daily topic on quantum mechanics

[Anonymous No.7176020]

guys,
if we define the projectors operators as ρ : | ψ>< ψ| /< ψ| ψ>

the equation of motion on the states in the schroendinger picture becomes :

i ħ d/dt ρ = [H;ρ]

is it correct to say that the equation of motion in the same formalism on the operators/observables A
is

i ħ d/dt A = [A;H]
?

from my point of view both ρ and A are operators and thus they should not be in different spaces ( the dual)

but an observable on a state should give us a number so it belongs to the dual, and that's would explain the minus sign in the second equation


guys pls

[Anonymous No.7176076]

i don't know any of those symbols. Try asking Jesus.

[Anonymous No.7176088]

>>7176076
that fucker is busy with that resurrection thing right now

[Anonymous No.7176102]

>>7176088

[Anonymous No.7176145]

>>7176020
In the schrodinger picture, the observable are time independent, and the state vector and it's density matrix are time dependent.

In the Heisenberg picture, the opposite is true.

The time dependence of the density matrix in the Schrodinger picture is NOT the same as the time dependence of the observables in the Heisenberg picture.

[Anonymous No.7176166]

>>7176020
Nice. You found Heisenberg's equation of motion.

[Anonymous No.7176167]

>>7176145
sorry but i didnt understand how this answer the question

we never used the density matrix but i understand the part about the time dependence

[Anonymous No.7176175]

>>7176167
>we never used the density matrix
The projection operator is the density matrix (for a pure state).

[Anonymous No.7176181]

>>7176166
b-b-but the heisenberg equation is ( at least on my book)

i ħ d/dt A = [H;A] on the observables

while ON the states, being positive linear functionals with few other features,
is
i ħ d/dt ρ = [ ρ;H]


fuuuck i am losing my sanity over this thing

[Anonymous No.7176186]

>>7176181
Then something's not right. If you're just playing around with the projection then it's probably something special that the projection has.

[Anonymous No.7176190]

>>7176181
>i ħ d/dt ρ = [ ρ;H]
This is wrong. It should be [H,ρ]

[Anonymous No.7176205]

first page

[Anonymous No.7176208]

second page

[Anonymous No.7176214]

>>7176190
but if i write the equation on the dual space i is replaced with -1 and the commutator has the property that
-[a,b] = [b,a]

[Anonymous No.7176218]

>>7176205
>>7176208
Whoever wrote that book must have fucked up. Their equations are wrong.

Heisenberg-picture observable:
i ħ ∂A/∂t = [A,H]

Density matrix in Schrodinger picture:
i ħ ∂ρ/∂t = [H,ρ]

[Anonymous No.7176227]

>>7176218

shit , it's already hard enough...

[Anonymous No.7176238]

>>7176214
>>7176227
Derivation of i ħ ∂ρ/∂t = [H,ρ] :

<span class="math">\rho = \displaystyle \left| \psi \right\rangle \left\langle \psi \right|[/spoiler]
<span class="math">\rho (t) = \displaystyle e^{-iHt/\hbar} \left| \psi_0 \right\rangle \left\langle \psi_0 \right| e^{+iHt/\hbar}[/spoiler]
<span class="math">\frac{d\rho}{dt} = \displaystyle -\frac{i}{\hbar}H e^{-iHt/\hbar} \left| \psi_0 \right\rangle \left\langle \psi_0 \right| e^{+iHt/\hbar} + e^{-iHt/\hbar} \left| \psi_0 \right\rangle \left\langle \psi_0 \right| e^{+iHt/\hbar} \frac{i}{\hbar}H [/spoiler]
<span class="math">\frac{d\rho}{dt} = \displaystyle \frac{i}{h}(- H\rho + \rho H )[/spoiler]
<span class="math">i\hbar \displaystyle \frac{d\rho}{dt} = (H\rho - \rho H )[/spoiler]
<span class="math">i\hbar \displaystyle \frac{d\rho}{dt} = [H,\rho] [/spoiler]

[Anonymous No.7176239]

>>7176218
btw wikipedia report
i ħ ∂A/∂t = [H,A]

https://en.wikipedia.org/wiki/Heisenberg_picture

[Anonymous No.7176244]

>>7176239
Wikipedia says
<span class="math">\frac{dA}{dt} = \frac{i}{\hbar} [H,A][/spoiler]
which is the same as
<span class="math">i\hbar\frac{dA}{dt} = [A,H][/spoiler]

[Anonymous No.7176258]

>>7176238
k but that's not the problem, the equation of motion on the projectors in the schroedinger picture give me no problems

the problem is that my teqcher said ,( truth is maybe i wrote it wrong ) that equation of motion on the generic operators in the schroedinger picture is :

i ħ ∂A/∂t = [A,H]

as if a generic operator is in the dual space of a projector .
even if the generic operator does not depend on time while projectors do, they should not be in different spaces however.

[Anonymous No.7176262]

>>7176258
>i ħ dA/dt = [A,H]
Yes, this is the correct equation for a Heisenberg-picture operator. What is wrong with it?

>as if a generic operator is in the dual space of a projector
This is nonsensical. Linear operators don't have dual spaces.

[Anonymous No.7176273]

>this thread
>I will probably never be close to understanding what any of this means
>mfw it doesnt bother me at all

[Anonymous No.7176275]

>>7176262
how he obtained it :

he took i ħ d/dt ρ = [H;ρ] with ρ : | ψ>< ψ| /< ψ| ψ>

which is an equation on states projectors

generalized it for a convex combination


then said " what is the related equation on the dual space ?"

he wrote

i ħ ∂A/∂t = [A,H]

where A is again an operator

for what i know if i have an equation on vectors and write the dual, i find an equation on the functionals and viceversa, how can he still be in the space of the operators ?

[Anonymous No.7176313]

>>7176275
"Dual space" doesn't make sense for linear operators.

A is an operator, A(t) is an operator-valued function of time, ρ(t) is an operator-valued function of time.
And for all t, all of the operators A, A(t), and ρ(t) live in the space of operators acting on the same Hilbert space.

[Anonymous No.7176321]

>>7176262
let's recap :

SCHROEDINGER :


i ħ ∂|A>/∂t = H |A> ON THE vectors of the hilbert space

i ħ ∂ρ/∂t = [H,ρ] on the projectors

on the operators : ???????? teacher said i ħ dA/dt = [A,H] but i dont' know why,maybe the ehrenfest theorem ?

HEISENBERG picture :

i ħ ∂A/∂t = [A,H] on the observables , now considered as elements of a C*algebra

i ħ ∂ρ/∂t = [H,ρ] on the functionals , ex- states in the schroedinger picture

am i right ?

[Anonymous No.7176325]

>>7176321
All those equations are correct.

>C*algebra
Sorry, I don't know much about the C*-algebra formulation of quantum mechanics.

[Anonymous No.7176336]

>>7176325
do you know how to justify the passage from eq 2 to eq 3 without using the heisenberg picture ?

[Anonymous No.7176380]

>>7176321
>am i right ?
yes. states and observables have that very important sign change in their evolution equations.