Equations Flashcards
Hamiltonian
H=p^2/2m+V(x)=-ℏ²/2m*∆²+V(X)
Schrodinger Equation for Infinite potential well
Hψ=-ℏ²/2m*d²/dx²Ψ=ΕΨ
General solution to wavefunction for infinite potential well
Ψ(x) = Asin(root(2mE/ℏ²)x²)+Bcos(root(2mE/ℏ²)x²)
Energy of infinite potential well
E=n²ℏ²π²/2ma²
Probability of finding particle at x
P(x) = |Ψ(x)|²
normalisation
<Ψ|Ψ>= ∫p(x)dx = ∫Ψ*(x)Ψ(x)dx =1
orthoganality
∫φ*ₙ(x)φₘ(x)dx = δₙₘ
δₙₘ=1 if n=m
δₙₘ= 0 if n≠m
uncertainty principle
ΔxΔp≥ℏ/2
Time dependent schrodinger equation
ĤΨ=iℏd/dtΨ (curly d)
Inner product of 2 states
∫Φ*(x)Ψ(x)dx = <φ|Ψ>
Time dependent Ψ
Ψ(x,t) = ΣcₙΨₙ(x)e^-iEt/ℏ
compact notation for energy eigenstates
Ĥ|Ψ>=E|Ψ>
<Φ|Ο|Ψ>
<Φ|Ο|Ψ> = ∫Φ*ΟΨdx
<Φ|Ο|Ψ>^† (Hermitian Conjugate)
<Φ|Ο|Ψ>^† = <Φ|Ο|Ψ>* = ∫ΦΟΨdx
Expectation value of operator
<Ο> = <Ψ|Ο|Ψ>/<Ψ|Ψ>
</Ο>
Resolution of identity
1 = Σ|aₙ><aₙ|
Dirac notation for finding expectation value of operator
<a> = <Ψ|A|Ψ> = Σ<aₙ|cₙAΣcₘ|aₘ> = Σcₙcₘ<aₙ|A|aₘ> (A|aₘ> = aₘ|aₘ>) = Σcₙcₘaₘ<aₙ|aₘ> = Σcₙcₘaₘδₙₘ = Σ|cₙ|^2aₙ</a>
When do 2 observables share eigenstates |aₙ>=|bₙ>
Eigenstates are shared when two observables are compatible. Compatible states always commute so eigenstates are also shared for commuting observables.
Show 2 compatible eigenstates commute
BA|aₙ> = Baₙ|aₙ>=aₙbₙ|aₙ>
AB|aₙ> = Abₙ|aₙ>=aₙbₙ|bₙ>
So (AB-BA)|aₙ> = 0
Commutator
[A, B] = AB-BA
Show 2 observables are compatible if they commute
[Â, B]|Ψ> = 0
so ÂB|bₙ> = BÂ|bₙ>= Âbₙ|bₙ>
so B(Â|bₙ>) = bₙ (Â|bₙ>)
so Â|bₙ ∝ |bₙ> so |bₙ> are eigenstates of Â
What is value of commutator of ladder operators â and â^†
[â, â†] = 1
affect of down ladder operator â on |Ψ₀>
â|Ψ₀> = 0
Hamiltonian for harmonic oscillator
Ĥ = P²/2m+1/2mωx^² (x and P operator)
Lowering operator
â = root(mω/2ℏ)(x+ip/mω) (x and p are operators)
Raising operator
↠= root(mω/2ℏ)(x-ip/mω) (x and p are operators)
x operator in terms of ladder operators
x = root(ℏ/2mω)(â+â†)
p operator in terms of ladder operators
x = -iroot(mℏω/2)(â-â†)
Is â hermitian?
NO! an operator is hermitian if H† = H. Here ↠≠ â therefore its not hermitian and therefore does not correspond to an observable
what is â?
Lowering operator which produces an eigenstate with energy lowered by ℏω
what is �
Raising operator which produces an eigenstate with energy raised by ℏω
Hamiltonian in terms of ladder operators
Ĥ = ℏω/2(ââ†+â†â) = ℏω(ââ†+1/2) = ℏω(â†â-1/2)
Angular momentum
L = r x p (cross product of radius and momentum)
(L=mvr)
commutator of Lₓ and Ly
[Lₓ, Ly] = [ŷpz-zpy, zpₓ-xpz] =[ŷpz, zpx]+[zpy, xpz] = ŷ[pz,z]px +py[z,pz]x = iℏLz
commutator of Ly and Lz
[Lx, Lz] = iℏLₓ
commutator of Lz and Lx
[Lz, Lₓ] = iℏLy
[A,BC]
[A,BC] = ABC-BCA=(AB-BA)C+B(AC-CA) = [A,B]C+B[A,C}
[L², Lz]
[L², Lz] = 0
Allows us to use L² and Lz simultaneously and use them to label angular momentum eigenvalues
Angular momentum raising operator
L₊=Lₓ+iLy
Raises eigenvalue of Lz
For S raising operator replace L with S
Angular momentum raising operator
L₋=Lₓ-iLy
Raises eigenvalue of Lz
For S lowering operator replace L with S
ml values
ml = -l, -l+1, …, l-1, l
Lz|Yₗ,ₘₗ>
Lz|Yₗ,ₘₗ> = mlℏ|Yₗ,ₘₗ>
L²|Yₗ,ₘₗ>
L²|Yₗ,ₘₗ> =l(l+1)ℏ²|Yₗ,ₘₗ>
Magnitude of angular momentum
|l| =ℏ√l(l+1)
L₊|Yₗ,ₘₗ>
L₊|Yₗ,ₘₗ> = Dₗ,ₘₗ|Yₗ,ₘₗ>
Dₗ,ₘₗ =ℏ√(l(l+1)-mₗ(mₗ+1))
L₋|Yₗ,ₘₗ>
L₋|Yₗ,ₘₗ> = Cₗ,ₘₗ|Yₗ,ₘₗ>
Cₗ,ₘₗ =ℏ√(l(l+1)-mₗ(mₗ-1))
S²|χ>
S²|χ> = s(s+1)ℏ|χ>
Sz|χ>
Sz|χ> = mₛℏ|χ>
|χ> (spin state)
|χ> = c₁²|↑| + c₂²|↓|
S₊|↑|
S₊|↑| = 0
S₊|↓|
S₊|↓| = ℏ|↑|
S₋|↓|
S₋|↓| = 0
S₋|↑| = 0
S₋|↑|= ℏ|↓|