Best way to find (ψ,(p ̂2/2M) ψ) for ψs with an awkward derivative?
(ψ,(p ̂2/2M) ψ)=(1/2M) (p ̂ψ,p ̂ψ)
How do you find the following integral:
∫(-∞)∞(xm/(1+(x/b)2 )n)dx
When m is even
A quick way to find the following integral:
∫ψ(x)(dψ(x)/dx)dx
∫ψ(x)(dψ(x)/dx)dx=1/2∫d/dx (ψ2 (x))dx=ψ2/2+C
Harmonic expansions of potential are usually done around the ________ point of the potential.
minimum
When finding bound states in a semi-infinite potential well, it is usually convenient to substitute ______ as u and ______ as v when considering boundary condition constraints.
kL=u
κL=v
where
k=(2ME)1/2/ℏ
wavenumber inside well (0 < x < L)
κ=(2M(V0-E))1/2/ℏ
wavenumber in semi-infinite barrier (X>L)
When finding bound states in a semi-infinite potential well, it is also convenient to subsitute:
u2+v2=_____ where b=_____
u2+v2=1/b2
where
b=ℏ/(L(2MV0)(1/2) )
How to find commutative properties of a commutator [A ̂,B ̂]?
Apply it to a general function, as with:
[x ̂,p ̂x ]f(x)=x ̂p ̂x f(x)-p ̂x (x ̂f(x))
p ̂x in the second term on the right had side also operates on x!
[m ̂,n ̂]=?
m,n∈{x ̂,y ̂,z ̂ }, position operators.
[m ̂,n ̂]=0
Position operators commute because they operate as multiplication by a scalar (x, y, z) of the function.
[m ̂,p ̂n]=?
m,n∈{x ̂,y ̂,z ̂ }, position operators.
[m ̂,p ̂n]=iℏδmn
As in [x ̂,p ̂x ]=iℏ, because p ̂x operates nontrivially on x ̂.
[p ̂m,p ̂n]=?
m,n∈{x ,y, z}
[p ̂m,p ̂n]=0
This happens because derivatives of independent coordinates are commutative (e.g. d/dx d/dy=d/dy d/dx and p ̂m=-iℏ(d/dm).
[L ̂m,L ̂n ]=?
[L ̂m,L ̂n ]=iℏL ̂l
Right had side obeys right hand rule.
e.g. [L ̂x,L ̂y ]=iℏL ̂z
When m≠n:
[L ̂m,p ̂n ]=?
[L ̂m,p ̂n ]=iℏp ̂l
Right had side obeys right hand rule.
e.g. [L ̂x,p ̂y ]=iℏp ̂z
When m≠n:
[L ̂m,n ̂ ]=?
m,n∈{x ̂,y ̂,z ̂ }
[L ̂m,n ̂]=iℏl ̂l
Right had side obeys right hand rule.
l∈{x ̂,y ̂,z ̂ }
e.g. [L ̂x,y ̂]=iℏz ̂
When m=n:
[L ̂m,p ̂m ]=?
m,n∈{x ̂,y ̂,z ̂ }
[L ̂m,p ̂m ]=0
When m=n:
[L ̂m,m ̂ ]=?
m,n∈{x ̂,y ̂,z ̂ }
[L ̂m,m ̂ ]=0
Given the hamiltonian operator:
H ̂=p ̂2/2M+V(x,y,z)
what are the conditions for [H ̂,L ̂ ]=0?
[p ̂2/2M, L ̂]=0 unconditionally.
[V ̂,L ̂ ]=0 when ∇V ̂∥e ̂r. This happens when V ̂=V(r) ( V ̂ is a central potential).
Do eigenstates of a H ̂ which satisfies:
H ̂(r ̂1,r ̂2,p ̂1,p ̂2)=H ̂(r ̂2,r ̂1,p ̂2,p ̂1)
satisfy:
ψ(r ̂1,r ̂2 )=ψ((r2 ) ̂,r ̂1)?
This is a H ̂ which describes a system of two identical particles, such as electrons in a He atom.
Not necessarily!
A good counterexample are eigenstates of a system of two identical particels in a 1D harmonic oscillator. These igenstates are not necessarily eigenstates of the permutation operator p ̂12.
Given that a H ̂ describing a system of 2 identical particles satisfies:
[H ̂,p ̂12 ]=0
Are all eigenstates of H ̂ symmetric\antisymmetric to permutation?
No! The system described by H ̂ can have degenerated eigenvalues.
This means that not all eigenfunctions of H ̂ are also eigenfunctions of p ̂12.
A good counterexample are eigenstates of a H ̂ of two identical particles in a harmonic oscillator.
What is a unitary operator?
A unitary operator U ̂ is an operator which satisfies, for every ψ1, ψ2:
(U ̂ψ1,U ̂ψ2 )=(ψ1,ψ2 )
An operator which conserves value of internal product. e.g. eiωt.
The hermitian conjugate of a unitary operator U satisfies…
U ̂=U(-1)
The absolute value |λ| of all eigenvalues λ of a unitary operator satisfy:
|λ|=1
Condition for orthonormality between Ylm and Yl’m’?
(Ylm,Y(l’ m) )=δ(ll’ ) δ(mm’)
Ylm are eigenfunctions of L2 and Lz which are both hermitian operators. This is why Ylm’s are orthonormal for different m values or l values!
Footnote
H ̂ for rotaring rigid body?
Such as a poly atomic molecule, in the non inertial system
H ̂RB=(L12)/2I1 + (L22)/2I2 + (L32)/2I3
H ̂ for symmetrical rotaring rigid body?
H ̂S=L ̂2/(2I1 )+1/2 (1/I3 -1/I1 ) L ̂32
I1=I2≠I3