1
I. What to recall about motion in a central potentialII. Example and solution of the radial equation for a particle trapped within radius “a”III. The spherical square well
(Re-)Read Chapter 12 Section 12.3 and 12.4
2
I. What to recall about motion in a central potentialV =V (|rr1 −
rr2 |) between m asses m 1 and m 2
1 24 4 34 4
Recall the tim e-independent Schrodinger Equation:H
p12
2m1+
p22
2m2+V (|rr1 −
rr2 |)⎡
⎣⎢⎢⎢
⎤
⎦⎥⎥⎥Ψ =EΨ
↓6 74 4 4 4 4 44 84 4 4 4 4 4 4{ Ψ =EΨ
To convert this into a separable PDE, define1M
=1m 1
+1m 2
M =m 1 + m 2
R =m 1
rr1 + m 2rr2
Mrr=rr1 −
rr2rP=
rp1 +rp2
rpm=
rp1m 1
−rp2
m 2
Then you get
H =p2
2M+
p2
2m+V (|rr |)
Ignore center of mass motion
Focus on this “Hμ”
rr1 −
rr2
rr1
rr2
m1 m2
O
3
Hm =p2
2m+V (r)
=−h2∇2
2m+V (r)
In spherical coordinates:
∇2 =1r2
r2∂∂r
⎛⎝⎜
⎞⎠⎟ +
1
r2 sinq
∂∂q
sinq∂∂q
⎛⎝⎜
⎞⎠⎟ +
1
r2 sin2q
∂2
∂f 2
1
r2−L2
h2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
1 24 4 4 4 4 44 34 4 4 4 4 4 4
Plug this ∇2 into Hm
Write out Hm |Ψ⟩ =E |Ψ⟩
Project into ⟨r,q,f | space:
−h2
2m1r2
∂∂r
r2∂∂r
⎛⎝⎜
⎞⎠⎟−
L2
h2r2⎡⎣⎢
⎤⎦⎥+V (r)
⎧⎨⎩⎪
⎫⎬⎭⎪⟨r,q,f |Ψ⟩ =E⟨r,q,f |Ψ⟩
Guess that ⟨r,q,f |Ψ⟩ =R(r)f(q,f )Separate the equation, the constant of separation turns out to be l(l+1).
f(q,f ) turns out to be Ψlm (q,f )
Then the R equation is:
1R
ddr
r2dRdr
⎛⎝⎜
⎞⎠⎟−
2mr2
h2 V (r)−E[ ] =l(l+1) "Form 1" of the radial equation
the angular momentum operator
4
You can get an alternative completely equivalent form of this equation if you derivem ≡rRThen you get
−h2
2md 2udr2
+ V +h2l(l+1)2mr2
⎡⎣⎢
⎤⎦⎥u=Eu "Form 2" of the radial equation
*Choose either Form 1 or Form 2 depending upon what V is--pick whichever gives and easier equation to solve*Rem em ber the boundary conditions on R:rR(r→ ∞)→ 0 BC1rR(r→ 0)→ 0 BC2Procedure for finding the total Ψ(
rr,t) for a system in a central potential:
(i) Get V (r)(ii) Plug it into the radial equation (either Form 1 or Form 2), solve for R and the energies Ei
(iii) Multiply that R by Ψlm (q,f ) and e
−iEith to get Ψ(r,t)=RΨe
−iEith
5
II. Example-Solution of the radial equation for a particle trapped within radius "a"
V= 0 for r < a∞ for r > a
⎧⎨⎩
This is also called a "spherical box"Recall the radial equation in Form 1 (without the substitution u=rR):
1r2
ddr
r2ddr
⎛⎝⎜
⎞⎠⎟ +
2mh2 E −V (r)−
h2l(l+1)2mr2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
Since V =∞ for r > a, the wave function cannot have any portion beyond r > a. So just solve the equation for r < a.Plug in V =0 (r < a )
1r2
ddr
r2ddr
⎛⎝⎜
⎞⎠⎟ +
2mEh2 −
l(l+1)r2
⎡⎣⎢
⎤⎦⎥R =0
expand this: call this "k2"
1r2
r2d 2
dr2+ 2r
ddr
⎛⎝⎜
⎞⎠⎟
d 2
dr2+2rddr
+ k2 −l(l+1)
r2⎡⎣⎢
⎤⎦⎥R =0
6
Define r ≡kr, so 1r=kr
Then ddr
=drdr
ddr
=kddr
d 2
dr2=
ddr
kddr
⎛⎝⎜
⎞⎠⎟ =
drdr
ddr
kddr
⎛⎝⎜
⎞⎠⎟ =k2 d 2
dr2
Plug this in:
k2 d 2
dr2 +2k2
rddr
+ k2 −l(l+1)k2
r2
⎡⎣⎢
⎤⎦⎥R =0
d 2
dr2 +2r
ddr
+ 1−l(l+1)r2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
The solution of this equation is R(r) = Cjl(r) + Dnl(r)
jl(r)≡p2r
⎛⎝⎜
⎞⎠⎟1/2
Jl+
12
(r)
"spherical Bessel function" "ordinary Bessel function of half-odd integer order"
exam ples: j0(r)=sinrr
j1(r)=sinrr2 −
cosrr
Normalization not yet specified
Spherical Neuman function, Irregular @ r=0, so get D=0
7
I. Particle in 3-D spherical well (continued)II. Energies of a particle in a finite spherical well
Read Chapter 13
8
Now apply the BC: the wave function R must = 0 @ r = a
jl (ka) =0 whenever a Bessel function = 0 its argum ent (here: (ka)) is called a "zero" of the spherical Bessel function and these colum ns are labelled by n n=1 n=2 n=3 etc.
S jl=0 for ka = 3.14 6.28 9.42 ...P jl=1 for ka = 4.49 7.73 ... ...
D jl=2 for ka = 5.76 9.10 ... ... F jl=3 ...etc.
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
Sum m arize:
Ψ3−Dcentral potential
tim e-independent = R ⋅Ψlm
spherical harm onic where R=C⋅jl
Because each jl has zeros at several n's, we have to specify n too, so R ≡R nl
In spectroscopy the rows are labelled by:
9
Because k2 ≡ 2mEh2 , the boundary condition that jl(ka) = 0 gives the allowed energies.
Recipe to find an allowed energy:(i) Pick the n, l levels that you want. (Exam ple: pick n = 2, l = 0)
(ii) Find the zero of that Bessel function kan =2jl =0
⎛⎝⎜
⎞⎠⎟=6.28
⎛⎝⎜
⎞⎠⎟
(iii) Plug into jl(ka) = 0
kan =2jl =0
⎛⎝⎜
⎞⎠⎟=6.28
use k2 ≡ 2mEh2
k2a2 = (6.28)2
2mEl=0
n=2
h2 a2 = (6.28)2
El=0n=2 =
(6.28)2h2
2ma2
10
I. Energies of a particle in a finite spherical square well
11
III. Energies of a particle in a finite spherical square well
Consider
V(r) = −V 0 for r < a 0 for r > a
⎧⎨⎪⎩⎪
This actuall looks very m uch like the potential between 2 nucleons in the nucleusSolve this analogously to 1-D square well procedure(i) Define regions I and II(ii) Plug in V into radial equation in each region
Pick Form 2, so solve for u=rR(iii) Apply BC @ r = 0, a, ∞ This leads to quantized E (we'll stop here)(iv) Norm alize if you want the exact form of Ψ Carry out this procedure, first for l = 0 only, then for general l:
V(r)
a
0
-V0
rI II
0
12
Case I: Find allowed energies of a particle in an l = 0 state of a finite spherical square well
(i) define regions I and II
Notice that the particle in the well will have E < 0 just because of the way we defined the potential*This is just a convention (i.e. a choice of the origin for the V scale), but since it is common we will use it.Notice this is a different convention that the one we used for the 1-D square well
That was:
but the forms of the solution inside and outside of the well (sinkx or e−kx) are unaffected by the choice of origin.
