since 2006
Help　Sitemap
< Forum >
< Problems >
< Quantum History Theory >
< Number Representation of Uda Equation >

The success of Uda Equation of Fermionic Mechanics @ Quantum History Theory @ Products made me want to make number representation of Uda equation.

X† = X, P† = P,
[X, P]- = ih,
H = P2/(2m) + (mω2/2)X2 - hω/2.

A ≡ {√[(mω)/(2h)]}X + [i/√(2mhω)]P,
A† = {√[(mω)/(2h)]}X - [i/√(2mhω)]P,
N ≡ A†A.

H = hωN.

Now I start Uda equation.

-∞dt{(ih/α)[dν(t)/dt][⊿/⊿ν(t)] + hων(t)}Φ[ν] = 0
-∞dt{i[dν(t)/dt][⊿/⊿ν(t)] + αων(t)}Φ[ν] = 0.

Next I start looking for a solution.

Φ[ν] = Πt=-∞δν(t),0.

Φ[ν] = exp[-iωα∫-∞dt tν(t)] Πt=-∞δν(t),n,
Φ[ν(□ - ε)] = exp[-iωα∫-∞dt tν(t - ε)] Πt=-∞δν(t-ε),n
= exp[-iωα∫-∞dt (t + ε)ν(t)] Πt=-∞δν(t),n
= exp[-iωεα∫-∞dt ν(t)] Φ[ν]
limε→0 (1/ε){Φ[ν(□ - ε)] - Φ[ν]} = -iωα[∫-∞dt ν(t)] Φ[ν]
ih limε→0 (1/ε){Φ[ν(□ - ε)] - Φ[ν]} = hωα[∫-∞dt ν(t)] Φ[ν]

[α∫-∞dt H] Φ[ν] = hωα[∫-∞dt ν(t)] Φ[ν]

Surely ih limε→0 (1/ε){Φ[ν(□ - ε)] - Φ[ν]} = [α∫-∞dt H] Φ[ν].

However this solution is abnormal at the following points.

Φ[ν] = exp[-iωα∫-∞dt tν(t)] Πt=-∞δν(t),n
= exp[-iωα∫-∞dt nt] Πt=-∞δν(t),n
∴ Φ[ν(□ - ε)] = Φ[ν].

When Πt=-∞δν(t),n ≠ 0, ∫-∞dt tν(t) is not well defined enough though it may be thought of as zero.

If total length of intervals in which ν(t) ≠ 0 is finite for the purpose of letting ∫-∞dt tν(t) be finite, Πt=-∞δν(t),n = 0.

These abnormal undesirable properties persistently appear when I try to solve various Uda equations.

An unentangled solution and Uda equation might be as if they were Dirac's delta function and the condition to define it.

That is to say, the mathematical contradiction might be mere pretense and unentangled solutions might be allowed physically.

Reserving such a pathological problem, let us switch our brains to looking for a healthy solution.

WRONG:
Φ[ν] = a exp[(-i/h)α∫-∞dt tν(t)],(2019/10/22/14:10JST)
Φ[ν(□ - ε)] = a exp[(-i/h)α∫-∞dt tν(t - ε)] = a exp[(-i/h)α∫-∞dt (t + ε)ν(t)],
ih limε→0 (1/ε){Φ[ν(□ - ε)] - Φ[ν]} = [α∫-∞dt ν(t)]Φ[ν].

CORRECT:
Φ[ν] = a exp[-iωα∫-∞dt tν(t)],(2019/10/23/13:57JST)
Φ[ν(□ - ε)] = a exp[-iωα∫-∞dt tν(t - ε)] = a exp[-iωα∫-∞dt (t + ε)ν(t)],
ih limε→0 (1/ε){Φ[ν(□ - ε)] - Φ[ν]} = [α∫-∞dt hων(t)]Φ[ν].

I want to find the corresponding solution in the old quantum mechanics.

|Ψ(t)＞ = Σn=0 a exp(-iωnt)|n＞,
ih(d/dt) |Ψ(t)＞ = Σn=0 nhωa exp(-iωnt)|n＞ = H|Ψ(t)＞.
This is it.

This let me notice that the following history also is a solution.

|Ψ(t)＞ = Σn=0 an exp(-iωnt)|n＞,
ih(d/dt) |Ψ(t)＞ = Σn=0 nhωan exp(-iωnt)|n＞ = H|Ψ(t)＞.

This let me notice the following solution of Uda equation.

Φ[ν] = a exp{-iωα∫-∞dt[tν(t) + bν(t)},(2019/10/23/14:13JST)
∀ n ∈ Z; bnC.
Any solution of this form represents a quantum history unentangled in time-like direction and can be represented also by the old grammar.

