H ψ=Eψ→Ψ=ψe−iEtℏ H=−ℏ2

2m∂2

∂ x2+V ( x ) p=− iℏ ∂

∂xσ x σ p≥

ℏ2

Dot product for overlapping a wavefunction with a state x∗Ψ=⟨ x|Ψ ⟩The ⟨ x|Ψ ⟩ term IS the wavefunction Ψ ( x )Probability of finding the particle between α and β at time t ∫

α

β

|Ψ (x , t )|2dx

Standard Deviation: σ=√ ⟨ j2 ⟩−⟨ j ⟩2

ρ(x ) is the probability densityProbability that x lies between α and β is:

Pαβ=∫α

β

ρ ( x )dx

Normalized probability to equal 1:∫−∞

∞

ρ ( x )dx=1

Average value of x:

⟨ x ⟩=∫−∞

∞

xρ (x )dx

Average value for f ( x ):

⟨ f ( x ) ⟩=∫−∞

∞

f ( x ) ρ ( x )dx

Momentum:

⟨ p ⟩=− iℏ∫(Ψ ¿ ∂Ψ∂x )dx

PProb ( x=8 )=|∫−∞

∞

(∂Dirac (x−8 ) )¿ ∙Ψ ( x )|2

=|∫−∞

∞

(∂Dirac ( x−8 ) ) ∙Ψ ( x )|2

=|Ψ (8 )|2

⟨ x ⟩=∫Ψ ¿ ( x ) XOPΨ ( x )dx

Kinetic energy:

T= p2

2m

⟨T ⟩=−ℏ2

2m∫Ψ ¿ ∂

2Ψ∂ x2

dx

Momentum, wavelength. DeBroglie formula:

p=hλ=2 πℏ

λ⟨H ⟩=E

Ψ ( x ,t )=∑ cnψn (x ) e−i E ntℏ

f ( x )=∑n=1

∞

cnψn ( x )=√ 2a∑n=1∞

cn sin( nπa x)

f p ( x )= 1√2πℏ

eipxℏ

p= kℏ

f p ( x )= 1

√2πℏeikx

k=2πλ

cn=∫ψn ( x )¿ f ( x )dx

cn=√ 2a∫0a

sin( nπa x )Ψ ( x ,0 )dx

~Ψ (k )= 12 π

∫−∞

∞

e−ikxΨ ( x )dx

(−ℏ22m∂2

∂ x2+V ( x ))Ψ E ( x )=EΨ E ( x )

Ψ n ( x ,t )=ψn ( x )e−i En t

ℏ =√ 2a sin( nπa x )e−i (n2π 2ℏ2ma2 )t

Particle energy above potential (easier to use first solution for free particles):Ψ ( x )=Ae ikx+Be−ikx=A sin (kx )+B cos (kx )

k=√2m (E−V )ℏ

Ψ (a )=A sin (ka )=0ka=nπ→n=1,2,3,4…

|Ψ n ( x )|2=2asin2( nπa x ) En=

n2π2ℏ2

2ma2k=nπ

a=√2mE

ℏ

Prob (7 t hatom ,at t=23 )=∫18

21

|Ψ ( x , t=23 )|2dxΨ ( x ,t )=ϕ ( x ) f ( t )

iℏ ddtϕ ( x ) f (t )=(−ℏ22m

∂2

∂ x2+V ( x ))ϕ (x ) f ( t ) iℏ d

dtf ( t )

f (t )=

(−ℏ22m∂2

∂x2+V ( x ))ϕ ( x )

ϕ (x )

(−ℏ22m∂2

∂ x2+V ( x ))ϕ (x )=c ϕ ( x )

H ϕ (x )=c ϕ ( x )→H ϕ (x )=E ϕ ( x )

iℏ ddtf ( t )=c f ( t )→E f ( t )

Ψ ( x ,t )=∑E

aE ϕE ( x )e−iEtℏ

∫−∞

∞

dxϕE'¿ ( x ) ∙Ψ ( x ,t=0 )=∫

−∞

∞

∑E

aEϕE ( x ) ∙ ϕE'¿ ( x )dx=∑

E

aE∫ ϕE ( x )ϕ E'¿ ( x )dx=aE'

aE=∫ ϕE¿ ( x )Ψ (x , t=0 )dx

P ( x , t )=|Ψ ( x ,t )|2=|ϕE ( x ) e−iEtℏ |

2

=|ϕE (x )|2 Ψ ( x ,t )=∑n

anΨ n ( x ) e−i En tℏ

Ψ ( x ,t=0 )=∂ ( x−7.5 Å )Ψ n ( x )=√ 2a sin ( nπa x) En=

ℏ2k 2

2m= ℏ2

2m ( nπa )2

an=∫−∞

∞

Ψ (x )¿ ∙ ∂ ( x−7.5 )dx

¿√ 2a sin( nπa ∗7.5)Ψ ( x ,t )=∑

n=1

∞

(√ 2a sin( nπa ∗7.5)) ∙(√ 2a sin( nπa x))e−i( ℏ22m ( nπa )