V(r)
a
0
-V0
rI II
0
0
+V0
13
Because E is intrinsically negative, we can write: E = -|E| when we wishRecall Form 2 of the radial equation:
−h2
2md 2udr2
+ V +h2l(l+1)2mr2
⎡⎣⎢
⎤⎦⎥u=Eu
when l = 0 this is:
−h2
2md 2udr2
+V u=Eu
This is exactly the sam e form as for the 1-D square well, so the wavefunctions "in" will have the sam e form as the "Ψ's" for the 1-D square well(ii) Plug in V :Make a table as we did in Chapter 4:
Region IV = -V 0
Region IIV = 0
General tim e-independent Radial equation:−h2
2md 2udr2
+V 0
⎡⎣⎢
⎤⎦⎥u=Eu
−h2
2md 2udr2
⎡⎣⎢
⎤⎦⎥u=Eu
Solve Radial equation to get:
where:
"allowed region solution:"uI = Asink1r+ Bcosk1r
k1 ≡2m(V 0 + E)
h
= 2m(V 0−|E |)
h
"forbidden region solution:"
uII = Ce+k2r + De−k2r
k2 ≡ 2m(−E)
h
= 2m |E |h
14
(iii) Apply BC's:
BC1: rR(r → ∞)→ 0 u(r→ 0)→ 0 So B=0BC2: rR(r→ ∞)→ 0 u(r→ ∞)→ 0 So C=0BC3: uI(r=a) = uII(r=a)
Asink1a = De−k2a "equation 1"
BC4: duI
dr(r=a) =
duII
dr(r=a)
k1Acosk1a = -k2De−k2a "equation 2"
Divide equation 1equation 2 :
k1 cotk1a = -k2 This has the sam e form (variation in som e factors or 2) as the Class 2 (or odd) solutions to the 1-D square well.Multiply both side by a: k1acotk1a = -k2a
-cotk1a =k2ak1a
15
Define l ≡2mV 0a
2
h2
Define y ≡ k1a = a 2m(V 0−|E |)
h Notice that
k2a = a 2m |E |
h=
(2mV 0a2 )−[2m(V 0−|E |)a2 ]
h2
= l −y2
-coty= l −y2
y Plot both sides versus y.
Intersection points are solution to the equation:
(use the identity that -cotx = +tan(p2+x))
y0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
LHS(-cot)
16
y0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
How the value of λ affects the plot:l −y2
y for large λ
l −y2
y for small λ
*Notice if l is very sm all there m ay be NO interactions
proportional to V 0a2
well depth (well width)2
To find E:Locate graphically, or com putationally, the y-coordinates of the points where:
-coty=l −y2
y
Call these yi Exam ple:
Then plug in the definition of y:
yi = ah
2m(V 0−|Ei |)
Plug in these values of a, u, V 0 , h, to solve for Ei
y1 y2 y3
17
Alternatively if you measure the Ei (for example the bound state energies of a deuteron),
you can work backwards to find V0a2 .
Recall Case 1 was for l = 0 only. Now,Case 2: Find the allowed energies for finite spherical square well for a particle in an arbitrary l state.
(i) Regions I II(ii) In this case Form 1 of the Radial equation is easier to solve:
1r2
ddr
r2 ddr
⎛⎝⎜
⎞⎠⎟ +
2mh2 E −V (r)−
h2l(l+1)2mr2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
expand the derivative and write E = -|E|
d 2
dr2+2rddr
+2mh2 −|E |−V (r)−
h2l(l+1)2mr2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
Make Table:
V(r)
a
0
-V0
rI II
0
18
Region IV = -V0
Region IIV = 0
General time-independent Radial equation:d 2
dr2 +2rddr
+2mh2 V 0−|E |−
l(l+1)r2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
d 2
dr2+2rddr
−2m |E |h2 −
l(l+1)r2
⎡⎣⎢
⎤⎦⎥R =0
kI ≡2m(V 0−|E |)
hrI ≡kIr (just like p.6 in notes)
d 2
drI2 +
2rI
ddrI
+ 1−l(l+1)rI
2
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥R =0
RlI(rI)=Aljl(rI)+ Bljl(rI)
kII ≡2m |E |h
rII ≡ikIIr
To solve this define:and
Then the radial equation becomes:
The solution is:
19
I. Energies of a particle in a finite spherical square well, continuedII. The Hydrogen Atom
20
If rII ≡ikIIr
Then 1r=ikIIrII
ddr
=drII
drd
drII
=ikIId
drII
d 2
dr2=
ddr
ikIId
drII
⎛⎝⎜
⎞⎠⎟ =
drII
drd
drII
ikIId
drII
⎛⎝⎜
⎞⎠⎟ =ikIIikII
d 2
drII2 =−kII
2 d 2
drII2
Plug these in: then the Radial Equation becom es
−kII2 d 2
drII2 +
2ikIIrII
ikIId
drII
−kII2 +
l(l+1)kII2
rII2
⎡⎣⎢
⎤⎦⎥R =0
Multiply through by (-1):
d 2
drII2 +
2rII
ddrII
+1−l(l+1)rII
2
⎡⎣⎢
⎤⎦⎥R =0
Again the solution is R lII(rII)=Cljl(rII)+ Dlnl(rII)
Since Region II does not include r=0, we do no thave to discard the Neum ann for nl.
Since we have not yet fixed C and D, we can rewrite R lII as a different form of linear com bination of jl and nl.
21
In particular we could define
h l(1) ≡jl + inl and
hl(2) ≡jl −inl "The spherical Hankel functions"
Then
R lII(rII)= ′Clhl
(1)(rII)+ ′Dlhl(2)(rII)
(iii) Now apply the boundary conditions:
BC1: rR(r→ 0)→ 0 This is satisfied by R lI =Ajl(rI)= Ajl(kIr)
BC2: rR(r→ ∞)→ 0 Exam ine the hl(r→ ∞):
jl(r → ∞)→1rcos r −
(l+1)p2
⎡⎣⎢
⎤⎦⎥
nl ≡(−1)l+1p2r
⎛⎝⎜
⎞⎠⎟12
J−l−
12
(r)
r→ ∞⏐ →⏐ ⏐1rsin r −
(l+1)p2
⎡⎣⎢
⎤⎦⎥
Plug these in:
22
I. Energies of a particle in a finite spherical square well (continued)II. The Hydrogen Atom
23
hl(1)(rII) r→ ∞⏐ →⏐ ⏐
1ikIIr
cos ikIIr−(l+1)p
2⎡⎣⎢
⎤⎦⎥+ i
1ikIIr
sin ikIIr−(l+1)p
2⎡⎣⎢
⎤⎦⎥
=1
ikIIrcos ikIIr−
(l+1)p2
⎡⎣⎢
⎤⎦⎥+ isin ikIIr−
(l+1)p2
⎡⎣⎢
⎤⎦⎥
⎧⎨⎩⎫⎬⎭
Ignore the phases for now
Recall eiq =cosq + isinq Here "q" is ikIIr
:1
ikIIrei(ikIIr){ }
:1
ikIIre−kIIr{ } This → 0 as r→ ∞ so it satisfies the BC
hl(2)(rII) r→ ∞⏐ →⏐ ⏐
1ikIIr
cos ikIIr−(l+1)p
2⎡⎣⎢
⎤⎦⎥−i
1ikIIr
sin ikIIr−(l+1)p
2⎡⎣⎢
⎤⎦⎥
:1
ikIIre+kIIr{ } This blows up as r→ ∞ so we m ust set its coefficient ′D =0
BC3: R I(r=a)=R II(r=a)
Ajl(kIa)= ′C hl(1)(kIIa)
BC4: ∂R I
∂r(r=a)=
∂R II
∂r(r=a)
Addr
jl(kIa)= ′Cddr
hl(1)(kIIa)
24
Divide BC4BC3
to eliminate the normal Bat. coefficients:
ddr
jl (kIa)
jl (kIa)=
ddr
h(1)l(kIIa)
h(1)l(kIIa)
For a specific V 0 , a, u, and l, one can solve this for the bound states Ei
I. The Hydrogen AtomWhat this m eans is "the eigen functions and eigen energiespossible for the electron in a one-electron atom "This is just the central potential problem again, now for
V =−Ze2
r
So we know that Ψelectron will have the form Ψ : R(r)Ψlm
because of this (l, m ), also subscript the ′Ψ :Ψlm
So the m ost general Ψ m ust be a linear com bination of all possible Ψlm's, so
Ψ = Ψlm = R(r)Ψl
m (q,f )l,m∑
l,m∑
# protons in nucleus
Charge of electron
energy levels
25
The Ylm's are standard functions so we can ignore them for now. We will find R and then just multiply by Yl
m later.To find R, recall the Radial Equation (Form 2):
−h2
2md 2udr2
+ V +h2l(l+1)2mr2
⎡⎣⎢
⎤⎦⎥u=Eu (where u≡rR)
Consider bound states, so se E=-|E|
Plug in V =−Ze2
r−h2
2md 2udr2
+h2l(l+1)2mr2
−Ze2
r+ |E |
⎡⎣⎢
⎤⎦⎥u=Eu
To sim plify the form of the equation, m ultiply through by −2mh2 and define
r ≡8m |E |h2
⎛⎝⎜
⎞⎠⎟
12
r
So
1r=
8m |E |h2
⎛⎝⎜
⎞⎠⎟
12 1r
and
ddr
=drdr
ddr
=8m |E |h2
⎛⎝⎜
⎞⎠⎟
12 ddr
Defined as E<0
26
and
d 2
dr2 =ddr
8m |E |h2
⎛⎝⎜
⎞⎠⎟
12 ddr
⎡
⎣⎢⎢
⎤
⎦⎥⎥
=drdr
ddr
8m |E |h2
⎛⎝⎜
⎞⎠⎟
12 ddr
⎡
⎣⎢⎢
⎤
⎦⎥⎥
=8m |E |h2
⎛⎝⎜
⎞⎠⎟
12 d 2
dr2
Plug these in:
8m |E |h2
⎛⎝⎜
⎞⎠⎟d 2
dr2 −l(l+1)r2
8m |E |h2
⎛⎝⎜
⎞⎠⎟ +
2mh2
Ze2
r8m |E |h2
⎛⎝⎜
⎞⎠⎟
12
−2m |E |h2
⎡
⎣⎢⎢
⎤
⎦⎥⎥u=0
divide through by 8m |E |h2
⎛⎝⎜
⎞⎠⎟ :
d 2
dr2 −l(l+1)r2 +
2mZe2
h2
8m |E |h2
⎛⎝⎜
⎞⎠⎟
12
1r−14
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥u=0
Ze2
hm
2 |E | call this "l"
d 2
dr2 −l(l+1)r2 +
lr−14
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥u=0
27
Procedure to solve: this is similar to the one for the analytic (non-a+a) solution of the harm onic oscillator.Recall the 2 radial BC's: BC1: u(r→ ∞)→ 0; BC2: u(r→ 0)→ 0.
