130_notes.dvi

[p,x]ψ(x) = ̄h i ( ψ(x) +x ∂ψ(x) ∂x −x ∂ψ(x) ∂x ) = ̄h i ψ(x) So, removing theψ(x) we used for computational purposes, we get th ...

6.6.2 Verify Energy Operator E(op) 1 √ 2 π ̄h ei(p^0 x−E^0 t)/ ̄h= 1 √ 2 π ̄h i ̄h −iE 0 ̄h ei(p^0 x−E^0 t)/ ̄h =E 0 1 √ 2 π ̄h ...

= i ̄hψ(x,t) + ( i ̄ht ∂ ∂t −i ̄ht ∂ ∂t ) ψ(x,t) = i ̄hψ(x,t) Removing the wave function, we have the commutator. [E,t] =i ̄h 6. ...

6.7.5 Commutator ofLxandLy Angular momentum is defined by ~L=~r×~p. So the components of angular momentum are Lz=xpy−ypx Lx=ypz− ...

O 1 ψ(x) = 1/ψ(x) O 2 ψ(x) =∂ψ∂x(x) O 3 ψ(x) =x^2 ψ(x) O 4 ψ(x) =−ψ(x+a) For a free particle, the total energy operator H is ...

7 The Schr ̈odinger Equation Schr ̈odinger developed adifferential equation for the time development of a wave function. Since t ...

Our wave function will be a solution of the free particle Schr ̈odingerequation providedE 0 = p (^20) 2 m. This is exactly what ...

If weintegrate if over some intervalinx ∫b a ∂P(x,t) ∂t dx=− ∫b a ∂j(x,t) ∂x ∂ ∂t ∫b a P(x,t)dx=j(x=a,t)−j(x=b,t) the equation s ...

Assume that we canfactorizethe solution between time and space. ψ(x,t) =u(x)T(t) Plug this into the Schr ̈odinger Equation. ( − ...

L(aψ+bφ) =aLψ+bLφ whereaandbare arbitrary constants andψandφare arbitrary wave-functions. A multiplicative constant is a simple ...

We have the equation. − ̄h^2 2 m ∂^2 ψ(x) ∂x^2 +V 0 ψ(x) =Eψ(x) ∂^2 ψ(x) ∂x^2 =− 2 m ̄h^2 (E−V 0 )ψ(x) Remember thatxis an indep ...

8 Eigenfunctions, Eigenvalues and Vector Spaces 8.1 Eigenvalue Equations The time independent Schr ̈odinger Equation is an examp ...

Since the eigenfunctions are orthogonal, we can easilycompute the coefficientsin the expansion of an arbitrary wave functionψ. α ...

Now taking the Hermitian conjugate ofA†. 〈 ( A† )† ψ|φ〉=〈Aψ|φ〉 ( A† )† =A If we take the Hermitian conjugate twice, we get back ...

8.4 Eigenfunctions and Vector Space Wavefunctions are analogous to vectors in 3D space. The unit vectors of our vector space are ...

〈φ|λ 1 ψ 1 +λ 2 ψ 2 〉=λ 1 〈φ|ψ 1 〉+λ 2 〈φ|ψ 2 〉 〈λ 1 φ 1 +λ 2 φ 2 |ψ〉=λ∗ 1 〈φ 1 |ψ〉+λ∗ 2 〈φ 2 |ψ〉 〈ψ|ψ〉is real and greater than ...

which automatically satisfies the BC at 0. To satisfy the BC atx=awe need the argument of sine to benπthere. un=Csin (nπx a ) Pl ...

Note that these states would have a definite parity ifx= 0 were at the center of the box. Theexpansionof an arbitrary wave funct ...

Together, these areexactly equivalent to the set of solutions we had with the boxdefined to be from 0 toa. Theu+n(x) have eigenv ...

satisfying thenormalization condition 〈p′|p〉=δ(p−p′) For afree particle Hamiltonian, both momentum and parity commute withH. So ...