A Derivation Of The Schrödinger Equation.

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LeconsdAnalyse

LeconsdAnalyse

Күн бұрын

Find the original derivation in:
"An Undulatory Theory Of The Mechanics Of Atoms And Molecules",
E. Schrödinger - Physical Review, Dec.1926 - APS.
(Received 3 September 1926; published in the issue dated December 1926).
REMARKS↓
1. Returning back to the abstract equation ∂ψ/∂t = Aψ, we identify from 0:49, Aψ=(iℏ/2m)Δψ + (-i/ℏ)V(x,t)ψ≐(-i/ℏ)ℍψ. Where ℍ is the Hamiltonian operator.
As it turns out the Schrödinger version of QM (There are three equivalent versions:
①. Heisenberg, ②. Schrödinger, and ③. Feynman) is viable only if ℍ is self-adjoint.
2. Functional analysis shows that the unique (& probability conserving) solution of the initial-value problem: ∂ψ/∂t = (-i/ℏ)ℍψ, ψ(0)= ψ₀, is given by ψ(t)≐e^[(-i/ℏ)∫₀ ᵗ ℍ(s)ds]ψ₀. Where ψ₀∈L²(ℝ³) (Meaning that measurable (i.e., "almost everywhere" continuous) function ψ₀:ℝ³→ℂ is such that the integral ∫ψ₀(x)ψ₀*(x)dμ(x) always exists). The integral is (always) the Lebesgue (pronounced: Lay-bez-guh) integral. ℍ may be bounded or unbounded, as long as it is self-adjoint.
3. Taking V(x,t)=V(x), if we assume ψ(x,t)≐φ(x)e^(-itE/ℏ), then ∂ψ(x,t)/∂t=(-iE/ℏ)ψ(x,t). So that
(iℏ)∂ψ/∂t=ℍψ becomes, ℍφ=Eφ (time dependence cancels out).
Note that assuming the particular solution ψ(x,t)=φ(x)e^(-itE/ℏ) is the same as taking
S(x, P, t) = - Et + S(x, P, 0) (Hamiltonian function is time independent) in ψ(x,t) = a(x,t)e^(iS(x,t)/ℏ) and redefining the amplitude. Viz., φ(x) = a(x,0)e^(iS(x,0)/ℏ).
Proof: S(x, P, t) = S(x, P, 0) + L·t + o(t) (t→0) with L = ∂S/∂t = -E (Hamilton-Jacobi equation). Therefore, ψ(x,t) = a(x,t)e^(iS(x,t)/ℏ) = a(x,t)e^(i[S(x, P, 0) + L·t + o(t)]/ℏ)
= a(x,t)e^(iS(x, P, 0)/ℏ)e^(i[ -tE + o(t) ]/ℏ)
~ a(x,0)e^(iS(x, P, 0)/ℏ)e^(-itE/ℏ) (t→0).
The more general (perturbation expansion) solution to the time-dependent Schrödinger equation (for certain potentials V(x,t)) was developed by Freeman Dyson.
4. Imagine a travelling wave with phase φ. Interference will not occur when the phase is equal to an integer n. That is, when φ=n, n∈ℤ.
The mechanical paths for a conservative system are the orthogonal trajectories of the surfaces S=constant. Where S is a particular solution of the Hamilton-Jacobi equation. Each mechanical path runs (or, flows) through the family of surfaces S=constant orthogonally.
Einstein's quantum condition (the old quantum condition) is S=nh, where h is Planck's constant (fundamental action). Therefore, φ=S/h upon eliminating n.
In OPTICS the condition S=constant represents surfaces of equal time for light paths, and in MECHANICS the condition S=constant represents surfaces of equal action for mechanical paths.
5. NOTATION:
①. ψ(x,t)≡ψ(t)(x), ②. ψ* is the complex conjugate of ψ.

Пікірлер: 14
@LeconsdAnalyse
@LeconsdAnalyse 8 жыл бұрын
Q: In (2), in the clip @0:12, we have h(x,∇S,t), i.e., we set p=∇S, why? A: Let S be the twice continuously differentiable solution of *Ṡ* = -h. Given *ṗ* = -∂h/∂x: ∂ *Ṡ* /∂x = -∂h/∂x = *ṗ*, so that ∂S/∂x = p, OR ∇S = p, unique up to constants.