(i) consider the case where r → ∞
This elim inates the 1r,
1r2 term s:
Then the equation is approxim ately:
d 2udr2 −
u4=0 (for u(r → ∞) only)
u : Ae−r2 + Be
+r2
Apply BC1: Recall a physically acceptable Ψ (or u) m ust → 0 for r→ ∞, r → ∞ so B m ust = 0.
(ii) Consider the case where r → 0
Then the 1r2 term dom inates the
1r and the
14 term s, so the equation is approxim ately:
d 2
dr2 −l(l+1)r2
⎡⎣⎢
⎤⎦⎥u=0
u=Cr−l + Drl+1
solution
solution
28
I. The Hydrogen Atom (continued)II. Facts about the principal quantum number nIII. The wavefunction of an e- in hydrogen
29
Apply BC2: u(r → 0)→ 0, soC=0
(iii) Now insist that the exact solution of the equation have a form which will join up with these asym ptotic cases sm oothly.
Assum e u(all r) = e−r2
⎛⎝⎜
⎞⎠⎟rl+1H(r)
H(r)= airi
i∑
H ≠a ham iltonian herePlug this into the full equation:
d 2
dr2 −l(l+1)r2 +
lr−14
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥e
−r2
⎛⎝⎜
⎞⎠⎟rl+1H(r)=0
ddr
e−r2
⎛⎝⎜
⎞⎠⎟ rl+1 dH
dr+ (l+1)rlH
⎡⎣⎢
⎤⎦⎥−12e
−r2
⎛⎝⎜
⎞⎠⎟rl+1H(r)
⎧⎨⎪⎩⎪
⎫⎬⎪⎭⎪
ddr
e−r2
⎛⎝⎜
⎞⎠⎟rl+1 dH
dr+ e
−r2
⎛⎝⎜
⎞⎠⎟(l+1)rlH −e
−r2
⎛⎝⎜
⎞⎠⎟rl+1 H
2
⎧⎨⎪⎩⎪
⎫⎬⎪⎭⎪
=e−r2
⎛⎝⎜
⎞⎠⎟ rl+1 d
2Hdr2 + (l+1)rl dH
dr⎡⎣⎢
⎤⎦⎥−12e
−r2
⎛⎝⎜
⎞⎠⎟rl+1 dH
dr
+ (l+1)e−r2
⎛⎝⎜
⎞⎠⎟ lrl−1H + rl dH
dr⎡⎣⎢
⎤⎦⎥−12e
−r2
⎛⎝⎜
⎞⎠⎟(l+1)rlH
−12e
−r2
⎛⎝⎜
⎞⎠⎟ (l+1)rlH + rl+1 dH
dr⎡⎣⎢
⎤⎦⎥+14e
−r2
⎛⎝⎜
⎞⎠⎟rl+1H
30
and divide out e−r2
⎛⎝⎜
⎞⎠⎟:
[rl+1 d2Hdr2 + (l+1)rl dH
dr−12rl+1 dH
dr
+ (l+1)lrl−1H + (l+1)rl dHdr
−12(l+1)rlH
− 12(l+1)rlH −
12rl+1 dH
dr+14rl+1H −l(l+1)rl−1H + lrlH −
14rl+1H] = 0
Collect term s:
rl+1 d2Hdr2 + (l+1)rl −
12rl+1 + (l+1)rl −
12rl+1⎡
⎣⎢⎤⎦⎥dHdr
+ (l+1)lrl−1 −12(l+1)rl −
12(l+1)rl +
14rl+1 −l(l+1)rl−1 + lrl −
14rl+1⎡
⎣⎢⎤⎦⎥H =0
Collect term s:
rl+1 d2Hdr2 + 2(l+1)rl −rl+1⎡⎣ ⎤⎦
dHdr
+ −(l+1)rl + lrl⎡⎣ ⎤⎦H =0
Divide through by rl :
rd 2Hdr2 + 2l+ 2 −r[ ]
dHdr
− l+1−l[ ]H =0
*Notice that for a given value of l, this is an eigenvalue equation for H with eigenvalue = l. Recall H here is part of a wavefunction and not a ham iltonian
31
Plug in H= airi∑
dHdr
= iairi−1∑
d 2Hdr2 = i(i−1)air
i−2∑
r i(i−1)airi−2∑ + (2l+ 2 −r) iair
i−1∑ −(l+1−l) airi∑ =0
i(i−1)airi−1∑ + (2l+ 2) iair
i−1∑ − (l+1−l + i)airi∑ =0
i(i−1+ 2l+ 2)airi−1∑ − (l+1−l + i)air
i∑ =0
i(i + 2l+1)airi−1∑ − (l+1−l + i)air
i∑ =0
Set coefficients of each power of r separately = 0:
32
Set coefficients of each power of r separately = 0:
r0 1(2l+2)a1 −(l+1−l + 0)a0 =0
r1 2(2l+2)a2 −(l+1−l +1)a1 =0
r2 3(2l+4)a3 −(l+1−l + 2)a2 =0
r3 4(2l+5)a4 −(l+1−l + 3)a3 =0
r4 5(2l+6)a5 −(l+1−l + 4)a4 =0
rn (n+1)(2l+ n+2)an+1 −(l+1−l + n)an =0
an+1 =l+1+ n( )−l⎡⎣ ⎤⎦
(n +1)(2l+ n + 2)an Recursive realtion for the ai
This relationship between the ai is like the one for an exponential function. To see this com pare:
The "H(r)" series "The er series", er =ri
i, so∑ ai =
1i!
an+1
an
=l+1+ n−l
(n +1)(2l+ n + 2)
an+1
an
=
1(n +1)!
1n!
=n!
(n +1)!=
1n +1
when n→ large when n→ large
≈n
n⋅n:1n :
1n
33
So if H(r) is not truncated, what we have so far is
u(r) = e−r2 rl+1H(r)
= e−r2 rl+1e+r
= e+r2 rl+1 r→ ∞⏐ →⏐ ⏐ ∞
To force u to be a physically acceptable wavefunction, truncate the series HPick som e i whose ai is the highest non-zero ai
call it im ax
Recall ai+1 =l+1+ i−l[ ]
(i +1)(2l+ i + 2)ai
when i=im ax
ai =ai m ax
ai+1 =0
0 = l+1+im ax −l
l=l+1+im ax
Since i and l and 1 are all integers, l m ust be an integer. Renam e it "n"
34
I. Facts about the principal quantum number n (continued)II. The wavefunction of an e- in hydrogenIII. Probability current for an e- in hydrogen
35
Facts about l=n1. "n" is called the Principal Quantum Num ber2. Recall the definition
n=l=Ze2
hm
2 |En |, so
En =−|En | = −Z2e4
h2
m2n2 These are the allowed bound state energies of the Coulom b potential.
