Download Easy Problems in Physics 130B

Easy Problems in Physics 130B
1. Find the (normalized) eigenvectors and eigenvalues of the Sx (matrix) operator for s = 1 in
the usual (Sz ) basis.
 √ 
2


2. A spin 1 system is in the following state in the usual Lz basis: χ = √15  1 + i . What is
−i
the probability that a measurement of the x component of spin yields zero? What is the
probability that a measurement of the y component of spin yields +h̄?
√ 3
1
1
3. A spin 2 system is in the following state in the usual Sz basis: χ = √5
. What is the
1+i
probability that a measurement of the x component of spin yields + h̄2 ?
4. At t = 0, an electron is the spin state
ψ(t = 0) =
2i
√
q3
1
3
!
.
A magnetic field B is applied in the z direction. a) Find the spin state of the particle as
a function of time. b) Find the expectation value of Sy as a function of time. c) What is
the probability to measure that the electron’s spin along the x direction is h̄2 as a function of
time? (10 points)
5. A spin 1 particle is in an ` = 2 state. a) Find the allowed values of the total angular
momentum quantum number, j. b) Write out the |j, mj i states for the largest allowed j
value, in terms of the |`, ml i|s, ms i basis. (That is give one state for every mj value.) c) If
the particle is prepared in the state |` = 2, ml = 0i|s = 1, ms = 0i, what is the probability to
measure J 2 = 12h̄2 ?
~ +S
~ but its not one for
a) This is a simple addition of angular momentum problem J~ = L
which we will know the answer. We have to do the calculation. The total angular momentum
quantum number j is in the range |` − s| ≤ j ≤ ` + s and takes integer steps between the
limits. In this case ` = 2 and s = 1 giving 1 ≤ j ≤ 3 or j = 1, 2, 3.
b) In this part of the problem we are writing the total angular momentum states in which
j, mj , `, s are good quantum numbers in terms of products of states for which `, m` , s, ms are
good quantum numbers. To do this we will first write down the only possible (product) state
for the j = 3, mj = 3 total angular momentum state. Remember that the z components of
the angular momentum add so that the only way to get mj = 3 is to have m` = 2 and ms = 1.
|j = 3, mj = 3i = |` = 2, m` = 2i|s = 1, ms = 1i
We will now economize in notation a bit and rewrite the same equation.
|3, 3i = |2, 2i|1, 1i
We now apply the lowering operator to get the other states for j = 3 which is all the problem
asks for; J− = L− + S− . The equations for the lowering operators are given in the formulas
of course.
J− |3, 3i = (L− + S− )|2, 2i|1, 1i = L− |2, 2i|1, 1i + S− |2, 2i|1, 1i
√
√
√
h̄ 12 − 6|3, 2i = h̄ 6 − 2|2, 1i|1, 1i + h̄ 2 − 0|2, 2i|1, 0i
s
4
|2, 1i|1, 1i +
6
s
2
|2, 2i|1, 0i
6
s
2
|2, 1i|1, 1i +
3
s
1
|2, 2i|1, 0i
3
|3, 2i =
|3, 2i =
Its good see the the state is normalized to check for arithmetic mistakes. We now lower the z
component of angular momentum again. We will skip some of the simple steps shown above.
s
2
J− |3, 2i = (L− + S− ) 
|2, 1i|1, 1i +
3
√
2√
6|2, 0i|1, 1i +
3
s
12
4
|2, 0i|1, 1i + 2
|2, 1i|1, 0i +
30
30
s
6
|2, 0i|1, 1i +
15
|3, 1i =
|3, 1i =

1
|2, 2i|1, 0i
3
s
10|3, 1i =
s
s
2√
2k2, 1i|1, 0i +
3
s
s
s
8
|2, 1i|1, 0i +
15
6
J− |3, 1i = (L− + S− ) 
|2, 0i|1, 1i +
15
s
s
s
s
1√
4|2, 1i|1, 0i +
3
s
1√
2|2, 2i|1, −1i
3
2
|2, 2i|1, −1i
30
1
|2, 2i|1, −1i
15
8
|2, 1i|1, 0i +
15
s