@LeconsdAnalyse
@LeconsdAnalyse 10 жыл бұрын
The function ψ(x,t) with x∈ℝ³, t∈ℝ is called the *wave function* which describes the *state* (a term originating from the word *_statistical_* ) by way of the integral ∫ψψ*dμ
@LeconsdAnalyse
@LeconsdAnalyse 9 жыл бұрын
a. On page 1050 of Schrödinger's paper (English translation) states, *"To explain the main lines of thought, I will take as an example of a mechanical system a material point, of mass m, moving in a conservative field of force V(x,y,z)."* Note the word *"example"*. In *general*, in classical mechanics, if a smooth change of coordinates, (*x*, *p*)→(*X*, *P*), can be found such that: 1. A given Hamiltonian function H(*x*, *p*, t) transforms into a function ℌ(*X*, *P*, t), and 2. ℌ also satisfies Hamilton's equations, and 3. There exists a function S(*x*, *P*, t) such that, ∂S/∂t = ℌ - H, then the change of coordinates is called *"canonical"*. In the *special case* that (*x*, *p*)→(*X*, *P*) transforms H→ℌ≡0, the equation ∂S/∂t = ℌ - H becomes ∂S/∂t = - H and is given the special name *Hamilton-Jacobi* equation. If the Hamiltonian function is time independent, H(*x*, *p*, t) = H(*x*, *p*) = E (total energy), then integrating the Hamilton-Jacobi equation (∂S/∂t)dt = - Edt gives S(*x*, *P*, t) = - Et + constant, OR S(*x*, *P*, t) = - Et + S(*x*, *P*, 0). THE POINT: In the derivation we should leave the Hamiltonian function in its most general time-dependent form. b. Compare (4) @0:10 in the clip with (9) on page 1054 of Schrödinger's paper (English translation). Note that, Ψ=A(x,y,z)sin(W/K)=ℐ[A(x,y,z)expi(W/K)]. Where the symbol ℐ denotes the imaginary part. The "W" used by Schrödinger is the same function as the S used above.
@LeconsdAnalyse
@LeconsdAnalyse 8 жыл бұрын
♦NOTES: The abstract equation must be first-order in ∂/∂t. So that the initial-value problem has solution, ψ(x, t) = e^(∫₀ ᵗ A(s)ds)ψ(x, 0). Furthermore, the condition ∫ψψ*dμ
@LeconsdAnalyse
@LeconsdAnalyse 7 жыл бұрын
Time Evolution: To illustrate how this works, suppose that the Hamiltonian is time-independent: ℍ=ℏωσₓ=ℏω(0 1/1 0)=(0 ℏω/ℏω 0) ω=constant with entries ℍ₁₁=0, ℍ₁₂=ℏω, ℍ₂₁=ℏω, ℍ₂₂=0. Therefore, exp(-itℍ/ℏ) = (2πi)⁻¹∮ₐexp(-itz/ℏ)(zI-ℍ)⁻¹dz (z∈ℂ) with "a" a simple closed curve enclosing ALL of the eigenvalues of ℍ. The eigenvalues are found by solving (secular equation): 0= det(ℍ-λI). They are λ=ℏω & λ= -ℏω. With I=(1 0/0 1)=id. Also, (zI-ℍ)⁻¹=adj(zI-ℍ)/det(zI-ℍ) is the inverse matrix of zI-ℍ, (zI-ℍ)⁻¹= (z(z-ℏω)⁻¹(z+ℏω)⁻¹ ℏω(z-ℏω)⁻¹(z+ℏω)⁻¹ / ℏω(z-ℏω)⁻¹(z+ℏω)⁻¹ z(z-ℏω)⁻¹(z+ℏω)⁻¹) So that, U(t)=exp(-itℍ/ℏ) = (½{exp-iωt+expiωt} ½{exp-iωt-expiωt} / ½{exp-iωt-expiωt} ½{exp-iωt+expiωt}) = ( cosωt -isinωt/ -isinωt cosωt). If, say, |ψ(0)>=(a b)ᵗ a,b∈ℂ, s.t., =|a|²+|b|²=1 then, |ψ(t)>=exp(-itℍ/ℏ)|ψ(0)>=(acosωt-bisinωt -aisinωt+bcosωt)ᵗ. Also, =(a* b*) (acosωt-bisinωt -aisinωt+bcosωt)ᵗ = |a|²cosωt-a*bisinωt-ab*isinωt+|b|²cosωt = cosωt-2ℜ(a*b)isinωt. So that, the probability density for the system at time t is, ||²=[]* = (cosωt-2ℜ(a*b)isinωt)(cosωt+2ℜ(a*b)isinωt) = cos²ωt + 4[ℜ(a*b)]²sin²ωt, where ℜ means the real part, and superscript t means transpose, and * means cc.