3. n = l+im ax +1, so n ≥l+1 (im ax =0,1,...) l≤n-1, or lm ax =n−14. Notice no m atter how large n is, its En will always be slightly < 0, so at lease weakly bound.
So this potential V =−Ze2
r has an infinite num ber of bound energy states (unlike the square well).
5. Energy degeneracies: (i) due to l→ The energy depends only on n, but for each n, there are n possible values of l
l = 0, 1,..., n-1 (ii) due to m → For each l, m can be -l, -l+1, ..., 0, ..., l-1, l (iii) total due to m and l is then
(2l+1)l=0
n−1
∑ =n2 (this ignores spin for now)
num ber m values
36
6. A (2l +1)-fold degeneracy of m-values is characteristic of a spherically symmertric potential.
II. The wavefunction of an e− in hydrogen
Recall we found that
Ψe inhydrogen
tim e-independent =R ⋅Ψlm
to solve the R equation we used Form 2,
so we defined r=8m |E |h2 r=2
2m |E |h2 r=2kr
and u=rR
R=1ru=
1r
e−r2 rl+1H(r)⎡
⎣⎤⎦
airi
i
im ax
∑
=2kr
e−r2 rl+1H(r)⎡
⎣⎤⎦
=e−r2 rl2k air
i
i
im ax
∑ These turn out to be fam ous functions
37
the associated Lguerre Polynomials, LqP (r)
H(r)=-Ln+l2l+1(r)
LqP(x)=
dP
dxP Lq(x)
Lq(x)≡exd q
dxq (xqe−x)
Exam ple of Lq's and LqP's:
L0(x)=1 ′L1(x)=−1L1(x)=1−x ′L2(x)=−4 + 2x
L2(x)=2 −4x + x2 ′L3(x)=2
Sum m ary:
R(r)=Ce−r2 rl -Ln+l
2l+1(r)( )
Norm alize:
1= |Ψ |2∫ dV 0L= |Ψlm |2 dW |R |2 r2 dr∫∫
autom atically=1 because the Ψ's are norm alized
Normalization not determined yet
38
| R(r)|2 r2 dr∫ recall r=2kr, so r2dr=
1(2k)3
r2dr
C2
(2k)3Ln+l2l+1(r)⎡⎣ ⎤⎦∫
2e−rr2lr2dr
2n (n + l)![ ]
3
(n −l−1)!
So C = (2k)3(n −l−1)!2n (n + l)![ ]
3
⎧⎨⎪⎩⎪
⎫⎬⎪⎭⎪
12
Plug in:
R=-(2k)3(n−l−1)!2n (n + l)![ ]
3
⎧⎨⎪⎩⎪
⎫⎬⎪⎭⎪
12
rle−r2 Ln+l
2l+1(r) where r=2kr=8m |E |h2 r
*It is com m on to derive a0 ≡−h2
me2, the Bohr radius.
Then r=8m |E |h2 r can be sim plified:
Recall E=−Z2e4mh22n2 , so |E|=
Z2e4mh22n2
r=8mh2
Z2e4mh22n2 r
39
I. Facts about Ψe in hydrogen
II. Probability current for an e- in hydrogen
Read 2 handoutsRead Chapter 14
40
r =2μ Ze2r
h2n=
2Zra0n
Specific Rnl (r)'s are listed in Goswami Eq. 13.23
In general, Rnl (r) =2Zna0
⎛⎝⎜
⎞⎠⎟
3(n − l −1)!
2n (n + l )![ ]3
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
12
2Zra0n
⎛⎝⎜
⎞⎠⎟
l
e−Zra0 n Ln+l
2l +1 2Zra0n
⎛⎝⎜
⎞⎠⎟
Facts about the Ψ e in hydrogen :
(i) They are totally orthogonal:
Ψ ′n ′l ′m Ψ nl m dVolume = δ ′n n δ ′l l δ ′m m∫
due to the eimϕ in the Ylm
due to the orthogonality of Legendre Polynomials Pl
recall Ylm =(Normalization constant) ⋅eimϕ ⋅Pl
m
The Laguerre Polynomials are orthogonal in n(ii) Recall what a complete set of ′l functions is: it can act as a basis for its space
i.e., any possible wavefunction in that space can be written as a linear combination of the elements of the basis.
41
The comlete set of eigen functions of the hydrogen atom look like:
So the bound states do not form a complete set by themselves
(iii) Notice all the Rnl : r l
So for l >0, Rnl (r =0)=0
No probability of finding the e− at the origin in these states for l=0, R nl(r=0)=constant
So the ground state e− has a spherical probability distribution which includes the
origin. So the ground state e− has finite probability to be found inside the nucleus.
(iv) To calculate the probability of finding the e− at a specific r, calculate
Probability(r)=r2 |R nl |2
This is sim ilar to P(x)=|Ψ(x)|2 . The r2 adjusts to spherical coordinates.
E>0 (scattering) states(we will study these in Chapter 23)
E<0 (bound) states
V =−Zer2
42
When you calculate r2 | R |2 you find
(v) to calculate the probability of finding the e− at a particular q,
calculate Probability(q)=|Ψlm (q,f )|2 sinqdqdf
since the f appears in eimf it will disappear from the probability
etc.
2p
2s3p
3s
1s(n =1,l=0)Th
e n=
1 sh
ell
The
n=2
shel
l
The
n=3
shel
l
43
You find:
Prob(l =0,m=0) = s-orbital
Prob(l =1) = p-orbital
Prob(l =2) = d-orbital
etc.
(vi) Recall Parity Rs the operation that inverts the Ψ through the origin
So RΨ(r,q,f )=R(r)(-1)lΨlm (p −q,p +f )
unchanged by R
So R(Ψe in hydrogen )=(−1)lΨ
44
Review SyllabusRead Goswami section 13.3 and chapter 14 plus the preceding chapter as necessary for reference
45
Read Chapter 14
I. Probability current for an e- in hydrogenII. The effect of an EM field on the eigen functions and eigen values of a charged particleIII. The Hamiltonian for the combined system of a charged particle in a general EM field
46
I. Probability current for an e− in Hydrogen
Recall the definition of probability current:vJ=
h2mi
Ψ*v∇Ψ−Ψ
v∇Ψ*( )
Recall this describes the spatial flow of probability→ NOT necessarily the m otion of the point of m axim um probability→ definitely not the m otion of a particle whose probability of location is related to Ψ
Calculate J for the e− in Hydrogen:
Plug in Ψnlm =Rnl(r)⋅Ψlm (q,f )⋅e
−iEnth
call this Q(q)⋅eimf , where Q(q) is pure real
and v∇spherical
coordinates=r
∂∂r
+q1r∂∂q
+ f1
rsinq∂∂f
you get:
J=h2mi
R*q*e−imf eiEt
h( )∇ Rqeimfe−iEt
h( )− Rqeimf e−iEt
h( )∇ R*q*e−imf eiEt
h( )⎡⎣⎢
⎤⎦⎥
Notice since R * =R
and q* = q,
that R*q*∇Rq −Rq∇R*q* =0so consider only the f part of the equation
47
So vJ=f
h2mi
R*q*e−imf eiEt
h( )1
rsinq∂∂f
Rqeimf e−iEt
h( )− Rqeimfe−iEt
h( )1
rsinq∂∂f
R*q*e−imf eiEt
h( )⎡⎣⎢
⎤⎦⎥
=fh2mi
R 2q 2 e−imf1
rsinq∂∂f
eimf −eimf1
rsinq∂∂f
e−imf⎛⎝⎜
⎞⎠⎟
=fhR 2q 2
2mie−imf
1rsinq
imeimf −eimf1
rsinq(−im )e−imf⎛
⎝⎜⎞⎠⎟
=fhR 2q 2mmrsinq
=f|Ψ |2 mhmrsinq
Note this current is(i) tim e-independent(ii) circulating around the z-axis (not f ) but rem aining sym m etric about it
(iii) NOT the sam e as an orbiting e−
This circulating vJ is related to the m agnetic dipole m om ent of the e−
To show this recall from EM:
48
If you have a physical charged current density vJe =
vIA