A stationary quantum history with n as an eigenvalue of energy is the following special case.
∀k; k ≠ n ⇒ ibk → +∞ ∈ R.

Any superposition of unentangled solutions is also a solution of Uda equation.
So, the following functional Φ is a solution of Uda equation.
Φ[ν] = ∫-∞da0-∞db0-∞da1-∞db1-∞da2-∞db2・・・
f(a0, b0, a1, b1, a2, b2, ・・・) exp{-iωα∫-∞dt[tν(t) + aν(t) + ibν(t)]}
,(2019/10/25/15:13JST)
∀ n ∈ Z; anR and bnR.
f is an arbitrary complex valued function of infinite number of real variables.
This solution represents an entangled solution but it may be not fully general.

To make ∫-∞dt[tν(t) + cν(t)] be finite, it is necessary that
limt→-∞ν(t) = limt→+∞ν(t) = 0 and c0 = 0.

Generalizing this condition, I noticed the following fact.
If limt→-∞ν(t) = limt→+∞ν(t) = n and cn = 0, ∫-∞dt{t[ν(t) - n] + cν(t)} is finite.

However, if Φ[ν] = exp[-iωα∫-∞dt{t[ν(t) - n] + cν(t)}],
ih limε→0(1/ε){Φ[ν(□-ε)] - Φ[ν]} = hωα∫-∞dt[ν(t) - n]Φ[ν].
This equation differs from the original one.

So, I can not avoid assuming that
a0 = b0 =0.
Φ[ν] = ∫-∞da1-∞db1-∞da2-∞db2・・・
f(a1, b1, a2, b2, ・・・) exp{-iωα∫-∞dt[tν(t) + aν(t) + ibν(t)]}
.
However, I came to be not allowed to assume that b0 → -∞.

f(a1, b1, a2, b2, ・・・) exp{-iωα∫-∞dt[tν(t) + aν(t) + ibν(t)]} covers almost all solutions of the old theory because general solution is linear combination of eigensolution in the old theory because initial state is linear combination of eigenstates.

Now I have got any superposition of almost all unentangled history as an entangled solution of Uda equation.
Is that all solutions of Uda equation?
If it is so, Uda equation does not excel the old theory very much, I am sad at it.
However, if it is so, the above result is enough as technical base, I am glad at it.

If there is some entangled solution which is not any superposition of unentangled solutions, Uda equation excels the old theory very much, I am glad at it.
If it is so, the above result is not enough as technical base, I am sad at it.

Unentangled histories can be basis of the vector space of all quantum histories.
However, I do not know if unentangled solutions can be basis of the vector space of all solutions.

I want to survey it below.

Let us consider the following general simplified problem.
(I started writing this term on 2019/10/30, and Shuri Okinawa fire happened on 2019/10/31.)

x|1>|1> + y|1>|2> + z|2>|1> + w|2>|2> = |1>(x|1> + y|2>) + |2>(z|1> + w|2>).
For making this vector unentangled, x|1> + y|2> must be parallel to z|1> + w|2>.
z/x = w/y or y/x = w/z.
On the other hand, solutions are as follows.
(y, z, w) = x(a, b, c) if space of solution is 1-dimensional.
(z, w) = (ax + by, cx + dy) if space of solution is 2-dimensional.
w = ax + by + cz if space of solution is 3-dimensional.
In 1-dimensional case, z/x = b, w/y = c/a, these do not have to be equal to each other.
In 2-dimensional case, w/z = (cx + dy)/(ax + by).
(cx + dy)/(ax + by) = y/x ⇔ [(x, y) is parallel to (ax + by, cx + dy)]
(x, y) = 0 if (a, b) is not parallel to (c, d).
In 3-dimensional case, for arbitrarily given z and w, ax + by = w - cz and y/x = w/z.
Such (x, y) always exists.
ax + b(w/z)x = w - cz.
x = (w - cz)/[a + b(w/z)] = z(w - cz)/(az + bw).
y = (w/z)(w - cz)/[a + b(w/z)] = w(w - cz)/(az + bw).
Only in 3-dimensional case, an unentangled solution is guaranteed to exist.
I must check if we can make three independent vectors as linear combinations of vectors of the form:
x|1>|1> + y|1>|2> + z|2>|1> + w|2>|2>.

(z, w) = (a - b, b - a) ⇒ x = -1 - c and y = 1 + c.
(z, w) = (a + b, a + b) ⇒ x = 1 - c and y = 1 - c.
(z, w) = (a + 2b, 2a + 4b) ⇒ x = 2 - c and y = 4 - 2c.