@LeconsdAnalyse
@LeconsdAnalyse 9 жыл бұрын
Geometrical Optics is to Wave Optics what Classical Mechanics is to Quantum Mechanics. The cycle between the four theories forms the so-called *Wave-Particle Duality*.
@LeconsdAnalyse
@LeconsdAnalyse 7 жыл бұрын
Example: P(∊)= (2πi)⁻¹∮ₐ (zI-ℋ(∊))⁻¹dz -(1) = -(2πi)⁻¹∮ₐ (ℋ(∊)-zI)⁻¹dz, with ℋ(∊)-zI=(∊-z 1/0 -z) so that, (ℋ(∊)-zI)⁻¹= adj(ℋ(∊)-zI)/det(ℋ(∊)-zI) =z⁻¹(z-∊)⁻¹(-z -1/0 ∊-z) =((∊-z)⁻¹ z⁻¹(∊-z)⁻¹/0 -z⁻¹) -(2). Now ℋ(∊)=(∊ 1/0 0)=(0 1/0 0) + ∊(1 0/0 0) has two discrete eigenvalues 0 and ∊. So if the simple closed curve "a" encloses ONLY the eigenvalue 0: P(∊)= - (2πi)⁻¹∮ₐ (ℋ(∊)-zI)⁻¹dz=(2πi)⁻¹∮ₐ z⁻¹(-z(∊-z)⁻¹ -(∊-z)⁻¹/0 1)dz =(0 -∊⁻¹/0 1). Careful with the signs (+ or -) and which factor is analytic, and which has a singularity. For the other eigenvalue ∊ ("a" now encloses only the eigenvalue ∊): P(∊)= - (2πi)⁻¹∮ₐ (ℋ(∊)-zI)⁻¹dz=(2πi)⁻¹∮ₐ (-(∊-z)⁻¹ -z⁻¹(∊-z)⁻¹/0 z⁻¹)dz =(2πi)⁻¹∮(z-∊)⁻¹( 1 z⁻¹/0 z⁻¹(z-∊))dz= ( 1 ∊⁻¹/0 0). REMARKS: 1. (Common knowledge) We can write the given operator as a sum of eigenprojections, each multiplied by its corresponding eigenvalue (non-degenerate case): Let the self-adjoint linear operator A have u discrete eigenvalues λ₁, λ₂, λ₃,...,λᵤ. Given Pᵨ ≐ (2πi)⁻¹∮ₐ₍ᵨ₎ (λI-A)⁻¹dλ (λ∈ℂ), with "a(ϱ)" a simple closed curve enclosing ONLY λᵨ. Let the simple closed curve 'a' enclose all of the simple closed curves a(ϱ), ϱ=1, 2, 3,...,u. That is, a(ϱ)⊂int(a) (the interior of a) for all ϱ=1, 2, 3,...,u. If A is an operator on X (say, a Hilbert space) then, X=Y₁⊕Y₂⊕...⊕Yᵤ, where PᵨX=Yᵨ for each ϱ=1, 2, 3,...,u. Next, A=A₁⊕A₂⊕...⊕Aᵤ, where Aᵨ is the restriction of the operator A to Yᵨ. With this aside, ∑ λᵨPᵨ= (2πi)⁻¹ ∑ ∮ₐ₍ᵨ₎ λᵨ(λI-A)⁻¹dλ = A₁⊕A₂⊕...⊕Aᵤ = A, where the sum is over ϱ=1, 2, 3,...,u. 2. Pᵨ ≐ (2πi)⁻¹∮ₐ₍ᵨ₎ (λI-A)⁻¹dλ. -(1) (2πi)⁻¹∮ₐ λᵐ (λI-A)⁻¹dλ= Aᵐ. (m=0, 1, ...) -(2). Therefore, summing over ϱ=1, 2, 3,...,u. ∑ Pᵨ = (2πi)⁻¹ ∑ ∮ₐ₍ᵨ₎ (λI-A)⁻¹dλ = (2πi)⁻¹ ∮ₐ (λI-A)⁻¹dλ = (2πi)⁻¹ (2πi)id (using (2) with m=0) = id 3. In Quantum Mechanics the Hamiltonian operator must be self-adjoint. Given a perturbed Hamiltonian, ℍ(∊)=ℍ₀ + ∊V, with ℍ₀ linear & self-adjoint, the question arises: Which conditions must a linear symmetric operator V satisfy so that ℍ(∊) will also be self-adjoint ? With ||u||:= √(u | u) the norm induced by the metric (u | u), and ψ an element of the domain of ℍ₀ (i.e., ψ ∈ D(ℍ₀)), and D(ℍ₀)⊂ D(V), a well known theorem (Kato-Rellich) states: ||Vψ|| ≤ a||ψ|| + b||ℍ₀ψ||; a,b ∈ ℝ; b
@bboyHarrypotter
@bboyHarrypotter 8 жыл бұрын
In many textbooks, the ansatz (4) is motivated by the Schrodinger equation, though that's not necessary at all. You should put the optical motivation for that ansatz.
