# Quantum mechanical analogue of conjugate momentum

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.

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.)

• 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