1
|2, 2i|1, −1i
15
√
s
s
√
6√
6
12|3, 0i =
6|2, −1i|1, 1i + 2
|2, 0i|1, 0i
15
15
s
s
s
8√
8√
1√
6|2, 0i|1, 0i +
2|2, 1i|1, −1i +
4|2, 1i|1, −1i
+
15
15
15
s
s
s
36
12
4
|2, −1i|1, 1i + 3
|2, 0i|1, 0i + 3
|2, 1i|1, −1i
180
180
180
s
36
|2, −1i|1, 1i +
180
s
3
|2, −1i|1, 1i +
15
s
9
|2, 0i|1, 0i +
15
s
6
|2, 0i|1, −1i +
15
s
8
|2, −1i|1, 0i +
15
|3, 0i =
|3, 0i =
|3, 0i =
|3, −1i =
s
2
|3, −2i =
|2, −1i|1, −1i +
3
|3, −3i = |2, −2i|1, −1i
s
s
s
108
|2, 0i|1, 0i +
180
s
36
|2, 1i|1, −1i
180
3
|2, 1i|1, −1i
15
s
1
|2, −2i|1, 1i
15
1
|2, −2i|1, 0i
3
c) Measuring J 2 = 12h̄2 means measuring that j = 3. Since the z components of angular
momentum add, mj = 0 is required. So we simply compute the probability given the state
χ = |` = 2, ml = 0i|s = 1, ms = 0i. We can read this from the 3, 0i state we computed in
part (b).
9
P = |hj = 3, mj = 0|χi|2 =
15
6. We want to find the eigenstates of total S 2 and Sz for two spin 1 particles which have an
S1 · S2 interaction. (S = S1 + S2 )
(a) What are the allowed values of s, the total spin quantum number.
(b) Write down the state of maximum ms for the maximum s state. Use |sms i notation and
|s1 m1 i|s2 m2 i for the product states.
(c) Now apply the lowering operator to get the other ms states. You only need to go down
to ms = 0 because of the obvious symmetry.
(d) Now find the states with the other values of s in a similar way.
~=S
~1 + S
~2 .
This is very similar to the previous problem. We are adding angular momentum S
The allowed values of s are given by |s1 − s2 | ≤ s ≤ s1 + s2 which means s = 0, 1, 2.
The s = 2, ms = 2i state is easy to write down in the notation specified.
|2, 2i = |1, 1i|1, 1i
√
S− |2, 2i = (S1− + S2− )|1, 1i|1, 1i
√
√
6 − 2|2, 1i =
2 − 0|1, 0i|1, 1i + 2 − 0|1, 1i|1, 0i
s
|2, 1i =
1
|1, 0i|1, 1i +
2
s
1
|1, 1i|1, 0i
2
s
1
S− |2, 1i = (S1− + S2− ) 
|1, 0i|1, 1i +
2
√
s
1√
2|1, −1i|1, 1i +
2
s
1
|1, −1i|1, 1i +
6
6|2, 0i =
|2, 0i =
s
s
s

