Insights The Classical Limit of Quantum Mechanical Commutator

[samalkhaiat]

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 \frac{\partial S}{\partial t} + H \left( x , \frac{\partial S}{\partial x}\right) = 0, shows up as the classical limit, $\hbar \to 0$ , of the Schrodinger wave equation i\hbar \frac{\partial \Psi}{\partial t} – H \left(x , – i \hbar \frac{\partial}{\partial x}\right) \Psi = 0.Textbooks also state, but without proof, that the equation of motion of a classical observable A(x,p),\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}is the classical limit of Heisenberg...

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[A. Neumaier]

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)!

 

[samalkhaiat]

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.

 

[samalkhaiat]

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) ..."

 

[A. Neumaier]

If you like, you can write $a(x,y)$ then write $A(x,p)$ for the Fourier amplitude.

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!

 

[A. Neumaier]

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.

 

[samalkhaiat]

You can (and should) make these corrections yourself!

I did it, thanks to Greg's help.

 

[samalkhaiat]

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}.​

 

[samalkhaiat]

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, xy – yx = \frac{ih}{2 \pi} \{x , p \}.

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 \hat{\pi} (f) \hat{\pi} (g) - \hat{\pi} (g) \hat{\pi} (f) = i \hbar \hat{\pi} \left( \{ f , g \}\right) , and \hat{\pi} (q) = Q , \ \ \hat{\pi} (p) = P , 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 3x^{2}p^{2} = \big\{x^{2}p , p^{2}x \big\} = \frac{1}{3} \big\{x^{3} , p^{3} \big\} .