) x (2) 0=p2/m mω2x2 p=mωx=mω¯h/2p p2/m=h¯ω/2. So
× − × ⇒ − ⇒ ⇒
p2 p2
H = 1 +(mωx)2 = = 1¯hω.
2 m m 2
(cid:18) (cid:19)
3.5 How many nodes are there in the wavefunction xn of the nth excited state of a harmonic
h | i
oscillator?
Soln: n nodes.
3.6 Show that in terms of a harmonic oscillator’s characteristic length ℓ ¯h/2mω the ladder
≡
operators can be written
x ∂ x ∂ p
A= +ℓ and A = ℓ . (3.4)
†
2ℓ ∂x 2ℓ − ∂x
Henceshowthatthewavefunctionofthesecondexcitedstateis x2 =constant (x2/ℓ2 1)e x2/4ℓ2
−
h | i × −
and find the normalising constant.
Soln: In the position representation,
ℓ x ℓ x ∂ x ∂
A= (mωx+ip)= +i p = +ℓ and A = ℓ
†
¯h 2ℓ ¯h 2ℓ ∂x 2ℓ − ∂x
(cid:18) (cid:19)
∂u x
Au =0 ℓ 0 = u u e x2/4ℓ2
0 0 0 −
⇒ ∂x −2ℓ ⇒ ∝
From 1= ∞ dx u0 2 we find that the normalizing constant is (2πℓ2)1/4.
| |
−∞
R u =A u = 1 1y ∂ e y2/4
1 † 0 (2πℓ2)1/4 2 − ∂y −
(cid:18) (cid:19) with y x/ℓ
= 1 ye y2/4 ≡
−
(2πℓ2)1/4
√2u2 =A†u1 = (2πℓ12)1/4 21y− ∂∂y ye−y2/4 = (2πℓ12)1/4 y2−1 e−y2/4
(cid:18) (cid:19)
So (cid:0) (cid:1)
1 x2
u (x)= 1 e x2/4ℓ2
2 (8πℓ2)1/4 ℓ2 − −
(cid:18) (cid:19)
3.7 Explainwhythe wavefunction xn oftheoscillator’snth stationarystatemusthavetheform
h | i
xn =H (x/ℓ)e x2/4ℓ2, (3.5)
n −
h | i
where H is an nth-order (‘Hermite’) polynomial. By casting the equations An = √nn 1 and
n
| i | − i
A n 1 =√nn in the x-representation, show that
†
| − i | i
x
H (x/ℓ)=√nH (x/ℓ) and √nH (x/ℓ)= H (x/ℓ) H (x/ℓ). (3.6)
n′ n−1 n ℓ n−1 − n′−1
and thus that
x
√nH (x/ℓ)= H (x/ℓ) √n 1H (x/ℓ). (3.7)
n n 1 n 2
ℓ − − − −
Problems 9
Figure 3.0 The wavefunctions hx|2i
and hx|40i of two stationary states of
a harmonic oscillator.
Given that H =(2πℓ2) 1/4 and H (y)=y/(2πℓ2)1/4, use this recurrencerelationto reproduce the
0 − 1
plots ofthe wavefunctions x2 and x40 showninFigure3.4. Explainthe physicalsignificanceof
h | i h | i
the vertical arrows. Why is the amplitude of x40 largest near the right arrow?
h | i
Soln: We obtain xn by applying A to e x2/4ℓ2 and by Problem 3.6 eachsuch applicationadds
† −
h | i
a factor x/2ℓ either by multiplication or from differentiation of the exponential, so it is clear that
xn will be a polynomial times the exponential. Now
h | i
x ∂
√nhx|n−1i=√nHn−1(x/ℓ)e−x2/4ℓ2 = 2ℓ +ℓ∂x Hn(x/ℓ)e−x2/4ℓ2 =Hn′(x/ℓ)e−x2/4ℓ2
(cid:18) (cid:19)
from which the first required relation follows. Similarly
x ∂
√n xn =√nH (x/ℓ)e x2/4ℓ2 = ℓ H (x/ℓ)e x2/4ℓ2
n − n 1 −
h | i 2ℓ − ∂x −
(cid:18) (cid:19)
x
= ℓHn−1(x/ℓ)−Hn′−1(x/ℓ) e−x2/4ℓ2
fromwhichthe secondrequiredrelationfollows.(cid:16)Using the firstrelationto el(cid:17)iminate H fromthe
n′ 1
second relation, we obtained the required recursion relation. The arrows are located at−the end of
the classically-alloweddomain x =√4n+2ℓ and the amplitude peaks there because classically
max
a particle lingers at its turning points. Here’s a routine to compute H (x/ℓ)
n
double Hermite(double x,int n) //returns normalised <x|n> less exponential
{
double tpi4=pow(2*acos(-1),.25), Hnm1=1/tpi4, Hn=x/tpi4;
for(int i=1; i<n; i++)
{
double Hsav=Hn;
Hn=(x*Hn-sqrt(i)*Hnm1)/sqrt(i+1);
Hnm1=Hsav;
}
return Hn;
}
3.8 Use
¯h
x= (A+A )=ℓ(A+A ) (3.8)
† †
2mω
r
to show for a harmonic oscillator that in the energy representation the operator x is
0 √1 0 0 ...
√1 0 √2 0
0 √2 0 √3
···
√3 ...
... ... ... ...
x =ℓ (3.9)
jk ... 0 √n 1 ...
√n 1 0− √n
− √n 0 √n+1
√n+1 0 ···
··· ··· ··· ··· ···