Consider a differentially sm all piece of it which is located at a distance vr from the origin.This piece form s an elem ent of a current loop which is at least m om entarily circulating relative to the origin.This is physically identical to a m agnetic dipole whose m agnetic dipole m om ent vm is given by:
vm ≡12
vr×volum ewhere Jecirculates
∫vJe(
vr)dV olume
we can convert our probability current vJ into a physical charged current
vJe by m ultiplying by the charge:
vJe =
−ec
⎛⎝⎜
⎞⎠⎟vJ
Then the m agnetic dipole m om ent of the e− is given by
vm ≡12
vr×volum e∫
−ec
⎛⎝⎜
⎞⎠⎟vJdV olume
Notice since vJ=|J|f , vm m ust be |vm |z
So we only need vr×vJ( )z =rJsinq
A I →
vr
O
vm
vrR × J
J
q
49
So we want
m=m z =−e2c
rJsinq dV olume∫
=−e2c
r|Ψ |2 mhmrsinq
sinq dV olume∫
=−emh2mc
|Ψ |2 dV olume∫ by norm alization
so m z =−eh2mc
⋅m
Since m h is the eigen value of Lz, m m ust be the eigen value of som e operator "−e2mc
Lz"
call this the z com ponent m z m agnetic m om ent operator M
Then M=−e2mc
L
angular momentum operator
magnetic momentum operator
quantum number “m”
This factor is called the Bohr magneton
Magnitude of the z-component of magnetic dipole moment of the e-
50
I. The effect of an EM field on the eigen functions and eigen values of a charged particleII. The Hamiltonian for the combined system of a charged particle in a general EM fieldIII. The Hamiltonian for a charged particle in a uniform, static B field
Read Chapter 15
51
The effect of and EM field on the eigen functions and eigen values of a charged particle
General plan:(i) We know the Hamiltonian of a free particle of momentum, vp:
H=p2
2m(ii) Find the H for that sam e particle with charge q in an EM field (
vE,
vB)
we will find that H :p2
2m+ [stuff] ⋅B+[stuf ′f]B2
we will study each kind separately
II. The Ham iltonian for the com bined system of a charged particle in a general EM fieldProcede to find H:
(i) Recall that classically, H ≡ pi&xi −i=1
3
∑ L
where the xi = canonical coordinatespi = canonical m om enta
⎧⎨⎪⎩⎪
(ii) Find L. Recall that equations of m otion m ust be obtainable from it via
∂L∂xi
−ddt
∂L∂&xi
⎛⎝⎜
⎞⎠⎟ =0
52
I. The Hamiltonian for a charged particle in an EM fieldII. The Hamiltonian for a charged particle in a uniform, static B fieldIII. The normal Zeeman effect
53
(iii) Plug in L to get H, then convert everything possible to operatorsCarry this out:To find L, recall that usuallyL=T-V
Here T=12mv2 (use m for m ass everywhere)
What is v? Recall the EM field has 2 kinds of potential:
scalar potential f and vector potential vA
How to com bine them ? (we can't just say "V =f +vA")
It turns out the L=12mv2 −qf +
qcvA⋅
vV
To dem onstrate this, we will show that
∂L∂xi
−ddt
∂L∂&xi
⎛⎝⎜
⎞⎠⎟ =0 (Lagrange's equation)
successfully produces the known Lorentz Force Law vF=q
vE+
qvvc
×vB
To plug into this, we need:∂L∂xi
=−q∂f∂xi
+qcvvg∂A∂xi
To find ∂L∂xi
, notice we can expand:
54
L =12m &xi
2∑ −qf +qc
Ai&xi∑
So ∂L∂&xi
=m&xi +qcAi [Note this is the definition of the canonical m om entum pi ]
Then ddt
∂L∂&xi
⎛⎝⎜
⎞⎠⎟ =m&&xi +
qcddt
Ai
∂A∂t
+∂xj
∂t∂A∂xjj
∑
∂A∂t
+ vv⋅∇vAi
Plug all this into Lagrange's Equation:
-q∂f∂xi
+ qvvc∂vA
∂xi=m&&xi +
qc
∂Ai
∂t+ vv⋅∇
vAi
⎛⎝⎜
⎞⎠⎟
Reorder:
m&&xi =-q∂f∂xi
−qc∂Ai
∂t+qvvc
∂vA
∂xi−qcvv⋅∇
vAi
This equation concerns com ponent i (i=1, 2, or 3) of a vector equation. Generalize to the full vector equation.
mva(=vF)=q −∇f −
1c∂A∂t
⎛⎝⎜
⎞⎠⎟ +
qc
v∇(vv⋅
vA)−(vv⋅∇)
vA⎡⎣ ⎤⎦
To understand this recall: vE
55
So qc
∇(vv⋅A)- (v⋅∇)A⎡⎣ ⎤⎦ u× (y× z)=y(u⋅z)−z(u⋅y)Plug in v ∇
vA
=∇(vv⋅A)- (v⋅∇)A
qc
v× (∇×A)⎡⎣ ⎤⎦
So we have vF=q
vE+
qcvv×
vB
So we know we have the right LagrangianNow find H= pi&xi −L∑= m&xi +
qcAi
⎛⎝⎜
⎞⎠⎟∑ &xi −
12m &x2
i −qf +qc
Ai&xi∑∑⎛⎝⎜
⎞⎠⎟
=m2∑ &x2
i + qf
=m&xi( )
2
2m∑i
+ qf
Now use again the definition of canonical m om enta
pi =m&xi +qcAi
m&xi( )2= pi −
qcAi
⎛⎝⎜
⎞⎠⎟∑2
∑
vp-qc
vA
⎛⎝⎜
⎞⎠⎟2
canonical momentum
56
H=
vp −qvAc
⎛⎝⎜
⎞⎠⎟2
2m+ qf
By convention, if the particle is: then we use:
an e− (negative) q=-e
e+ (positive) q=+e
II. The Ham iltonian for a charged particle in a uniform static vB field
Recall general H=12m
vp−qvAc
⎛⎝⎜
⎞⎠⎟2
+ qf
=12m
vp2 −qc
vA⋅vp+ vp⋅
vA( )+
q2
c2vA2⎛
⎝⎜⎞⎠⎟+ qf
we want to sim plify thisRecall:
[f(x),p]=ih∂f∂x
(see next page)
Generalize to 3D: [vf(vr),p]=ih
v∇⋅
vf
Plug in A, then:[A,p]= ih
v∇⋅
vA
now recall that for a static vB field,
v∇⋅
vA=0
Show this:
Recall in general B can be produced by (i) a uniform current and (ii) a ∂E∂t
.
Consider the static case, so A=f(I) only
57
Recall if a current vI flows along a segment d ′l of path C, then the vector potential
vA at distance R from C is
vA(x)=
m0
4pId ′lR
this is perm iability, not m ass
vAdue to
total C
(x)=m0
4pId ′vlRC
—∫
v∇⋅
vA=
v∇
m0
4pId ′vlRC
—∫⎡⎣⎢
⎤⎦⎥=
m0I4p
v∇d ′vlRC
—∫ Recall ∇⋅u
vV( )=
vV ⋅∇u+u
v∇⋅
vV( )
1R d ′vl
d ′vl ⋅
v∇
1R
⎛⎝⎜
⎞⎠⎟ +
1R
v∇d ′
vl
0 since ∇ operating on unprim ed coordinates only, and d ′vl is constructed of prim ed coordinates
v∇
1R
⎛⎝⎜
⎞⎠⎟ =
v∇
1
(xi′∑ −xi)
⎛
⎝⎜⎜
⎞
⎠⎟⎟ =−
v∇
1R
⎛⎝⎜
⎞⎠⎟
Rewrite:v∇⋅
vA=
−m0I4p
v′∇
1R
⎛⎝⎜
⎞⎠⎟d ′
vl
C—∫
Recall Stoke's Theorem : for any vector vv,
vvdvl
C—∫ =
v∇× vv( )dArea
surface enclosedby C
∫
vx
O ′vx
R =v′x −vx d ′l vA
C ∇=
∂∂x
etc. not ∂∂ ′x
58
I. The Hamiltonian for an e− in a uniform , static B field (continued)II. The norm al Zeem an effect
III. Response to the e− to the B2 term
IV . Sum m ary of e− response to static uniform BzV . The discovery of spin
59
v∇⋅
vA=
−m0 I4p
′∇ × ′∇1R
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
⎡⎣⎢⎢
⎤⎦⎥⎥⋅dArea
S∫
but the curl of a divergence always = 0So,
v∇⋅
vA=0 for static B
Return to A, p⎡⎣ ⎤⎦=ihv∇⋅
vA
0
So vA and vp com m ute.