1
|1, 1i|1, 0i
2
1√
2|1, 0i|1, 0i +
2
4
|1, 0i|1, 0i +
6
s
s
1√
2|1, 0i|1, 0i +
2
s
1√
2|1, 1i|1, −1i
2
1
|1, 1i|1, −1i
6
Unlike the previous problem, this problem asks us to find the states with lower values of s.
We write down the state orthogonal to |2, 1i that has m = 1 then lower it to get |2, 0i. Then
we find the |0, 0i state by picing the m = 0 state orthogonal to |2, 0i and |1, 0i.
s
1
|1, 0i|1, 1i −
2
s
1
|1, −1i|1, 1i −
2
s
1
|1, 1i|1, −1i
2
s
1
|1, −1i|1, 1i −
3
s
1
|1, 1i|1, −1i +
3
|1, 1i =
|1, 0i =
|0, 0i =
s
1
|1, 1i|1, 0i
2
s
1
|1, 1i|1, −1i
3
7. A hydrogen atom is in the state ψ = R43 Y30 χ+ . A combined measurement of of J 2 and of Jz
is made. What are the possible outcomes of this combined measurement and what are the
probabilities of each? You may ignore nuclear spin in this problem.
This is again an addition of angular momentum problem. We can read the quantum numbers from the hydrogen state ψ = Rn` Y`m` χ+ = ψ = R43 Y30 χ+ . We see that ` = 3 and
m` = 0. Since this is hydrogen the electron has s = 21 and the state χ+ means ms = 12 . Since
~ + S,
~ we have |` − s| ≤ j ≤ ` + s or 5 ≤ j ≤ 7 or
mj = m` + ms , we know mj = 12 . For J~ = L
2
2
j = 52 , 72 . so this gives us the two possible outcomes of the combined measurement.
h̄2 and Jz = mj h̄ = 12 h̄
J 2 = j(j + 1)h̄2 = 63
4
or J 2 = j(j + 1)h̄2 = 35
h̄2 and Jz = mj h̄ = 12 h̄
4
We have solved the problem of adding spin
j = ` + 21 given in the formulas.
s
ψj,mj = ψ`+ 1 ,m+ 1 =
2
2
1
2
to some ` in general and have the state with
`+m+1
Y`m χ+ +
2` + 1
s
`−m
Y`,m+1 χ−
2` + 1
Comparing this to ψ = R43 Y30 χ+ , we see that m = 0 in the formula. We get the probability
to get j = 72 by taking the dot product.
s
3+0+1
Y30 χ+ +
P 7 = |h
2
7
s
3−0
4
Y31 χ− |Y30 χ+ i|2 =
7
7
We can use the other similar formula to read of that the probability to measure j =
P 7 = 37 .
5
2
is
2
8. Two (identical) electrons are bound in a Helium atom. What are the allowed states |j`s`1 `2 i
if both electrons have principal quantum number n = 1? What are the states if one has n = 1
and the other n = 2?
If an electron is in an n = 1 state, it can only have ` = 0 since ` < n for hydrogen states.
So the only angular momenta we can add is the two spins. Adding spin one-half to spin
one-half gives us a spin 1 state that is symmetric under interchange and a spin-0 state that
is antisymmetric under interchange. Remember that the highest angular momentum state
is symmetric. But the overall wavefunction must be antisymmetric and the spatial states
are both the same (n = 1, ` = 0) so the spatial state is symmetric and we must choose the
antisymmetric spin state. So we get only one possibility j = 21 , ` = 0, s = 0, `1 = 0, `2 = 0.
If an electron is in the n = 2 state, we can have ` = 0, 1. With the spatial states being
different, we can have a symmetric or antisymmetric spatial state and therefore there is no
constraint on the spin.
So for `1 = 0, `2 = 0, we must have ` = 0. s can be 0 or 1 and j = s.
For `1 = 0, `2 = 1, we have ` = 1. If s = 0 then j = 1. If s = 1 then j = 0, 1, 2.
9. Assume an electron is bound to a heavy positive particle with a harmonic potential V (x) =
1
mω 2 x2 . Calculate the energy shifts to all the energy eigenstates in an electric field E (in the