2)tℏ

Probability Flow:

j ( x ,t )= ℏ2mi (Ψ ¿ ∂Ψ

∂ x−Ψ ∂Ψ ¿

∂ x ) I=∑k

e ∙ jk→Landauer FormulaI=(e∗V ) (e )( mL

πℏ2k )∗ jΨ

Tunneling:Particle energy above potential:Ψ=A eikx+Re−ikx

k=√ 2m ( E−V 0 )ℏ2

Particle energy below potential:Ψ ( x )=C eκx+De−κx

κ=√2m (V 0−E )ℏ

Match the Ψ and the d Ψ /dx at the region boundariesSet A=1

Fraction transmitted:jtjb

=|T|2

|r|2+|t|2=1

⟨ p ⟩=md ⟨ x ⟩dt

f ( x )= 1√2π ∫−∞

∞

Φ (k ) eikx dk

vgroup=dωdk

→ω=ℏk2

2m

SHO:

V ( x )=12k x2 a=√mω2ℏ x+i √ 1

2mωℏp a†=√mω2ℏ x−i √ 1

2mωℏp H= ωℏ ( a† a+ 12 )

H (a†Ψ n )=λ∗(a†Ψ n ) ℏω (a† a+ 12 )∗a†Ψ n=(En+ ωℏ ) (a†Ψ n )The energy of the eigenstate is En+ ωℏa† is the raising operator

ℏω (a† a+ 12 ) (aΨ n )=(En−ℏω) aΨ n

There must exist a lower bound Ψ 0 such that aΨ 0=0

E0=12ℏω

H Ψ 0=E0Ψ 0E1=E0+ ωℏ

En= ωℏ (n+ 12 )

Bra-Ket Notes∂∂ t

(|Ψ ( t ) ⟩ )= 1iℏ

H|Ψ ( t ) ⟩ ddt

(⟨Ψ ( t )|)=−1iℏ

⟨Ψ ( t )|H ∂∂ t

⟨Q ⟩= iℏ

⟨Ψ ( t )|H Q|Ψ (t ) ⟩+⟨ ∂ Q∂t ⟩−⟨Ψ (t )|Q iℏH|Ψ ( t )⟩

3D

iℏ ∂∂ t

|Ψ ( t ) ⟩=(−ℏ22m ( ∂2

∂x2+ ∂2

∂ y2+ ∂2

∂ z2 )+V ( r ))|Ψ ( t ) ⟩V ( x , y , z )=V (x )+V ( y )+V ( z )

(−ℏ2m ( ∂2

∂ x2+ ∂2

∂ y2+ ∂2

∂z2 )+V ( x )+V ( y )+V ( z ))X ( x )Y ( y )Z ( z )=E X ( x )Y ( y )Z ( z )

(−ℏ22mX ' ' ( x )X ( x )

+V ( x ))=c xX ( x ) (−ℏ22mY ' ' ( y )Y ( y )

+V ( y ))=c yY ( y ) (−ℏ22mZ ' ' ( z )Z ( z )

+V ( z ))=c zZ (z )

H=−ℏ2

2m∇2 Ψ x ( x )=Ae ikx Ψ ( r )= 1

√Lx1

√L y

1

√Lze i(k x x+k y y+k z z )= 1

√Lx L y Lzei k ∙ r k=k x x+k y y+k z z

Ψ ( r )=Ψ xΨ yΨ zΨ x=√ 2Lx

sin( nx πLx x ), Ex=ℏ2 kx

2

2mk n=k x x+k y y+k z y=

nx π

Lxx+

ny π

Ly

y+nz π

Lz

z

Ψ ( x )free=ei kx x

Ψ ( x )bound=sin( nπΔ x x ) E free=ℏ2 kx

2

2mEbound=

ℏ2π 2n2

2m (Δ z )2

Angular Momentum

(−ℏ22m ( 1r2 ∂∂r (r 2 ∂∂r ))+^L2

2mr2+V (r ))Ψ (r , θ ,ϕ )=EΨ (r , θ ,ϕ ) KE=1

2mv2= 1

2mp2

KE=12I ω2= L2

2 I= L2

2mr2

L+¿=Lx+i Ly→raisingoperator ¿ L−¿=Lx−i Ly→lowering operator ¿¿ ¿

L z¿ L+¿¿ raises the eigenvalue by ℏ

L+¿|l , m⟩=bm+ℏ¿ L+¿2|l , m⟩=bm+2ℏ¿

L+¿n|l , m⟩=bm+nℏ¿ L z|l ,m ⟩= mℏ |l ,m ⟩ L2|l ,m ⟩=ℏ2 l ( l+1 )|l ,m ⟩ Ψ (r , θ ,ϕ )=R (r )Y lm (θ ,ϕ )