So (vA⋅vp+vp⋅
vA)=2
vA⋅vp
Then He in static B =12m
p2 −qc2vA⋅vp+
q2
c2A2⎡
⎣⎢
⎤⎦⎥+ qf
Now further restrict this static (no tim e dependence) B field to be also uniform:call it
vB=|B|z
constant, 1 directional, no position dependenceIn general for any B,vB=
v∇×
vA
Bxx + Byy+ Bzz =∂Az
∂y−∂Ay
∂z
⎛⎝⎜
⎞⎠⎟x + +
∂Ax
∂z−∂Az
∂x⎛⎝⎜
⎞⎠⎟y+
∂Ay
∂x−∂Ax
∂y
⎛⎝⎜
⎞⎠⎟z
When vB=|B|z, this reduces to:
0 + 0 + Bz =0+0+∂Ay
∂x−∂Ax
∂y
⎛⎝⎜
⎞⎠⎟z
60
There is more than 1 solution to this. One is:
Ax =−12B⋅y
Ay =12B⋅x
Ax =0
⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
This can be written precisely as:vA=
12
vB× vr
Plug this into vA⋅vp:
12
vB× vr( )⋅
vp
Recall the vector identity: vu⋅(vy⋅vz)=(vu⋅vy)⋅vz12
vB⋅ vr× vp( )
vL= angular m om entum
Now we have:
He in static B =12m
p2 −qc212
vB⋅
vL+
q2
c2A2⎡
⎣⎢
⎤⎦⎥+ qf
Plug in A2 =12
vB× vr( )
⎡⎣⎢
⎤⎦⎥2
61
Recall vB × vr=Brsinq and
vB⋅vr=Brcosq
SovB× vr( )
2=B2r2sin2q =B2r2(1−cos2q)=B2r2 −
vB⋅vr( )
2
So
A2 =14
B2r2 −vB⋅vr( )
2⎡⎣⎢
⎤⎦⎥
Plug this into H
He in uniform static B =p2
2m−
q2mc
vB⋅
vL+
q2
8mc2B2r2 −
vB⋅vr( )
2
( )+ qf
note qvL
2mc is the m agnetic m om ent
vM of the particle
Since we choose the coordinate system so that vB=Bz,
vB⋅
vL=BLz
and r2B2 − vr⋅vB( )
2=(x2 + y2 + z2 )B2 −(x2Bx
2 + y2By2 + z2Bz
2)=(x2 + y2 )B2
Then
He in uniform static B =p2
2m−
q2mc
BLz +q2
8mc2B2(x2 + y2 )( )+ qf
For an electron, q=-e, so
He in uniform static B =p2
2m−eBLz
2mc+e2B2
8mc2(x2 + y2 )−ef
"m" is the reduced m ass of the system *If you want the answer in m ks units, set c=1
62
I. The normal Zeeman effectII. Response to the e− to the B2 term
III. Sum m ary of e− response to static uniform BzIV . The discovery of spin
63
We will study the effect of each term separately upon the e's wavefunction and energy.
II. The Normal Zeeman Effect
Recall He in uniform static B in z =p2
2m+eLzB2mc
+e2B2(x2 + y2)
8mc2−ef
= "H0"+"H1" + "H2" -ef
com pare relative sizes of H1 and H2
H2
H1
=
e2B2(x2 + y2)8mc2
eLzB2mc
=eB(x2 + y2)
4cLz
Plug in e = 1.6x10−19 C
B = 1 tesla = 104 gauss
(x2 + y2)~(5x10−11)2 m 2 (Bohr radius)2
c = 1 (unitless) to convert Gaussian→ MKS units
Lz ~mh~h = 1x10−34 Joule-seconds
Then,H2
H1
~(1.6x10−19)(1 tesla)(5x10−11)2m 2
4(1)(1x10−34J−sec)=10−6
So H2 = H1 for B<105 −106T
m agnetic field at earth's surface is ~0.5x10−4 T superconducting m agnets ~ 10 TSo, in a norm al situation:
64
He in uniform static B in z ≈p2
2m+eLzB2mc
−ef
Suppose f=zer, so
-ef=−ze2
rThen the e is in the Coulom b Field of its own nucleus
This gives it En =−mz2e4
2h2n2
plus the extra vB field
Recall each energy level "n" is degenerate, all of its l and m levels have the sam e energyHow does this H1 affect the e's energy levels?
H1Ψnlm =eLzB2mc
Ψnlm =eB2mc
⎛⎝⎜
⎞⎠⎟LzΨnlm
m hΨnlm quantum num ber "m" NOT m ass
call this collection of constants "wL",
the Larm or frequency
65
So the presence of H1 means that the different "m" levels are no longer degenerate;
each has its own energy given by:
Em =−mz2e4
2h2n2+w Lmh
So without B with B
−mz2e4
2h2n2+w Lh m=+1
−mz2e4
2h2n2 m=0
E of all levels −mz2e4
2h2n2−wLh m=-1
with sam e n m=-2 etc.
The fact that a m agnetic field can cause the levels designated by "m" change energy causes "m" to becalled "The m agnetic quantum num ber"
II. Response of the e- to the ~B2 term
Recall H = p2
2m+eBLz
2mc+e2B2(x2 + y2)
8mc2−ef
=pz2
2m+eBLz
2mc+px2 + py
2
2m+e2B2(x2 + y2 )
8mc2−ef
ignore for now (let→ 0)this is identical to the Harm onic Oscillator:
H2DHo
=px2
2m+
py2
2m+12k1x
2 +12k2y
2 where k1 =k2 =e2B2
4mc2
E
66
From Chapter 9 (2-D systems)H2 D
HOΨ =(nx + ny+1)hwΨ
w=km
=eB
2c m
1
m=
eB2cm
=wLarm or
III. Sum m ary of e response to static uniform BzSo far we found that
H = pz2
2m+eBLz
2mc+px2 + py
2
2m+e2B2(x2 + y2)
8mc2−ef
Recall: a particular Ψ is sim ultaneously an e function of 2 operators vΨ and
vZ only if [
vΨ,Z]=0.
Notice [pz,Lz] =0 since Lz =xpy−ypx
[pz,H2DHO
] =0
[Lz,H2DHO
] =0
So all the term s of H have the sam e e function
call it "Ψnmk"
Plug it in:
Set = 0 for nowH2D, HO
67
I. The Discovery of SpinII. Filtering particles with a Stern-Gerlach apparatusIII. Experiments with filtered atoms
68
HΨnmk =pz2
2mΨnmk +
eBLz
2mcΨnmk + H2D
HOΨnmk
Recall: pzΨ =hkΨ
LzΨ =mhΨ
H2DHOΨ =(nx + ny+1)hw LΨ
HΨnmk =h2k2
2m+ mhwL + (nx + ny+1)hw L
⎡⎣⎢
⎤⎦⎥Ψnmk
But in general HΨ =EΨ, so this m ust be "E"
Enmk =h2k2
2m+ (m + nx + ny+1)hw L
III. The discovery of spin
Suppose you wanted to m easure the total angular m om entum of a particle
call it vJ as in Ch. 11 (Note: this J is not a current)
W e showed in Chapter 13 that angular m om entum ∝ m agnetic m om ent
−e
vL
2mc =
vM
Now call this "vJ" to be general, i.e. to allow for m ore than just orbital angular m om entum
69
So vJ=
−2mcvM
e
So we want to design an apparatus to m easure vM
Recall from E&M that if a m agnetic dipole vM is in a m agnetic field
vB it feels a force on it
which depends on the relative orientation of vM and
vB:
vF=
v∇(
vM⋅
vB) (stored energy e=
vM⋅
vB then
vF=
v∇e)
Expand vM=Mxx + Myy+ Mzz
Then vF=
v∇ MxBx +M yBy+MzBz
⎡⎣ ⎤⎦
Design and apparatus in which B=Bz is in the z direction only
Then vF=
v∇ MzBz⎡⎣ ⎤⎦
Since M is a fundam ental property of a particle, it has no dependence on z
To get v∇ MzBz
⎡⎣ ⎤⎦≠0, m ust have Bz = f(z)
Then F=Mz
∂B∂z
Now if a particle with m om ent Mz is in the apparatus, it will feel a force ∝ Mz.
If the particle is m oving through the apparatus, the force will deflect its path from a straight lineand its deflection will m easure Mz
deflection ∝ Mz z
F ∝ Mz
particle
apparatus
70
An apparatus like this is called a Stern-Gerlach ExperimentNotice that if you pass 1 particle through the SG, you find out its specific M z .
