In summary: Please be more explicit. They are not well-known enough that I'd know them.Please be more explicit. They are not well-known enough that I'd know them.
  • #1
samalkhaiat
Science Advisor
Insights Author
1,802
1,200
The Classical Limit of Commutator (without fancy mathematics)
Quantum mechanics occupies a very unusual place among physical theories: It contains classical mechanics as a limiting case, yet at the same time it requires this limiting case for its own formulation.
Many textbooks on elementary QM show you how the Hamilton-Jacobi [tex]\frac{\partial S}{\partial t} + H \left( x , \frac{\partial S}{\partial x}\right) = 0,[/tex] shows up as the classical limit, ##\hbar \to 0## , of the Schrodinger wave equation [tex]i\hbar \frac{\partial \Psi}{\partial t} – H \left(x , – i \hbar \frac{\partial}{\partial x}\right) \Psi = 0.[/tex]Textbooks also state, but without proof, that the equation of motion of a classical observable [itex]A(x,p)[/itex],[tex]\begin{equation}\frac{d}{dt}A(x,p) = – \big\{ H , A \big\}(x,p) = \frac{\partial H}{\partial p}\frac{\partial A}{\partial x} – \frac{\partial H}{\partial x}\frac{\partial A}{\partial p},\end{equation}[/tex]is the classical limit of Heisenberg...

Continue reading...
 
Last edited by a moderator:
  • Like
  • Informative
Likes atyy, bhobba, kith and 5 others
Physics news on Phys.org
  • #2
1. It is slightly confusing to use the same symbol ##A## for ##A(u,v)## with two spatial arguments and for its Fourier transform ##A(y,p)## with respect to ##v##.

2. ''If you work hard, you can do what Heisenberg himself was unable to do.'' ... on Oct 23rd 1925, the date of his letter to Pauli.

But Dirac could do it on Nov 7th 1925, two weeks later, without using integrals (see Section 4 of the reference)!
 
  • #3
A. Neumaier said:
1. It is slightly confusing to use the same symbol ##A## for ##A(u,v)## with two spatial arguments and for its Fourier transform ##A(y,p)## with respect to ##v##.
If you like, you can write ##a(x,y)## then write ##A(x,p)## for the Fourier amplitude.

2. ''If you work hard, you can do what Heisenberg himself was unable to do.'' ... on Oct 23rd 1925, the date of his letter to Pauli.

But Dirac could do it on Nov 7th 1925, two weeks later, without using integrals (see Section 4 of the reference)!
There are well known technical issues with Dirac's reasoning.
 
  • #4
Few corrections:
The RHS of equation (2) should be ##\frac{i}{\hbar}(\hat{H}\hat{A} - \hat{A}\hat{H})##.
The line after equation (15) should read "..., let's rewrite (15) in ...".
The line after equation (18) should read "... by equating the Fourier amplitudes in (18) and (16) ..."
 
  • #5
samalkhaiat said:
If you like, you can write ##a(x,y)## then write ##A(x,p)## for the Fourier amplitude.
samalkhaiat said:
Few corrections:
The RHS of equation (2) should be ##\frac{i}{\hbar}(\hat{H}\hat{A} - \hat{A}\hat{H})##.
The line after equation (15) should read "..., let's rewrite (15) in ...".
The line after equation (18) should read "... by equating the Fourier amplitudes in (18) and (16) ..."
You can (and should) make these corrections yourself!
 
  • #6
samalkhaiat said:
There are well known technical issues with Dirac's reasoning.
Please be more explicit. They are not well-known enough that I'd know them.
 
  • #7
A. Neumaier said:
You can (and should) make these corrections yourself!
I did it, thanks to Greg's help.
 
  • #8
A. Neumaier said:
Please be more explicit. They are not well-known enough that I'd know them.
Dirac was perfect but not absolutely perfect, and one should not take everything Dirac wrote as being “carved in stone”.
In the “reference paper” and in his famous book, Dirac makes the same wrong assumption: (in the quote bellow, I am changing Dirac’s notation for the Poisson bracket from [x , y] to {x , y})
Dirac writes in section 4
We make the fundamental assumption that the difference between the Heisenberg products of two quantum quantities is equal to ih2πih2π times their Poisson bracket expression. In symbols,

xy–yx=ih2π{x,y}. (11)xy–yx=ih2π{x,y}. (11)
This is not correct as you can see from equation (19) in the insight, there are very complicated higher order (in ℏℏ) terms. This was highlighted by the following very well-known no-go theorem due to Groenewold and van Hove:
There is no map f→^π(f)f→π^(f) from polynomials on R2R2 to self-adjoint operators on L2(R)L2(R) satisfying

^π(f)^π(g)−^π(g)^π(f)=iℏ ^π({f,g}),π^(f)π^(g)−π^(g)π^(f)=iℏ π^({f,g}),

and


π(q)=Q, π(p)=P,π(q)=Q, π(p)=P,

for any Lie subalgebra of the functions on R2R2 larger than the subalgebra of polynomials of degree ≤2≤2.