@LeconsdAnalyse
@LeconsdAnalyse 8 жыл бұрын
+bboyHarrypotter Re: Classical Mechanics & Optics analogy, ∂S/∂t = - h is analogous to the eikonal equation, ∂φ/∂t=±c|∇φ| and, ψ = ae^(iS/ℏ) is analogous to u = ae^(iφ/λ) (λ
@LeconsdAnalyse
@LeconsdAnalyse 8 жыл бұрын
+bboyHarrypotter Namely, ∂φ/∂t=±c|∇φ| or, (∂φ/∂t)² = c²|∇φ|². Given u = ae^(iφ/λ) we find, for λ
@LeconsdAnalyse
@LeconsdAnalyse 7 жыл бұрын
♣ *The same concept when viewed in a different scale.* ♣ The Quantum Mechanical analysis of the free (of external fields), neutral, relativistic, Hydrogen atom shows that its energy eigenvalues are given by, E(n,k) = mc²{1 + F(α, n, k)}^(-½) -➊ With: F(α, n, k) = α²{n + √(k² - α²)}⁻² and, 1. α is the Sommerfeld (Arnold) fine structure constant, α=e²/ℏc, e being the electron charge, and 2. For the "reduced mass", μ= mM/(m+M) = m/(1+m/M) ~ m (m
@LeconsdAnalyse
@LeconsdAnalyse 7 жыл бұрын
Rayleigh-Schrödinger Perturbation (non-degenerate case): Let the quantum system described by, ℍ=ℏωσₓ=ℏω(0 1/1 0)=(0 ℏω/ℏω 0), suffer a small perturbation, ∊V=∊ℏϖ(1 0/0 0) with real ∊>0 small, ϖ=constant. So consider, ℋ(∊)=ℍ+∊V=(∊ℏϖ ℏω/ℏω 0). Next, (zI-ℋ(∊))⁻¹ = (z(z-λ₊)⁻¹(z-λ₋)⁻¹ ℏω(z-λ₊)⁻¹(z-λ₋)⁻¹ / ℏω(z-λ₊)⁻¹(z-λ₋)⁻¹ z-∊ℏϖ(z-λ₊)⁻¹(z-λ₋)⁻¹ ) with ( z-λ₊ )( z-λ₋ ) ≡ z²-∊ℏϖz-ℏ²ω² ( λ₊>λ₋ ). Let ψ denote the normalized eigenvector of ℍ corresponding to, say, the eigenvalue ℏω. Then P(∊)ψ=ψ(∊)≠0 is an eigenvector of ℋ(∊) with eigenvalue λ(∊) i.e., ℋ(∊)ψ(∊)=λ(∊)ψ(∊) with, P(∊)= (2πi)⁻¹∮ₐ (zI-ℋ(∊))⁻¹dz (z∈ℂ) (Riesz), and with the simple closed curve "a" enclosing ONLY the eigenvalue ℏω, and traversed in the positive (ccw) sense. Therefore, ℋ(∊)ψ(∊) = λ(∊)ψ(∊) or, ℋ(∊)P(∊)ψ = λ(∊)P(∊)ψ so that, = or, λ(∊) = and, λ(∊) = /{} (dividing through by ) = /{} = /{} + ∊/{} (∊>0) ~ + ∊ (∊
@LeconsdAnalyse
@LeconsdAnalyse 7 жыл бұрын
♦Re: Time dependent Schrödinger equation. See... METHODS OF MODERN MATHEMATICAL PHYSICS II: FOURIER ANALYSIS. SELF-ADJOINTNESS, MICHAEL REED - BARRY SIMON (1975).... Section X.12 - Time-dependent Hamiltonians, page 282.
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