2
x direction).
This is a time independent perturbation theory problem. The unperturbed problem is simply
a one dimensional harmonic oscillator with En = (n + 21 )h̄ω. The perturbing potential is
H1 = eEx for a constant electric field. The first order pertubation theory result gives zero
since hn|x|ni = 0. This can be seen either from the raising and lowering calcularion since x
will raise or lower the state and the dot product will then give zero or from the calculation
using an integral since the integral over x will be odd. So we need to go to second order
perturbation theory. There are no degenerate states.
The perturbation can be written in terms of the raising and lowering operators.
s
H1 = eEx = eE
h̄
(A + A† )
2mω
The energy shift to the nth energy eigenstate to second order is.
En(2)
=
X |hk|H1 |ni|2
(0)
k6=n
(0)
En − Ek
√
√
e2 E 2 h̄ X |hk|A + A† |ni|2
e2 E 2 h̄ X | nδk,n−1 + n + 1δk,n+1 |2
=
=
2mω k6=n (n − k)h̄ω
2mω k6=n
(n − k)h̄ω
En(2)
e2 E 2 h̄ n
n+1
e2 E 2 h̄
e2 E 2
=
−
=−
=−
2mω h̄ω
h̄ω
2mωh̄ω
2mω 2
10. Calculate the fine structure energy shifts (in eV!) for the n = 1, n = 2, and n = 3 states of
Hydrogen. Include the effects of relativistic corrections, the spin-orbit interaction, and the socalled Darwin term (due to Dirac equation). Do not include hyperfine splitting or the effects
of an external magnetic field. (Note: I am not asking you to derive the equations.) Clearly list
the states in spectroscopic notation and make a diagram showing the allowed electric dipole
decays of these states.
This problem just asks you to show the results of the fine structure calulations that are given
in the formulas.
1
3
1
)
∆E12 = − 3 α4 mc2 (
1 −
2n
4n
j+2
1
1
The answer only depends on n and j. The states to be calculated are the 1S 2 , 2S 2 which
1
3
1
will have the same energy as the 2P 2 , the 2P 2 , the 3S 2 which will have the same energy as
1
3
3
5
the 3P 2 , the 3P 2 which will have the same energy as the 3D 2 , and the 3D 2 . Let me just do
and the n = 2 example here.
∆E12 = −
First for j = 12 .
∆E12 = −
α4 mc2 1
(
16 j +
1
2
3
− )
8
α4 mc2 5
5α4 mc2
( )=−
= −0.0000566eV
16 8
128
Now for j = 23 .
∆E12 = −
α4 mc2 1 3
α4 mc2
( − )=−
= −0.0000113eV
16 2 8
128
See the class notes for the diagram.
11. Calculate and show the splitting of the n = 3 states (as in the previous problem) in a weak
magnetic field B. Draw a diagram showing the states before and after the field is applied.
This problems asks you to use the calculation we did of splitting of Hydrogen in a weak B
field. The formula is given in the Hydrogen section.
∆EB =
1
eh̄B
(1 ±
)mj
2mc
2` + 1
for j = ` ± 21 . This is to be added to the fine structure splitting from the problem above.
Each j state splits into 2j + 1 states of different mj . The energy shift from the formula is the
familiar gL µB Bmj where gL is the Lande g factor which multiplies the basic energy shift of
eh̄B
m = µB Bmj . The Lande g factors are 2 for the 3S 1 , 23 for the 3P 1 , 43 for the 3P 3 , 45 for
2mc j
2
2
2
the 3D 3 , and 65 for the 3D 5 . See the class notes for a diagram.
2
2
12. The energies of photons emitted in the Hydrogen atom transition between the 3S and the 2P
states are measured, first with no external field, then, with the atoms in a uniform magnetic
