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In classical mechanics, we define the concept of canonical momentum conjugate to a given generalised position coordinate. This quantity is the partial derivative of the Lagrangian of the system, with respect to the generalised velocity.

My first question is as follows:

Given a quantum mechanical system (specified fully by a quantum mechanical Hamiltonian), and given a generalised "position" operator $Q$ (which may not necessarily be a simple $x$ or $y$ coordinate), is there a systematic process for deriving the quantum-mechanical operator $P$, analogous to the canonical conjugate momentum to $Q$?

My second question is:

Assuming that $P$ exists for a given $Q$, is it true that $[Q, P] = i\hbar$? In the same way that it is true for the special case of the linear momentum operator and the linear position operator?

I keep reading that $[q, p] = i \hbar$ is supposed to be postulate of quantum mechanics. However, if there DID exist a systematic process for deriving P from a given $Q$, then - given a specific $P$, $Q$ (and $H$, if relevant) - you should be able to verify that this is the case.

Thank you in advance for any help you guys can offer. These questions have been completely driving me insane.

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1 Answer

up vote 2 down vote accepted

Brief explanation:

  1. When going from classical Lagrangian (say, non-relativistic point-)mechanics to quantum mechanics, there is an intermediate step known as classical Hamiltonian mechanics.

  2. To reach the intermediate step, one has to perform a Legendre transformation $(q,\dot{q}) \longrightarrow (q, p)$, where $(q, p)$ are (generalized) canonical phase space variables.

  3. Note in particular, that while the generalized momentum is defined as $p_j = \frac{\partial L}{\partial \dot{q}^j}$ in Lagrangian mechanics, the generalized momentum is a free variable in Hamiltonian mechanics (as long as the Legendre transformation is not singular).

  4. In the classical Hamiltonian formalism the $q^i$ and $p_j$ satisfy the canonical Poisson bracket relations $\{q^i,p_j\}=\delta^i_j.$

  5. In the quantization process, the Poisson bracket relations get replaced by canonical commutation relations $[\hat{q}^i,\hat{p}_j]=i\hbar {\bf 1}\delta^i_j$. (This part of what is known as the correspondence principle between classical and quantum mechanics.)

  6. In the position representation $\hat{q}^i=q^i$ and $\hat{p}_j= \frac{\hbar}{i}\frac{\partial}{\partial q^j}$. (This representation is known as the Schrödinger representation. See also Stone-von Neumann Theorem.)

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Also, I believe that you cant just take any set of canonical coordinates for this procedure -- quantization is not that simple. I mean, it makes a difference, unless you do it in a smarter way. –  Peter Kravchuk Apr 20 '13 at 16:29
Yes, in general, there are also e.g. operator ordering ambiguities and topological issues. And if the Legendre transformation is singular, we get into the realm of constrained dynamics. This answer is meant as a brief introduction (rather than a complete explanation) to a huge topic. –  Qmechanic Apr 20 '13 at 16:40
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