Today’s Outline - October 09, 2013csrri.iit.edu/~segre/phys405/13F/lecture_13.pdf · Today’s Outline - October 09, 2013 Exam #1 solutions Tips for success The 3D Schr odinger

Today’s Outline - October 09, 2013

• Exam #1 solutions

• Tips for success

• The 3D Schrodinger equation

Homework Assignment #06:Chapter 3: 15,17,22,24,27,32due Monday, October 14, 2013

C. Segre (IIT) PHYS 405 - Fall 2013 October 09, 2013 1 / 10

























Problem 1

Prove the Jacobi identity:[AB, C

]= A

[B, C

]+[A, C

]B

and use it to evaluate the commutator[T , x

]


Problem 2

Suppose there is a potential given by

V =

{0 x ≤ 0

V0 x ≥ 0

where V0 is positive. Calculate the reflectivity and proba-bility of finding a particle of energy E < V0 in the regionx > 0. (NOTE: Do not try to normalize the overall wave-function, leave a symbolic normalization constant in yourcalculations). Are these calculations consistent?


Problem 3

For a quantum harmonic oscillator, the raising and lower-ing operators are given by

a+ =1√

2mω~(mωx − ip)

a− =1√

2mω~(mωx + ip)

(a) Express the p, x , and T operators in terms of theraising and lowering operators,

(b) Compute the expectation values of these operatorsfor the harmonic oscillator state |n〉.


Problem 4

Consider an infinite square well between 0 < x < a.

V (x) =

{0 0 ≤ x ≤ +a

∞ elsewhere

with a particle prepared in a wave packet of constant mag-nitude B =

√2/a from x = 0 to x = a/2.

(a) Derive the stationary states and the energies for theinfinite square well.

(b) Using the “Fourier trick” discussed in the book,compute the first 4 non-zero coefficients of theinfinite sum which represents the wavepacket.

(c) Using this partial sum, calculate the expectationvalue of the energy for this wavepacket.


Tips for success

1 Do the reading assignments before lecture, you willunderstand them better.

2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook.

TAKE NOTES!

3 Ask questions in class, it’s likely that others have thesame ones.

4 Go through the derivations yourself, kill some trees!

5 Do the homework the “right” way, only use the solutionsmanual as a last resort.

Struggling is good and helps youlearn!

6 Come to office hours with questions, I’ll be less lonelyand it will help you too!


Tips for success


2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook.

TAKE NOTES!







Tips for success


2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook. TAKE NOTES!







Tips for success









Tips for success









Tips for success









Tips for success





5 Do the homework the “right” way, only use the solutionsmanual as a last resort. Struggling is good and helps youlearn!



Tips for success





5 Do the homework the “right” way, only use the solutionsmanual as a last resort. Struggling is good and helps youlearn!



3D Schrodinger equation

The three-dimensionalSchrodinger equation,written in Cartesiancoordinates is

but

pn =~i

∂

∂n

but the term in parenthe-ses is simply the Lapla-cian, ∇2

with volume elementd3~r = dx dy dz , we have

if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

Ψn(~r , t) = ψn(~r)e−iEnt/~




but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1





but

pn =~i

∂

∂n



if V = V (~r) only

i~∂Ψ

∂t=

p2

2mΨ + VΨ

=1

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1



Spherical coordinates

Most common potentials in 3-dimensions are central and thus only dependon the distance from the origin. Thus spherical coordinates are mostuseful.

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2

)the time-independent Schrodinger equation is thus

− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

This is difficult to solve unless we can find separable solutions,ψ(r , θ, φ) = R(r)Y (θ, φ) which give

− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY




∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2

)

the time-independent Schrodinger equation is thus

− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ


− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY




∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2


− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ


− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY




∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2


− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ


− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY




∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2


− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ


− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY




∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

(∂2

∂φ2


− ~2

2m

[1

r2∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ


− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY


Separation of variables

− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E

multiplying by −2mr2/~2 and bringing the E over[1

R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0

finally, we regroup the radial and angular parts



− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E


R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0




− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E


R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0




− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E

multiplying by −2mr2/~2 and bringing the E over

[1

R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0




− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E


R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0




− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E


R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0

finally, we regroup the radial

and angular parts



− ~2

2m

[Y

r2∂

∂r

(r2∂R

∂r

)+

R

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

2m

[1

Rr2∂

∂r

(r2∂R

∂r

)+

1

Yr2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Yr2 sin2 θ

(∂2Y

∂φ2

)]+V (r) = E


R

d

dr

(r2dR

dr

)+

1

Y sin θ

d

dθ

(sin θ

dY

dθ

)+

1

Y sin2 θ

(d2Y

dφ2

)]− 2mr2

~2[V (r)− E ] = 0

finally, we regroup the radial and angular partsC. Segre (IIT) PHYS 405 - Fall 2013 October 09, 2013 9 / 10

The radial and angular equations{1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}+

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= 0

Since the two parts depend on different variables, the only way they cancancel is to each be equal to a separation constant, which we will take tobe l(l + 1). This gives two independent equations:{

1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

We will first solve the angular equation



R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}+

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= 0

Since the two parts depend on different variables, the only way they cancancel is to each be equal to a separation constant, which we will take tobe l(l + 1). This gives two independent equations:

{1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)




R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}+

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= 0


1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)




R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}+

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= 0


1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)




R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}+

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= 0


1

R

∂

∂r

(r2∂R

∂r

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

1

Y

{1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)



Documents

Today’s Outline - October 09, 2013csrri.iit.edu/~segre/phys405/13F/lecture_13.pdf · Today’s Outline - October 09, 2013 Exam #1 solutions Tips for success The 3D Schr odinger