field B. Explain in detail the spectrum of photons before and after the field is applied. Be
sure to give an expression for any relevant energy differences.
This is a problem that again employs the two things we calculated in the fine sturcture chapter,
the fine structure correction and the splitting in a weak magnetic field. The energy eigenstates
are the total angular momentum states with j = ` ± 12 . Of course the S state can only have
j = 21 . We therefore need these corrections and splittings for the 3S 1 2P 1 and 2P 3 states.
2
∆E12 = −
∆EB =
1 4 2 1
α mc (
2n3
j+
1
2
−
2
2
3
)
4n
eh̄B
1
(1 ±
)mj
2mc
2` + 1
for j = ` ± 21 .
For the 3S 1 state,
2
E=−
α2 mc2
1
3
eh̄B
− α4 mc2 ( ) + 2
mj
18
54
4
2mc
For the 2P 1 state,
2
E=−
α2 mc2
1
5
2 eh̄B
− α4 mc2 ( ) +
(2)mj
8
16
8
3 2mc
For the 2P 3 state,
2
α2 mc2
1
1
4 eh̄B
E=−
− α4 mc2 ( ) +
(2)mj
8
16
8
3 2mc
The energies of the photons emitted will be the energy differences between the states E =
E3S − E2P . For the transitions to the 2P 1 state,
2
5α2 mc2
5
1
eh̄B
2
E=
+
−
α4 mc2 +
(2) 2mj1 − mj2
72
128 72
2mc
3
You should do the energy for transions to the 2P 3 too.
2
13. Calculate the energy shifts to the four hyperfine ground states of hydrogen in a weak magnetic field. (The field is weak enough so that the perturbation is smaller than the hyperfine
splitting.)
This is the calculation we did in the section on hperfine splitting. We did it for strong and
intermediate B fields too but this is the weak B field case. The four states refered to in the
problem are the hyperfine states with total spin f = 1 and f = 0. In the weak B field case
we assume the f = 0 state is not degenerate with the others. We find the perturbation gives
a diagonal matrix for the f = 1 states so the energy shifts are just the simple first order
perturbation results. The calculation is done in the section “Hyperfine Splitting in a Weak B
Field” in the notes.
14. Write down the ground state (in spectroscopic notation) for the element Oxygen (Z = 8).
This exercises the use of Hund’s rules from atomic phyiscs which I should give you in the
formulas. The 1S and 2S take the first 4 electrons and there are 4 electrons in the 2P state.
We can think of this as 2 holes in the 2P state. The first rule is maximum s which means
s = 1 for the two holes. The next rule is maximum ` which means ` = 2 for the two holes.
The shell is more than half full so we couple these to give maximum j so this is j = 3. The
state would be labeled 3 D3 .
15. A hydrogen atom is in a uniform electric field in the z direction which turns on abruptly at
t = 0 and decays exponentially as a function of time, E(t) = E0 e−t/τ . The atom is initially
in its ground state. Find the probability for the atom to have made a transition to the 2P
state as t → ∞. You need not evaluate the radial part of the integral. What z components
of orbital angular momentum are allowed in the 2P states generated by this transition?
This is just a time dependent pertubation problem. It is not a harmonic potential. We use
the most basic time dependent perturbation forumla.
t
1 Z
0
dt0 ei(En −Ei )t /h̄ hφn |V (t0 )|φi i
cn (t) =
ih̄
0
The perturbing potential is V (t) = eE0 e−t/τ z = eE0 e−t/τ r cos θ = eE0 e−t/τ r