If you accumulate a large number of identical particles, you can find out whatare all the possible M z values that they can have.
Recall m can take only quantized values. But are there any restrictions on M z ?
∝ the apparent orientation of the object Mz =m cosq
particle's m agnetic dipole
Stern and Gerlach m ad a beam of silver atom s.For each atom g the nucleus has
vM ≈0
g all e− but the outerm ost one are paired, so their cum ulative vM ≈0
g the outerm ost e was in the s-suborbital of its shell
l=0, m=0
g the atom s were cooled so that it was unlikely the outerm ost e− could acquire
enough therm al energy to m ove a higher -l
higher m
⎧⎨⎪⎩⎪
⎫⎬⎪⎭⎪ sub-orbital
When the atom s passed through the SG, they all deflected, but each ended up in one of only 2 possible spots:
z q M z
M
71
I. The Discovery of SpinII. Filtering particles with a Stern-Gerlach apparatusIII. Experiments with filtered atoms
Read handout from Feynman lectures and Chapter 15
72
spot 1
spot 2
smear that didn't appear
⎛⎝⎜
⎞⎠⎟
What this meant:
1. the vM cannot have arbitrary orientation: otherwise instead of 2 spots there would have been a continuous
smear reflecting that all possible M z states were present
2. each outer e− had a non-zero Mz which was not related to its orbital angular m om entum
that had been arranged to have l=m=0 call this "non-orbital angular m om entum " = spin. Its quantum num ber is s and its "orbital angular m om entum " m = m s
3. Recall for orbital angular m om entum , the possible values m can take are -l, l+1, ..., 0, ..., l−1, l so the num ber of possible values of m is (2l+1)4. here, experim entally it was found that the num ber of possible m s values is 2
so (2s+1) = 2
s = 12
*Conclusion: Every particle has, in addition to orbital angular m om entum , another property which is m athem atically like angular m om entum but which is not due to any kind of rotation. This new kind of angular m om entum is called spin.
73
Recall that regular angular momentum is quantized in the direction its allowed to have, so that
Lz = (integer m )*h
the possible num ber of orientations is 2l+1
m sz is also quantized, but since s for an e− is always 1
2,
the num ber of possible m s orientations is always only 2:
+ 12 and - 1
2
I. Filtering particles with a Stern-Gerlach apparatus
Recall if you have a collection of atom s with different m z values,
if you pass them through a Stern-Gerlach device, they will separate into different beam s, one beam for each value.
Notice if you obstruct all but 1 output path, you can produce a beam that ispurely com posed of particles with 1 definite m z value.
z 2h
1h
0h
−1h etc.
quantum # m
if their l =0 this isdue to their spin z
mz =+1
mz =0
mz =−1
etc.
particles with allpossible m z
74
pure mz =+1
Example:
pure mz =+1
When a beam is put into a definite state like this, it (the output of the SG) is called a "prepared beam"or a "filtered beam" or a "polarized beam"
Now make a slightly modified SG that can return the polarized beam to the original axis of travel
modified SG device
this barrier is moveable so wecould select any m z value
75
Make up a symbol for the modified SG device:
+1 0 −1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ this shows what is blocked
S this gives a particular device a nam e in case m ore than 1 is in series
Make up sym bols for the prepared states:
what com e out of S|+ >|0 >|- >Now im agine placing several SG's in series.Exam ple:
beam → +1 0 −1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 −1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
S ′S
If this S is the initial state we are forcing the particles to have (labelled |+ >, etc.) then this ′Sis the final state we are checking to see IF they have label final states with bras.This one is < -1| Other possibilities for this system are < 0 | or < +1|
*Note S can represent both the device thatprepares a particle's state, or the state itself
76
I. Filtering particles with a SG (continued)II. Experiments with filtered atomsIII. SG in seriesIV. Basis states and interference
77
Examples of some diferent possible results of putting 2 SG's in series:
Configuration: Result exiting ′S A sym bolic way to represent this:
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ all the pure m z =+1 exit final
stateinitialstate
=fraction S that pass S
S ′S +1 +1 =1
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ nothing exits −1 +1 =1
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ all the pure m z =−1 exit −1−1 =1
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ nothing exits 0 −1 =1
*Notice we draw the S, ′S in the order in which the beam reaches them , but we order final initial from right to left.
W e could sum m arize all possibilities in a m atrix as we have done before:
initial state:
final state +1 0 −1
+1 1 0 0
0 0 1 0
−1 0 0 1
beam →
beam →
beam →
beam →
78
All these examples have
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ S ′S prepares 1 "analyzes" state state
Suppose S could prepare several states with definite fractions, so
initial =a + + b 0 + c −
Then the am plitude for having a particular final state exit would befinalinitial =a final+ + b final0 + c final−
= a if final = +1
= b if final = 0
= c if final = −1
Then the probability of observing a particular final state is
finalinitial2= a
2, b
2, or c
2
We always assum e that finalinitial2
final∑ =1
a2+ b
2+ c
2=1
(This is norm alization.)
particles with unknown states
79
II. Experiments with filtered atoms*The purpose of these examples is to show you how different basis sets could actually be realized in nature.*Suppose we put 2 SG filters in series, but one is tilted with respect to the other:
+S ≠ +T , so +T +S ≠1
However, since +S is not orthogonal to +T , it is also not true that +T +S =0
It turns out that +T +S = som e am plitude "a" where 0 ≤a≤1 and a=f(a)
There are also specific am plitudes for all of the following possibilities:
+T +S +T 0S +T −S
0T +S 0T 0S 0T −S
−T +S −T 0S −T −S
Note: for norm alization, the square of the 1 in each colum n m ust sum to 1.
*Keep in m ind that the m atrix of possibilities does not have to be 3x3. It is in general nxn, where n is the num ber of states the beam particles can have.*
beam →
S call this "T"
a
x
y
z
80
III. SG filters in series
The message of this section is,once a particle goes through a filter it loses all information about the orientation of previous filters it passes through.That is not the same as saying, "each filter analyzes, or measures the state of the particle, and the measurementprocess places the particle in and eigenstate of that measurement"
aligned along one of its basis states of that SG filter
To see this, consider 3 consecutive SG filters:
Suppose the not only have relative angles, but also have their blocking pads in different places:
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
Notice S and ′S represent the sam e basis (which has 3 states), and T is a different one (which also has 3 states).
beam →
S T
a
′S
81
I. SG filters in series (continued)II. Basis states and interferenceIII. Describing a measurement matrix
82
You might guess that a particle got to here, S T S
it would get to here
with 100% probability, because it would "remember" that is had been +S earlier. It does NOT.
The T filter places it into a 0T state, which does not have 100% overlap with a +S .
Dem onstrate that the fraction of particles that pass through T and ′S depends only on T and ′S (not S)
Com pare:
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ S T ′S S T ′S
Am plitude to exit ′S is:
+ ′S 0T 0T 0S 0 ′S 0T 0T 0S
Ratio of am plitudes is:
LHSRHS
=+ ′S 0T 0T 0S
0 ′S 0T 0T 0S
independent of state of S.
83
I. Basis states and interferenceII. Describing a measurement with a matrixIII. Sequential measurements
Read Goswami 11.2
84
So the presence of S affects the absolute number of particles that get to T (and then have the chance to reach ′S ),but once they are at T, having passed through S does not affect their chance to pass through ′S .
IV . Basis states and interference
Consider several experim ents:
Experim ent 1: +1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
N particles survive⏐ →⏐ ⏐ ⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ aN⏐ →⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ b∝N⏐ →⏐ ⏐
S T ′S
Experim ent 2: +1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N ⏐ →⏐ ⏐ ⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ aN⏐ →⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ g∝N⏐ →⏐ ⏐
S T ′S
Experim ent 3:
repeat Exp. 1 but
rem ove blocks from T
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N ⏐ →⏐ ⏐ ⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N⏐ →⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N⏐ →⏐
S T ′S
Experim ent 4:
repeat Exp. 2 but
rem ove blocks from T
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N ⏐ →⏐ ⏐ ⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ N⏐ →⏐
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ 0⏐ →⏐
S T ′S
b = + ′S 0T
2
g = 0 ′S 0T
2
a = 0T +S
2
85
Conclusions:g Experiment 2 produces more final state particles than Experiment 4:Inserting blocks in T must eliminate destructive interference (or produce constructive interference) in this case.g Experiment 1 produces less than Experiment 3:Inserting blocks in Tmust produce destructive interference in this case.