( 1R ∂∂r

r2∂∂r

R−2mr2

ℏ2(V (r )−E ))+(−1Y ( L

2

ℏ2Y ))=0 −ℏ2

2m∂2

∂r2U+(V (r )+

ℏ2 (l ( l+1 ) )2mr2 )U=EU

Ψ (E|M ) (r , θ ,ϕ )=U (r )r

Y lm (θ ,ϕ) (−ℏ22m∂2

∂ x2+V eff ( x ))U ( x )=EU ( x )

V=0 ,inside ballV=∞,outside ball

−ℏ2

2m∂2

∂r2U+(V (r )+

ℏ2 (l ( l+1 ) )2mr2 )U=EU

−ℏ2

2m∂2

∂r2U+(ℏ

2 (l (l+1 ) )2mr2 )U=EU k=√ 2mEℏ2

E=ℏ2k 2

2m∂2

∂ r2U kl (r )=( l (l+1 )

r2−k2)U kl (r )

General solution:U kl (r )=Ar jl (kr )+Br nl (kr )

jl (kr ) spher Besselnl (kr ) spher Neumann

Ψ klm (r , θ ,ϕ )=A jl (kr )Y lm (θ ,ϕ ) This is the solution inside Outside is Ψ=0

Hydrogen Atom

V (r )= −ze2

4 π ε0rH=−ℏ2

2m∇2+( −z e2

4 π ε0 r )Ψ Elm (r , θ ,ϕ )=R (r )Y (θ ,ϕ )

R (r )=U (r )r

−ℏ2

2m∂2

∂r2U (r )+( −z e2

4 π ε0 r+ℏ2 (l (l+1 ) )2mr2 )U (r )=EU (r ) ρ=√−2mE

ℏ ρ0=z e2

2 π ε0ℏ √ m−2E

∂2

∂ ρ2U ( ρ )=(1− ρ0

ρ+

(l (l+1 ) )ρ2 )U ( ρ )

U ( ρ )=A e−ρ+Beρ=Ae− ρ

U ( ρ )=A ρl+1+B ρ−l=A ρl+1U ( ρ )=V ( ρ )e− ρ ρl+1

V ( ρ )=∑j=0

∞

a j ρj a j=

A2 j

j !V ( ρ )=A∑

j

(2 ρ ) j

j !=Ae2ρ

U ( ρ )=A ρl+1 eρ

a jmaxim+1=

2 ( jmaxim+l+1 )−ρ0

( jmaxim+1 ) ( jmaxim+2l+2 )=0

E jmax ,l=−m2ℏ ( z e2

4 π ε0 )2

( 1

( jmaxim+l+1 )2 ) E jmaxim=−13.6 eV∗z2

( jmaxim+l+1 )2

Spin

μ=IA=−eτπ r0

2 I=−eτ τ=d

v=2 πr 0v

s=angmoment=mv r0 μ= −e2me

s

[ Sx , S y ]=iℏSz[S2 , Sz ]=0

S+¿=Sx+i S y ¿

S−¿=S x−i S y¿

Ψ=|s ,ms ⟩S+¿|s ,ms ⟩=A+¿|s ,ms +1⟩ ¿ ¿

S−¿|s ,m s ⟩=A−¿|s ,ms−1⟩ ¿¿

A+¿=ℏ√s ( s+1)−m (m+1 )¿

A−¿=ℏ√s ( s+1 )−m (m−1) ¿

Ψ= 1

√Le ikx (ab)

S2|s ,m ⟩=ℏ2 s (s+1 )|s ,m ⟩ Sz|s ,m ⟩= mℏ |s ,m ⟩

Stern-Gerlach

μ=−em

s F=−∇U U=− μ∙ B F=−∇ (− μ ∙ B ) F=μ ∇B Bcenter=B z z

∇Bcenter=∂B z

∂ z<0 F=μ ∇B=μz ∇B z=(−e

msz)( ∂B z

∂ z )= em|∂B z

∂ z |sz z sz=ℏ2,−ℏ2

Pauli MatricesS=Sx x+S y y+Sz zS2=Sx

2+S y2+Sz

2 S2 :(S112 S12

2

S212 S22

2 ) S2 :34ℏ2(1 00 1) Sz:

ℏ2 (1 00 −1) Sx :

ℏ2 (0 11 0) Sy :

ℏ2 (0 −ii 0 )

σ x=Sxσ y=Sy

σ z=(1 00 −1) S=ℏ

2(σ x i+σ y j+σ z k )→Pauli spinmatrices

∂|ψ (r )|2

∂t+∇ ∙ j (r )=0

j= iℏ2m

(Ψ ∇Ψ ¿−Ψ ¿ ∇Ψ )