# The commutator of position and momentum - interpretation?

1. Sep 9, 2012

### Kontilera

Hello friends!
Im trying to get an geometric interpretation of QM and am now confused about the commutation relation between operators.
Lets take momentum and position... sure, the fact that they dont commute show that we can not diagonalize them simultanesly.
But what is the interpretation of the matrix that we get we we take the commutator? (i hbar I.)

Is it the difference we get when we change the order of a translation in position space and a translation i momentumspace?
(I cannot help to try to relate this commutator relation to the fact that P is used when defining the unitary operator for position translation etc.)

All the best!
/ Kontilera

2. Sep 9, 2012

### Kontilera

Sakurai refers to classical mechanics to explain why we should expect momentum to be the generator of translation in position space, is there any nice mathematical way to see this? Or is it a physical postulate that I should accept to work?

3. Sep 9, 2012

### Dickfore

The semi-classical form of the wave function of a particle is:
$$\Psi(\mathbf{x}, t) \equiv \langle \mathbf{x} \vert \Psi, t \rangle \sim \exp \left( \frac{i}{\hbar} S(\mathbf{x}, t) \right)$$
This should be accepted as a kind of a postulate of Quantum Mechanics, which is the mathematical formulation of Bohr's Correspondence Principle, and a starting point in deriving the WKB approximation to the Schroedinger equation.

Now, as you said, the action of an infinitesimal translation operator is:
$$\hat{T}(\delta \mathbf{a}) \, \left\vert \mathbf{x} \right\rangle \equiv \left\vert \mathbf{x} + \delta\mathbf{a} \right\rangle \approx \left[ \hat{1} - \frac{i}{\hbar} \, \delta \mathbf{a} \cdot \hat{\mathbf{P}} \right] \, \left\vert \mathbf{x} \right\rangle$$
Multiplying from the left by a state bra $\langle \psi \vert$, we get:
$$\langle \psi \vert \mathbf{x} + \delta\mathbf{a} \rangle \approx \langle \psi \vert \mathbf{x} \rangle - \frac{i}{\hbar} \delta\mathbf{a} \cdot \langle \psi \vert \hat{\mathbf{P}} \vert \mathbf{x} \rangle$$
Taking the hermitian adjoint of the above equality and using Taylor's formula:
$$\langle \mathbf{x} + \delta\mathbf{a} \vert \psi \rangle \approx \langle \mathbf{x} \vert \psi \rangle + \delta\mathbf{a} \cdot \nabla \langle \mathbf{x} \vert \psi \rangle$$
we get the following expression for the action of the generator of translations in x-representation:
$$\left\langle \mathbf{x} \right\vert \hat{\mathbf{P}} \left\vert \psi \right\rangle = \frac{\hbar}{i} \, \nabla \, \left\langle \mathbf{x} \right\vert \psi \left. \right\rangle$$
On the other hand, the gradient of the semiclassical wave function is:
$$\nabla \Psi(\mathbf{x}, t) \sim \frac{i}{\hbar} \nabla S(\mathbf{x}, t) \, \Psi(\mathbf{x}, t)$$
Thus, we have:
$$\left\langle \mathbf{x} \right\vert \hat{\mathbf{P}} \left\vert \Psi, t\right\rangle \sim \nabla S(\mathbf{x}, t) \, \left\langle \mathbf{x} \right\vert \Psi, t \left. \right\rangle$$

But, in classical mechanics, we know that the gradient of the action w.r.t. to the endpoint of the trajectory $\nabla S(\mathbf{x}, t)$ is the momentum $\mathbf{p}$ of the particle. Therefore $\hat{\mathbf{P}}$ is indeed the linear momentum operator in Quantum Mechanics.