Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Okay, I know that in quantum mechanics the quantum observable is obtained from the classical observable by the prescription

$$ X \rightarrow x,\quad P \rightarrow -i\hbar\frac{\partial}{\partial x} $$

in the position basis. Now my question is, what if $x$ or $p$ appears in the denominator in a classical expression? How to promote this to a quantum expression? What would be the meaning of division by an operator?

Edit: Thank you for your responses. My expression likely contains a mixture of x and p. For eg., it could contain terms like $$\frac{p}{x^2}$$ or $$\frac{xp}{(x^2 + a^2)^{3/2}}$$. How to resolve products of non-commuting operators like x,p in a satisfactory way?

share|improve this question
    
Weyl quantization is e.g. discussed in this Phys.SE post. –  Qmechanic Jun 24 '13 at 8:38
add comment

2 Answers 2

The general problem of converting classical expressions to quantum operator ones is in general unsolvable because classical mechanics is an approximation to quantum mechanics and not the other way around. There is always an ambiguity in how to order noncommuting operators. You have to handle it on a case by case basis, and there are a number of "quantization" schemes out there. In general these can lead to different quantum theories which have to be distinguished experimentally.

Anyway, in your case it is probably fine just to use the eigenvalue decomposition:

$$ \frac{1}{x} \to \frac{1}{\hat{x}} \equiv \int \mathrm{d}x\ |x\rangle \frac{1}{x} \langle x |, $$

$$ \frac{1}{p} \to \frac{1}{\hat{p}} \equiv \int \mathrm{d}p\ |p\rangle \frac{1}{p} \langle p |, $$

etc., where $|x\rangle,|p\rangle$ are the orthonormal eigenvectors of position and momentum resp. You can use $\langle x|x'\rangle=\delta(x-x')$ to show that $\frac{1}{\hat{x}}$ has the desired action on position eigenstates. You can also clearly generalise this sort of thing, e.g. $\sqrt{p}\to\sqrt{\hat{p}}\equiv\int \mathrm{d}p\ |p\rangle \sqrt{p} \langle p |$. To give a real example, the following operator, called the resolvent, is very important in quantum scattering theory:

$$ \hat{R}(z) = \frac{1}{\hat{H}-z} = \sum_n \frac{|n\rangle\langle n|}{E_n - z}, $$

where $z$ is a complex number.

You'll have ambiguities if the classical expression is something like $p/x$ or $\sqrt{px}$ or whatever, since $\hat{p}$ and $\hat{x}$ don't commute.

share|improve this answer
2  
@guru : The logic of the answer of Michael Brown can be extended to any function $f(x)$ or $g(p)$, which is the associated with the operators $f(\hat x)$ and $g(\hat p)$. For mixed (x,p )quantities, the most popular quantization procedure is Weyl Quantization. –  Trimok Jun 21 '13 at 9:12
    
One thing one has to consider though is that e.g. $x$ and $p$ don't have a common base (yielding the uncertainty principle), so if you have an expression depending on both of them you may have to find a different Eigenbase first –  Tobias Kienzler Jun 21 '13 at 14:14
add comment

A different approach from what Michael Brown wrote is to use a Taylor expansion in order to turn your function of the operator into a polynomial. You can then - in principle - evaluate the action of each term on your states and then contract the expression again. This effectively leads to the same expressions as Michael Borwn's approach but you might be more comfortable with it.

share|improve this answer
1  
...only that in this case worrying about the convergence of the expansion, both for values and operators inserted, may give you your headache back... –  Tobias Kienzler Jun 21 '13 at 12:22
1  
True. It's still closer to a physicist's intuition imo –  Neuneck Jun 21 '13 at 14:02
    
Agreed. Plus this way one might notice issues with non-commuting operators –  Tobias Kienzler Jun 21 '13 at 14:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.