From my previous question the objective was to prove eqn. $(9)$ below:
$$\left(A_c \ast B_c\right)(x, p)=A_c\left(x,p\right)B_c\left(x,p\right)-i\hbar\frac{\partial A_c}{\partial p}\frac{\partial B_c}{\partial x}+\mathcal{O}\left(\hbar^2\right)\tag{9}$$
For now, taking eqn. $(9)$ for the moment to be true, I now need to prove the next question:
Use the above result to prove Dirac's famous relation between commutators and Poisson brackets:
$$\lim_{\hbar \to 0}\frac{1}{i\hbar}\left([\hat A, \hat B]\right)_c \to \left[\left[A_c, B_c\right]\right]_{\mathrm{P.B}}\tag{A}$$where the commutator is $\left[\hat A, \hat B \right]=\hat A \hat B- \hat B \hat A$ and $\left([\hat A, \hat B]\right)_c$ denotes the c-number associated with the commutator,
$$\left([\hat A, \hat B]\right)_c=\left(A_c \ast B_c\right)-\left(B_c \ast A_c\right)\tag{B}$$
and the Poisson bracket is defined by
$$\left[\left[A, B\right]\right]_{\mathrm{P.B}}=\frac{\partial A}{\partial x}\frac{\partial B}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial B}{\partial x}\tag{C}$$
Dirac’s identification of Poisson brackets with commutators is the basis of the canonical quantization framework and its extension for Gauge theories, the Dirac Quantization procedure. In short, it states that to quantize a theory, you write down the classical equations of motion in Poisson bracket language, and replace variables with operators and the Poisson brackets with commutators. This is still the easiest way to think about quantizing field theories, strings and gravity.
At first sight, this looks incredibly straightforward with just two applications of eqn. $(9)$ and subtracting the difference:
So,
$$\left(A_c \ast B_c\right)-\left(B_c \ast A_c\right)=\left([\hat A, \hat B]\right)_c$$
$$=A_c\left(x,p\right)B_c\left(x,p\right)-i\hbar\frac{\partial A_c}{\partial p}\frac{\partial B_c}{\partial x}+\mathcal{O}\left(\hbar^2\right)$$
$$-B_c\left(x,p\right)A_c\left(x,p\right)+i\hbar\frac{\partial B_c}{\partial p}\frac{\partial A_c}{\partial x}-\mathcal{O}\left(\hbar^2\right)$$
The first and fourth terms cancel as they are just classical numbers and commute, from this I find that,
$$\left([\hat A, \hat B]\right)_c=i\hbar\bigg(\frac{\partial A_c}{\partial x}\frac{\partial B_c}{\partial p}-\frac{\partial A_c}{\partial p}\frac{\partial B_c}{\partial x}\bigg)=i\hbar\left[\left[A, B\right]\right]_{\mathrm{P.B}}\tag{D}$$
In $(\mathrm{D})$ the $\mathcal{O}\left(\hbar^2\right)$ cancel and rearranging this equation leads to
$$\frac{1}{i\hbar}\left([\hat A, \hat B]\right)_c=\left[\left[A, B\right]\right]_{\mathrm{P.B}}\tag{E}$$
Taking the limit $\hbar \to 0$ gives
$$\lim_{\hbar \to 0}\frac{1}{i\hbar}\left([\hat A, \hat B]\right)_c=\left[\left[A, B\right]\right]_{\mathrm{P.B}}\tag{F}$$
While this gives the correct relation to be proved, $(\mathrm{A})$, does it do so for the right reasons?
This is what the authors' handwritten solution says:
So I have two questions regarding this solution above:
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Why is there still a $\mathcal{O}\left(\hbar^2\right)$ when they clearly cancel when you take the difference (boxed in red)?
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If $\hbar \to 0$ then the LHS of $(\mathrm{F})$, (or the last line in the author's solution) is undefined, so how can one possibly take the limit that $\hbar \to 0$ when it is in the denominator?
Best Answer
The $O(\hbar^2)$ denotes an error term, which is some (convenient) abuse of notation. The difference of two $O(\hbar^2)$ error terms is again of order $O(\hbar^2)$; we cannot say it is zero.
The limit $\hbar \to 0$ of a function (or an expression) in $\hbar$ can be defined even if the function itself is not defined at the limit value (here $\hbar=0$); see any textbook on limits. Here one can easily do the division for $\hbar\neq 0$ and take the limit $\hbar \to 0$, using the expression in the next-to-last line in the author's solution.