The proof uses the fact that the same (phase space) function can have two different expression as Poisson bracket. For instance

3x2p2={x2p,p2x}=13{x3,p3}.3x2p2={x2p,p2x}=13{x3,p3}.​
 
Last edited:
  • #9
I am posting this again for the forum section only because for some reason the equations have gone funny.
Dirac was perfect but not absolutely perfect, and one should not take everything Dirac wrote as being “carved in stone”.
In the “reference paper” and in his famous book, Dirac makes the same wrong assumption: (in the quote bellow, I am changing Dirac’s notation for the Poisson bracket from [x , y] to {x , y})
Dirac writes in section 4
We make the fundamental assumption that the difference between the Heisenberg products of two quantum quantities is equal to ##\frac{ih}{2 \pi}## times their Poisson bracket expression. In symbols, [tex]xy – yx = \frac{ih}{2 \pi} \{x , p \}.[/tex]
This is not correct as you can see from equation (19) in the insight, there are very complicated higher order (in ##\hbar##) terms. This was highlighted by the following very well-known no-go theorem due to Groenewold and van Hove:
There is no map ##f \to \hat{\pi} (f)## from polynomials on ##\mathbb{R}^{2}## to self-adjoint operators on ##L^{2}(\mathbb{R})## satisfying [tex]\hat{\pi} (f) \hat{\pi} (g) - \hat{\pi} (g) \hat{\pi} (f) = i \hbar \hat{\pi} \left( \{ f , g \}\right) ,[/tex] and [tex]\hat{\pi} (q) = Q , \ \ \hat{\pi} (p) = P ,[/tex] for any Lie subalgebra of the functions on ##\mathbb{R}^{2}## larger than the subalgebra of polynomials of degree ##\leq 2##.
The proof uses the fact that the same (phase space) function can have two different expression as Poisson bracket. For instance [tex]3x^{2}p^{2} = \big\{x^{2}p , p^{2}x \big\} = \frac{1}{3} \big\{x^{3} , p^{3} \big\} .[/tex]
 
  • Like
Likes bhobba, Spinnor, arivero and 2 others

1. What is the classical limit of quantum mechanical commutator?

The classical limit of quantum mechanical commutator refers to the point at which the principles of classical mechanics, which describe the behavior of macroscopic objects, can be applied to the behavior of quantum mechanical systems. This occurs when the quantum commutator, which describes the non-commutativity of certain physical quantities in quantum mechanics, approaches zero.

2. Why is the classical limit of quantum mechanical commutator important?

The classical limit of quantum mechanical commutator is important because it allows us to bridge the gap between the behavior of microscopic particles described by quantum mechanics and the behavior of larger objects described by classical mechanics. It also helps us to understand the behavior of quantum systems in the macroscopic world.

3. How is the classical limit of quantum mechanical commutator calculated?

The classical limit of quantum mechanical commutator is calculated by taking the limit of the quantum commutator as the Planck's constant, h, approaches zero. This is known as the correspondence principle, which states that the laws of quantum mechanics should approach the laws of classical mechanics in the limit of large quantum numbers or small values of h.

4. What are some examples of the classical limit of quantum mechanical commutator?

One example of the classical limit of quantum mechanical commutator is the Heisenberg uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. In the classical limit, this uncertainty disappears and we can know both the position and momentum of a particle with arbitrary precision.

Another example is the commutation relation between position and momentum operators, which becomes the Poisson bracket in the classical limit. This allows us to use the Hamiltonian formalism of classical mechanics to describe the dynamics of quantum systems.

5. What are the implications of the classical limit of quantum mechanical commutator?

The implications of the classical limit of quantum mechanical commutator are vast and have led to many important developments in physics. It allows us to understand the behavior of quantum systems in the macroscopic world, and has applications in fields such as quantum computing, quantum cryptography, and quantum metrology. It also helps us to reconcile the seemingly contradictory principles of quantum mechanics and classical mechanics, leading to a deeper understanding of the fundamental laws of nature.

Similar threads

Replies
7
Views
458
  • Quantum Physics
Replies
6
Views
962
Replies
3
Views
282
  • Quantum Physics
Replies
9
Views
787
Replies
3
Views
774
Replies
17
Views
1K
Replies
5
Views
773
Replies
4
Views
1K
Replies
15
Views
2K
Replies
2
Views
1K
Back
Top