initial state is the 1S state and the final state is the 2P state.
eE0
c2p (∞) =
ih̄
s
eE0
=
ih̄
s
eE0
ih̄
s
c2p
c2p =
c2p
c2p
4π
Y .
3 10
The
∞
4π Z
0
0
dt0 ei(En −Ei )t /h̄ e−t /τ hψ21m |rY10 |ψ100 i
3
0
s
c2p
q
4π
3
s
∞
Z
1 Z
0
∗
∗
dt0 e[iω21 −1/τ ]t r2 drdΩR21
Y1m
rY10 R10
4π
0
h
i Z
1
1
0 ∞
∗ 3
e[iω21 −1/τ ]t
drR21
r R10 δm0
0
3 [iω21 − 1/τ ]
s
Z
eE0 1 [0 − 1]
1 1 3/2 −r/2a0 3
1
=
dr
e
r 2
ih̄ 3 [iω21 − 1/τ ]
3 2a0
a0
Z
−1
1
eE0
drr3 e−3r/2a0
= √
3 2ih̄ [iω21 − 1/τ ] a30
−1
1 3!
eE0
= √
3 2ih̄ [iω21 − 1/τ ] a30 3 4
3/2
e−r/a0
2a0
c2p
c2p
P2p
ieE0
1
6 · 16
= √
a0
81
3 2h̄ [iω21 − 1/τ ]
1
32
ieE0
a0
= √
2 [ih̄ω21 − h̄/τ ] 81
512e2 E02 a20
=
2
812 [( 38 α2 mc2 )2 − h̄τ 2 ]
16. State the selection rules for radiative transitions between hydrogen atom states in the electric
dipole approximation. These are rules for the allowed changes in l, m, s, and parity. They can
be easily derived from the matrix element given on the front of the test. Draw an energy level
diagram (up to n = 3) for hydrogen atoms in a weak B field. Show the allowed E1 transitions
from n = 3 to n = 1 on that diagram.
h̄ = 1.05 × 10−27 erg sec
c = 3.00 × 1010 cm/sec
1eV = 1.602 × 10−12 erg
α=
1 Å= 1.0 × 10−8 cm
1 Fermi = 1.0 × 10−13 cm
a0 =
mp = 938.3 MeV/c2
mn = 939.6 MeV/c2
me = 9.11 × 10−28 g = 0.511 MeV/c2
kB = 1.38 × 10−16 erg/◦ K
ge = 2 +
µBohr =
R∞
−8
eh̄
2me c
= 0.579 × 10
eA =
n=0
An
n!
R∞
√ 1 e
2πσ 2
α
π
h̄
αme c
= 0.529 × 10−8 cm
gp = 5.6
R∞
−ax2
q
=
dx e
sin θ =
−x2 /2σ 2
h̄c = 1973 eV Å= 197.3 MeV F
dx f (x) δ(g(x)) =
∞
P
n=1,3,5...
dr rn e−ar =
π
a
n−1
θn
(−1) 2
n!
n!
an+1
0
use
∂
∂a
1
dg
|
| dx
−∞
−∞
∞
P
P (x) =
= 1/137
eV/gauss
R∞
dx f (x) δ(x − a) = f (a)
−∞
e2
h̄c
e = 1.602 × 10−19 coulomb
f (x)
g(x)=0
for other forms
∞
P
n
n
θ
(−1) 2
cos θ =
n!
n=0,2,4...
√
E = m 2 c4 + p2 c2
GENERAL WAVE MECHANICS
E = hν = h̄ω
∆p ∆x ≥
h̄
2
R∞
√1
2πh̄ −∞
ψ(x) =
λ = h/p
p = h̄k
∆A ∆B ≥ h 21 [A, B]i
∆A =
dp φ(p) eipx/h̄
hA2 i − hAi2
R∞
√1
2πh̄ −∞
φ(p) =
∂
Eop = ih̄ ∂t
∂
xop = ih̄ ∂p
Huj (x) = Ej uj (x)
ψj (x, t) = uj (x)e−iEj t/h̄
−h̄2 ∂ 2 ψ
2m ∂x2
ψ(x) continuous
dψ
dx
pop =
h̄ ∂
i ∂x
q
∆ dψ
=
dx
2mλ
ψ(a) for
h̄2
R∞
∗
hφ|ψi =
−∞
φ=
P
dx ψ(x) e−ipx/h̄
+ V (x)ψ = ih̄ ∂ψ
∂t
continous if V finite
V (x) = λδ(x − a)
dxφ (x)ψ(x) hui |uj i = δij
P
|ui ihui | = 1
i
ai = hui |φi
ai u i
ψ(x) = hx|ψi
i
hφ|A|ψi = hψ|A|φi∗ = hφ|Aψi = hA† φ|ψi
φ(p) = hp|ψi
1
~ 2 + V (~r)]ψ(~r) = Eψ(~r)
[ 2m
(~p + ec A)
Hψ = Eψ
[px , x] =
h̄
i
ψi = hui |ψi
[Lx , Ly ] = ih̄Lz
[L2 , Lz ] = 0
Aij = hui |A|uj i
dhAi
dt
= h ∂A
i + h̄i h[H, A]i
∂t
HARMONIC OSCILLATOR
H=
p2
+ 12 mω 2 x2 = h̄ωA† A + 12 h̄ω
2m
un (x) =
A=(
q
∞
P
ak y k e−y
k=0
mω
x
2h̄
A† |ni =
q
2 /2
p
+ i √2mh̄ω
)
ak+2 =
A† = (
En = (n + 12 )h̄ω
2(k−n)
a
(k+1)(k+2) k
q
(n + 1) |n + 1i A |ni =
mω
x
2h̄
q
y=
q
n = 0, 1, 2...
mω
x
h̄
p
− i √2mh̄ω
) [A, A† ] = 1
(n) |n − 1i
1
u0 (x) = ( mω
) 4 e−mωx
h̄π
2 /2h̄
ANGULAR MOMENTUM
[Li , Lj ] = ih̄ijkLk
[L2 , Li ] = 0
L2 Y`m = `(` + 1)h̄2 Y`m
Lz Y`m = mh̄Y`m
L± = Lx ± iLy
Y22 =
L± Y`m = h̄ `(` + 1) − m(m ± 1) Y`,m±1
q
√1
q4π
15
32π
Y00 =
Y11 = −
q
e2iφ sin2 θ
Y21 = −
h
∂2
∂r2
+
2 ∂
r ∂r
i
3
8π
15
8π
eiφ sin θ
eiφ sin θ cos θ
∗
Y`(−m) = (−1)` Y`m
Y`` = ei`φ sin` θ
−h̄2
2µ
−` ≤ m ≤ `
q
Rn` (r) + V (r) +
`(`+1)h̄2
2µr 2
~µ =
Si = h̄2 σi
0 1
σx =
1 0