This interference of amplitudes is similar to what happens in a double slit experiment with light:
light⏐ →⏐ ⏐ light⏐ →⏐ ⏐
2 slits 1 slit(like "no blocks" in T) (loss of one slit like adding a block in T)gives m inim um gives m axim um
output at 20o output at 20o
20o 20o
86
Write the amplitudes: Condense the rotation:Experiment 4: 0 ′S +T +T +S 0 ′S +T +T +S
all T∑ =0
+ 0 ′S 0T 0T +S
+ 0 ′S −T −T +S
0
Experim ent 3: + ′S +T +T +S + ′S +T +T +Sall T∑ =1
+ + ′S 0T 0T +S
+ + ′S −T −T +S
1
Facts about all of this:1. Experim ent 3 would have the sam e result ifT is present but all open or T is not present at all
0 ′S T T +ST∑ = 0 ′S +S
⇒ So T TT∑ =1 If T includes all possible im term ediate states
it is a basis, a com plete set
87
2. Experiment 3 would have the same result if T were replaced by some other filter "R" tipped atan angle other than a, as long as R were also unblocked.Now
⇒ So a choice of basis is not unique.
3. All of this only works if the states within a basis are orthogonal, for exam ple Ti Tj =d ij
To see this, go back to
0 ′S T T +ST∑ = 0S +S
Note that ′S and S are really the sam e basis, so delete the prim e
0S T T +ST∑ = 0S +S
Renam e +S = j generic
0S = c generic
T T = i i any basis
beam →
S T
a
′S
beam →
S R
d
′S
88
c i i ji∑ = c j
Now since j is generic, it could be a m em ber of the basis set, j
c i i ji∑ = c j
This can only be true if i j =d ij
4. Revising the order of a process (i.e. exchanging the initial and final states) is the sam e as takingthe com plex conjugate of its am plitude.Show this:2 colum ns:Colum n 1: If a particle starts in som e state, it m ust end up in one of the possible final states (i.e. it cannot get lost). So, for exam ple:
+T +S2+ 0T +S
2+ −T +S
2=1
Expand:
+T +S*+T +S + 0T +S
*0T +S + −T +S
*−T +S =1 "Equation A"
Colum n 2:
Now also recall that if a state (say +S ) is norm alized, for exam ple:
+S +S =1
we can insert T TT∑ =1
89
I. Describing a measurement with a matrixII. Sequential measurementsIII. Relating matrix notation and Dirac notationIV. Spinors
90
Example:
Suppose:+S = an electron
A = and interaction or m easurem ent
+R = and up quark
Finding +R A+S tells us the probability that this interaction
converts e → w, which is fundam ental inform ation about the interaction.
91
+S T T +ST∑ =1
+S +T +T +S + +S 0T 0T +S + +S −T −T +S =1 "Equation B"
Com pare Equation A and Equation BBoth have RHS=1, so their LHS's m ust be equal.This can only be true if
Si Tj = Tj Si
*
V . Describing a m easurem ent by a m atrix
Com m on question in physics:
g A system begins in som e initial state, say +S (The full set of possible states is "S")
g Som ething happens to it (a m easurem ent--call it "A" or and interaction force)
g What is the probability that it will end up in any particular final state, say +R (the full set of possible final states is "R")
So we want +R A+S . (see previous page)
How to calculate this?
Here +S and +R are bras and kets in Hilbert space, it is hard to calculate with them without choosing a basis.
But what basis is best for them ? What if calculating with each are easier if they are in different basis?
(This could happen, for exam ple, if +S is a state with rectangular sym m etry and +R is a state with spherical sym m etry.)
92
How to handle this:
We have:
+R A+S = +1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
A
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ S R
Inserting an unblocked T
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ anywhere has no effect:
So
+R A+S = +1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
A
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+1 0 -1
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ S T T R
Make a table:The am plitude for going from +S→ T T→ A→ T T→ +Ris given by
Ti +S Tj ATii∑ +R Tj
j∑
93
I. Describing a measurement matrix (continued)II. Sequential measurementsIII. Relating matrix notation to Dirac notation
94
Reordering:
+R A+S = +R Tj Tj ATi Ti +Sij∑
Renam e +S = j generic state
+R = c generic state
Ti , Tj → i , j m em bers of any basis
Then
c Aj = c j jA i i ji, j∑
What this m eans:
Suppose that j and c can be written in term s of bases with 3 basis states.
Then i=(1,2,3) and j=(1,2,3)
So there are only 9 possible am plitudes jA i
For exam ple: i→ + 0 -j↓
+ + A+ + A0 + A−
0 0 A+ 0 A0 0 A−
- − A+ − A0 − A−
(Notice order the colum ns and rows in descending order of the eigenvalues of the quantum num ber involved.)
95
And there are only 3 amplitudes i j
And there are only 3 am plitudes c j
So a total of 9+3+3=15 pieces of inform ation are required.
Once they are plugged into the sum of the RHS, you get the LHS, which is a very general peice of
inform ation: "How does the m easurem ent A relate the states j and c ?"
operator
V I. Sequential m easurem entsSuppose "m easurem ent A" really involves "first m easure B, then C"
Exam ple: to find out the m ass of a fundam ental particles you could m easure first its vp, then its vv, then
calculate m=pv
Get p from tracking the curvature of its path in a vB field:
curvature k ∝Bp
Then get vv by putting it through a "speed trap": m easure its tim es t1 and t2 crossing 2 points
separated by length l, then com pute v=l
t2 −t1
So procedurally the m easurem ent would be:
{ } → A{ } → { } = { } → B{ } → C{ } → { }
j c j c
96
{ } A{ } { } = { } B{ } { } C{ } { }
j c j T c
Sym bolically:
c Aj = c C Tj Tj B jTj
∑ notice generator order is right-to-left
since Tj unblocked
each of these is a m atrix in which the ket=initial state labels colum ns bra= final state, labels rows
The sum over the Tj Tj represents the norm al procedure for m atrix m ultiplication.
Show this:In norm al m atrix m ultiplication the (c,j)th elem ent of m atrix A is the sum of the elem ent-by-elem entproduct of the m atrix elem ents in the c-th row of B and the j-th row of C:
Exam ple: for c=2, j=3
A =B⋅C
97
(1,1) (1,2) (1,3) ... (1,5)(2,1) (2,2) (2,3) ... (2,5)
MM
(5,1) (5,2) (5,3) ... (5,5)
⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
=(2,1) (2,2) (2,3) (2,4) (2,5)
⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
g
(1,3)
(2,3)
(3,3)
(3,4)
(3,5)
⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
m atrix A m atrix B m atrix CA23 =B21C13 + B22C23 + B23C33 + B24C43 + B25C53 = B2Tj
CTj3Tj
∑
I. Relating m atrix notation and Dirac notationRecall we have
initial state = i
finale state = f
operator = AThen
f A i m eans:
g som ething ("A") is close to state i
g that event changes the state to som ething else (call is state Ai )
g we want to know how Ai com pares to the state f
g the overlap of them is given by f Ai
c
f A i
Another way to say this is "what is the probablility that Ai is identical
to f ?"
98
Now recall that since the elements a of a basis are com plete,
a aa∑ =1
This is true for any basis so also true for the basis b
b bb∑ =1
This allows is to write
f A i = f b b Aa a ib∑
a∑
Exam ple: Suppose that i and f can each take on 3 values. (Ex: +1, 0, -1)
Then
g f A i is a 3x3 m atrix
g if the a 's span the space in which i exists there need be only 3 values of a
g if the b 's span the space in which f exists there need be only 3 values of b
Recall that these are matrices
Recall that this symbol means "Hilbert space state i
projected into the a basis"
Recall f b = b f*
so this is the com plex conjugate
of Hilbert space state f
projected into the b basis.
99
I. SpinorsII. The matrices and eigenspinors of Sx and Sy
100
So the equation looks likef A i = f b
a ,b∑ b Aa a i
Norm al rules of m atrix m ultiplication state that the vector to the right of the m atrixm ust be a column vectorand the vector to the left of the m atrix m ust be a row vector
So translating from Dirac notation to m atrix notation gives:
⎛
⎝⎜⎜
⎞
⎠⎟⎟ = ( )
⎛
⎝⎜⎜
⎞
⎠⎟⎟
⎛
⎝⎜⎜
⎞
⎠⎟⎟
3x3
3-component vector
3x3 3-component vector
f A i
f b
b Aa
a i
a i
f b