Quantum Mechanics for Mathematicians
In chapter 5, for finite dimensional unitary representations of a Lie group Gwe found corresponding Lie algebra representations ...
Definition(Momentum operators). For a quantum system with state space H=L^2 (R^3 )given by complex-valued functions of position ...
10.3 The energy-momentum relation and the Sch- r ̈odinger equation for a free particle We will review this subject in chapter 40 ...
This is an easily solved simple constant coefficient second-order partial differ- ential equation. One method of solution is to ...
The time-dependent state will be |ψ(t)〉= ∑ k ckeik·qe−it~ |k|^2 2 m Since each momentum eigenstate evolves in time by the phase ...
Chapter 11 Fourier Analysis and the Free Particle The quantum theory of a free particle requires not just a state spaceH, but al ...
While this deals with the problem of eigenvectors not being inH, it ruins an important geometrical structure of the free particl ...
are rotations of the circle counterclockwise by an angleθ, or if we parametrize the circle by an angleφ, just shifts φ→φ+θ By th ...
Theorem 11.1(Fourier series).Ifψ∈L^2 (S^1 ), then |ψ〉=ψ(φ) = +∑∞ n=−∞ cneinφ= ∑+∞ n=−∞ cn|n〉 where cn=〈n|ψ〉= 1 2 π ∫ 2 π 0 e−inφ ...
as the eigenvector equation. This has an orthonormal basis of solutions|n〉, with E= ~^2 n^2 2 m The Schr ̈odinger equation is fi ...
the analysis relatively easy, withH =L^2 (S^1 ) and the self-adjoint operator P =−i~∂φ∂ behaving much the same as in the finite ...
In others, the factor of 2πmay appear instead in the exponent of the complex exponential, or just in one ofForF ̃and not the oth ...
ψa(q) =ψ(q+a) then ψ ̃a(k) =√^1 2 π ∫+∞ −∞ e−ikqψ(q+a)dq = 1 √ 2 π ∫+∞ −∞ e−ik(q ′−a) ψ(q′)dq′ =eikaψ ̃(k) Sincep=~k, one can ea ...
The linear functional that takes a function to its Fourier transform atk: f∈S(R)→f ̃(k) The linear functional that takes a fu ...
Heuristically (ignoring problems of interchange of integrals that don’t make sense), the Fourier inversion formula can be writte ...
Tψ, one has FtTψ[f]≡Tψ[Ff] =Tψ [ 1 √ 2 π ∫+∞ −∞ e−ikqf(q)dq ] = 1 √ 2 π ∫+∞ −∞ ψ(k) (∫+∞ −∞ e−ikqf(q)dq ) dk = ∫+∞ −∞ ( 1 √ 2 π ...
In order to have the standard derivative when one identifies functions and dis- tributions, one defines the derivative on distri ...
to agree with the opposite sign conventions for spatial and time translations in the definitions of momentum and energy. One fin ...
we find that δ(E− k^2 2 m ) = √ m 2 E (δ(k− √ 2 mE) +δ(k+ √ 2 mE)) and ψ ̃E(k) =c+δ(k− √ 2 mE) +c−δ(k+ √ 2 mE) (11.10) The two c ...
Chapter 12 Position and the Free Particle Our discussion of the free particle has so far been largely in terms of one observ- ab ...
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