0 √12
 1
√
0
Sx = h̄ 
 2
0 √12
[σi , σj ] = 2iijk σk
0 −i
σy =
i 0

−i
0 √
2
 i
√
0
Sy = h̄ 
 2
0 √i2

√1
2



0
q
3
4π
cos θ
Y20 =
q
5
16π
(3 cos2 θ − 1)
Y`m (π − θ, φ + π) = (−1)` Y`m (θ, φ)
Rn` (r) = ERn` (r)
~
H = H0 − ~µ · B
0
Y10 =
−e ~
L
2mc
~µ =
0

−i
√
2



0
−ge ~
S
2mc
{σi , σj } = 2δij
1 0
σz =
0 −1


1 0 0


Sz = h̄  0 0 0 
0 0 −1
HYDROGEN ATOM
p2
H=
2µ
−
Ze2
r
n = nr + ` + 1
Rn` (ρ) = ρ`
∞
P
ak ρk e−ρ/2
En = − Z
n = 1, 2, 3, ...
` = 0, 1, ..., n − 1
ak+1 =
k=0
3
R10 = 2( aZ0 ) 2 e
−Zr
a0
3
µ=
4
H1 = − 8mp3 c2
e2 gp ~
S
3mMp c2
HB =
eB
(Lz
2mc
3
~
· I4πδ
(~r)
e2
~
S
2m2 c2 r 3
∆E3 =
∆EB =
+ 2Sz )
ρ=
Zr
)
2a0
−Zr
e 2a0
m1 m2
m1 +m2
H2 =
H3 =
k+`+1−n
a
(k+1)(k+2`+2) k
R20 = 2( 2aZ0 ) 2 (1 −
Rn,n−1 ∝ rn−1 e−Zr/na0
2 α2 µc2
ψn`m = Rn` (r)Y`m (θ, φ)
~
·L
= − 13.6
eV
n2
−8µE
r
h̄2
−Zr
3
√1 ( Z ) 2 ( Zr ) e 2a0
2a
a0
3
0
2
2
Zα2 mc2
hψn`m | er |ψn`m i = nZe
2a =
n2
0
R21 =
∆E12 = − 2n1 3 α4 mc2 ( j+1 1 −
2
2gp mα4 mc2
(f (f
3Mp n3
eh̄B
(1
2mc
q
2n2
±
+ 1) − I(I + 1) − 43 )
1
)mj
2`+1
for j = ` ±
1
2
ADDITION OF ANGULAR MOMENTUM
~ +S
~
J~ = L
ψjmj `s =
L ~· S = 12 (J 2 − L2 − S 2 )
|` − s| ≤ j ≤ ` + s
P
m` ms
C(jmj ; `m` sms )Y`m` χsms =
ψj,mj = ψ`+ 1 ,m+ 1 =
q
ψj,mj = ψ`− 1 ,m+ 1 =
q
2
2
2
2
q
`+m+1
Y`m χ+
2`+1
+
`−m
Y χ
2`+1 `m +
q
−
P
m` ms
hjmj `s|`m` sms iY`m` χsms
`−m
Y
χ
2`+1 `,m+1 −
for s =
1
2
and any `
`+m+1
Y`,m+1 χ−
2`+1
for s =
1
2
and any `
3
)
4n
PERTURBATION THEORY AND RADIATIVE DECAYS
En(1) = hφn |H1 |φn i
1
ih̄
cn (t) =
Rt
En(2) =
2π
h̄
Γrad
m→k
=
ΓE1
m→k =
ˆ · r̂ =
(1)
cnk =
0
0
R Q
k
α
2πm2 c2
α
2πc2
R
Γi→f =
2π
|hψf |V
h̄
|ψi i|2 ρf (E)
3
V d pk
2 3
( (2πh̄)
3 ) |Mf i | δ (momentum conservation) δ(Energy conservation)
R
~
dΩp ωkm |hφm |e−ik·~r ˆ · p~|φk i|2
3
dΩp ωkm
|hφm |ˆ · ~r|φk i|2
∆l = ±1, ∆s = 0
q
I(ω) ∝
hφk |H1 |φn i
(0)
(0)
En −Ek
dt0 ei(En −Ei )t /h̄ hφn |V (t0 )|φi i
Fermi’s Golden Rule:
Γi→f =
|hφk |H1 |φn i|2
(0)
(0)
k6=n En −Ek
P
+iy
y
4π
(z Y10 + −x√+i
Y11 + x√
Y1−1 )
3
2
2
q
Γ/2
3
Γcollision = P σ mkT
(ω−ω0 )2 +(Γ/2)2
ˆ · ~k = 0
( ∆ω
)
=
ω Dopler
q
kT
mc2
ATOMS AND MOLECULES
Hund: 1) max s
Erot =
`(`+1)h̄
2I
2
≈
2)max ` (allowed)
1
2000
eV
Evib = (n + 12 )h̄ω ≈
3) min j (≤
1
50
eV
1
2
shell) else max j