(-1 - c, 1 + c, a - b, b - a) and (1 - c, 1 - c, a + b, a + b) and (2 - c, 4 - 2c, a + 2b, 2a + 4b) are linear independent from each other.

Memo:
(-2c, 2, 2a, 2b), (-2, 2c, -2b, -2a), (1, 3 - c, b, a + 3b).
(-2c, 2, 2a, 2b), (1, -c, b, a), (1, 3 - c, b, a + 3b).
(-2c, 2, 2a, 2b), (1, -c, b, a), (0, 3, 0, 3b).

In 3-dimensional case, basis of solution space is guaranteed to consist of unentangled solutions.

In general problem, an unentangled solution is not guaranteed to exist.
However, basis of solution space may be guaranteed to consist of unentangled solutions if there are infinite number of unentangled solutions.

In this survey, I see that perhaps Uda equation does not excel the old theory very much and the above result perhaps is enough as technical base.

I should try applying Stacked Daruma Game Formula to an entangled quantum history given as a general solution in this page.

By considering applying Daruma formula, I came to recognize necessity of normalization of state vector at each time.
This means that a0 = b0 = 0 is impossible.
Quantum state at time t is Σn=0exp(-iωnt + an + ibn)|n＞ = Σn=0cnexp(-iωnt)|n＞
where cn ≡ exp(an + ibn).
Normalization condition is that Σn=0exp(2an) = 1.
∴ an ＜ 0 for all n.
-iωtν(t) + ibν(t) ～ 0 (t → ±∞).
bν(t)/ν(t) ～ ωt (t → ±∞).
This condition restricts ν.
If ∀L; ∃n; bn/n ＞ L, such ν may exist.
aν(t) → 0 (t → ±∞).
If ∀ε ＞ 0; ∃n; -ε ＜ an ＜ 0, such ν may exist.

ν'(t') = ν(t') if t' ＜ a.
ν'(t') = ν(a) if a ≦ t' ＜ a + ε.
ν'(t') = ν(t' - ε) if t' ≧ a + ε.

ν''(t'') = ν(t'') if t'' ＜ b.
ν''(t'') = ν(b) if b ≦ t'' ＜ b + ε.
ν''(t'') = ν(t'' - ε) if t'' ≧ b + ε.

WRONG:
＜ν(b) |ν(a)＞Daruma
= [Πk=-∞Σn(k)=0] exp[iωαε∫abdt ν(t)] Φ[ν'']* Φ[ν']. (2019/11/04/18:41JST)(FTP19:18JST)

ε = 1/α.
b = a + Nε.
ν(t) = nk if a + kε ≦ t ＜ a + (k + 1)ε.

Φ[ν'] = Σs exp{α∫-∞dt'[-iωt'ν'(t') + a(s)ν'(t') + ib(s)ν'(t')]}
= Σs exp{α∫-∞adt'[-iωt'ν(t') + a(s)ν(t') + ib(s)ν(t')]
+ α∫aa +εdt'[-iωt'ν(a) + a(s)ν(a) + ib(s)ν(a)]
+ α∫a +εdt'[-iωt'ν(t' - ε) + a(s)ν(t' -ε) + ib(s)ν(t' -ε)]}
= Σs exp{α∫-∞adt[-iωtν(t) + a(s)ν(t) + ib(s)ν(t)]
+ [-iω(a +ε/2)ν(a) + a(s)ν(a) + ib(s)ν(a)]
+ α∫a dt[-iω(t +ε)ν(t) + a(s)ν(t) + ib(s)ν(t)]}
= exp[-iωαε∫a dt ν(t)]
×Σs exp[-iω(a +ε/2)ν(a) + a(s)ν(a) + ib(s)ν(a)] exp{α∫-∞dt[-iωtν(t) + a(s)ν(t) + ib(s)ν(t)].

Φ[ν''] = Σr exp{α∫-∞dt''[-iωt''ν''(t'') + a(r)ν''(t'') + ib(r)ν''(t'')]}
= Σr exp{α∫-∞bdt''[-iωt''ν(t'') + a(r)ν(t'') + ib(r)ν(t'')]
+ α∫bb +εdt''[-iωt''ν(b) + a(r)ν(b) + ib(r)ν(b)]
+ α∫b +εdt''[-iωt''ν(t'' - ε) + a(r)ν(t'' -ε) + ib(r)ν(t'' -ε)]}
= Σr exp{α∫-∞bdt[-iωtν(t) + a(r)ν(t) + ib(r)ν(t)]
+ [-iω(b +ε/2)ν(b) + a(r)ν(b) + ib(r)ν(b)]
+ α∫b dt[-iω(t +ε)ν(t) + a(r)ν(t) + ib(r)ν(t)]}
= exp[-iωαε∫b dt ν(t)]
×Σr exp[-iω(b +ε/2)ν(b) + a(r)ν(b) + ib(r)ν(b)] exp{α∫-∞dt[-iωtν(t) + a(r)ν(t) + ib(r)ν(t)].

