Non-Commutative Nature of QM

1. Mar 14, 2012

mysearch

Hi,
I am trying to get a few fundamental concepts sorted out in my mind associated with the maths of non-commutative operators and the physical implications on QM. As such, I am simply looking for confirmation, clarification or corrections to any of the following issues.

One of the most fundamental issues in QM appears to be the Heisenberg Uncertainty principle (HUP). In mathematical terms this often seems to be described in terms of the non-commutative nature of position [x] and momentum [p]. However, before addressing this specific case, I wanted to first confirm my ‘basic’ understanding of operators in a generic example, where [A] and are both operators, while [Q] is some quantity on which they operate:

[1] $\hat A = x$; $\hat B = \frac {d}{dx}$;

$\left[ \hat A, \hat B \right]Q = \hat A \left (\hat B Q \right) = x \frac {dQ}{dx} = xQ’$

$\left[ \hat B, \hat A \right]Q = \hat B \left (\hat A Q \right) = \frac {d(xQ)}{dx} = x \frac {dQ}{dx} + Q \frac {dx}{dx} = xQ’ + Q$

The operators work from left to right, such that the result of [A,B]Q is different from [B,A]Q as [1] tries to illustrate. It is also highlighted that the expansion of the [B,A] case, in [1], is linked to the differentiation product rule:

[2] $\frac {d(uv)}{dx} = u \frac {dv}{dx} + v \frac {du}{dx}$

$\frac {d(xQ)}{dx} = x \frac {dQ}{dx} + Q \frac {dx}{dx}$

If we now use the normal non-commutative expansion, we get a non-zero result:

[3] $\left[ \hat A, \hat B \right]Q = \left [AB – BA \right]Q = x \frac {dQ}{dx} - \left( \frac {d(xQ)}{dx} \right)$

$= x \frac {dQ}{dx} - \left(x \frac {dQ}{dx} + Q \frac {dx}{dx} \right) = -Q$

What [1,2,3] seems to illustrate is that the non-commutative property stems from operator being defined in terms of a first derivative, which in itself has nothing to do with quantum mechanics, i.e. the idea of quantization. Of course, the idea of non-commutative operators seem to be important to QM, because the momentum operator [p] can be described in terms of [d/dx], if we derive its form from a wave function, e.g.

[4] $\psi = e^{ 2 \pi i \left ( \frac {x}{\lambda} - \omega t \right)} = e^{ i \left( \kappa x - \omega t \right)} = e^{ \frac {i}{\hbar} \left( px - Et \right) }$

$\frac {\partial \psi}{\partial x} = \frac {ip}{\hbar} e^{ \frac {i}{\hbar} \left( px - Et \right) } = \frac {ip}{\hbar} \psi$

$\hat p = \frac {1}{i} \hbar \frac {\partial }{\partial x} = -i \hbar \frac {\partial }{\partial x}$

Therefore, if we replace [A,B]Q for [x,p]Ψ in [1,2,3] we get:

[5] $\left[ \hat x, \hat p \right] \psi = \left [xp – px \right]\psi = x \left( -i \hbar \frac {\partial}{ \partial x} \right) \psi - \left( -i \hbar \frac {\partial}{ \partial x} \right) x\psi$

$= x \left( -i \hbar \right) \frac {\partial \psi}{ \partial x} - \left( -i \hbar \right) \left( x \frac {\partial \psi}{ \partial x} + \psi \frac {\partial x}{\partial x} \right) = -i \hbar \psi$

$\left[ \hat x, \hat p \right] = i \hbar$

So my first real question is whether the logic above is basically correct? My next question then relates to the wave-particle duality issues. The reason we appear to end up with the non-commutative result in [5] seems to be predicated on the wave nature defined in [4]? As such, is the wave nature fundamentally more descriptive of the ‘true’ nature of quantum ‘particles’? Is there an equivalent approach that can come to the same non-commutative result without any wave assumption? For example, it is assumed that Heisenberg came to his conclusion about HUP based on his own matrix formulation, however I don’t really know if this approach still has a basic wave assumption built into the matrix formulation? Thanks

2. Mar 14, 2012

A. Neumaier

Your notation is somewhat strange. You should use $$\psi$$ in placce of Q which is the standard typographical distinction in QM for wave functions your operators act on. And you should write $$(AB)\psi=A(b\psi)$$ and not use $[A,B]$ as the latter denotes a commutator and means AB-BA. This will improve communication with the rest of the world.