Φ[ν'']* Φ[ν']
= exp[-iωαε∫a bdt ν(t)]ΣsΣr φr(ν(b), b +ε/2)* φs(ν(a), a +ε/2) Πk=-∞c(r)n(k)* c(s)n(k).

WRONG:
＜ν(b) |ν(a)＞Daruma
= [Πk=-∞Σn(k)=0sΣr φr(ν(b), b +ε/2)* φs(ν(a), a +ε/2) Πk=-∞c(r)n(k)* c(s)n(k)
= ΣsΣr φr(ν(b), b +ε/2)* φs(ν(a), a +ε/2) (δrs)
= Σs φs(ν(b), b +ε/2)* φs(ν(a), a +ε/2)
= ＜ν(b)|exp[(-i/h)(b +ε/2)H]|us* ＜ν(a)|exp[(-i/h)(a +ε/2)H]|us
= Σs ＜ν(a)|exp[(-i/h)(a +ε/2)H]|us＞＜us|exp[(i/h)(b +ε/2)H]|ν(b)＞
= ＜ν(a)|exp[(-i/h)(a +ε/2)H]exp[(i/h)(b +ε/2)H]|ν(b)＞
= ＜ν(a)|exp[(-i/h)(a - b)H]|ν(b)＞
= Gold(ν(a), a; ν(b), b). (2019/11/04/18:41JST)(FTP19:18JST)

|us＞ = Σn=0c(s)n|n＞.
{ |us＞ | ∃s } is a basis of the state space.
＜us | ur＞ = δrs.

|v(k)s＞ = Σn=0c(k, s)n|n＞
{ |v(k)s＞ | ∃s } is a basis of the state space for each k.
＜v(k)s | v(k)r＞ = δrs for each k.

| |φ> | = 1 and | |ψ> | = 1 and not ∃c ∈ C; |φ> = c|ψ> ⇒ | ＜φ|ψ＞ | ＜ 1.

WRONG:
If j ≠ k ⇒ ∀r; ∀s; not ∃c ∈ C; |v(j)r＞ = c|v(k)s＞,
Φ[ν] = ΣkΣs exp{α∫-∞dt[-iωtν(t) + a(k, s)ν(t) + ib(k, s)ν(t)]}
⇒ ＜ν(b) |ν(a)＞Daruma = (number of k) × Gold(ν(a), a; ν(b), b). (2019/11/04/18:41JST)(FTP19:18JST)

c(k, s)n ≡ exp(a(k, s)n + ib(k, s)n).

CORRECT:
＜ν(b) |ν(a)＞Daruma
= [Πk=-∞Σn(k)=0] exp[-iωαε∫abdt ν(t)] Φ[ν''] Φ[ν']*
. (2019/11/05/08:46JST)

CORRECT:
＜ν(b) |ν(a)＞Daruma
= ＜ν(b)|exp[(-i/h)(b - a)H]|ν(a)＞
= Gold(ν(b), b; ν(a), a). (2019/11/05/08:48JST).

CORRECT:
If j ≠ k ⇒ ∀r; ∀s; not ∃c ∈ C; |v(j)r＞ = c|v(k)s＞,
Φ[ν] = ΣkΣs exp{α∫-∞dt[-iωtν(t) + a(k, s)ν(t) + ib(k, s)ν(t)]}
＜ν(b) |ν(a)＞Daruma = (number of k) × Gold(ν(b), b; ν(a), a). (2019/11/05/08:49JST)

If we think of |Φ[ν]|2 as something like probability that the history is ν, physical meaning of α is as follows.
Φ[ν] = exp{α∫-∞dt[-iωtν(t) + aν(t) + ibν(t)]}
⇒ [⊿/⊿ν(a)] |Φ[ν]|2 = 2α[aν(a)+1 - aν(a)] |Φ[ν]|2　・・・ (2019/11/05/09:07JST)
= αln[|cν(a)+1|2/|cν(a)|2]　・・・ (2019/11/05/09:44JST)

Author Yuichi Uda, Write start at 2019/10/14/15:01JST, Last edit at 2019/11/05/09:44JST