Canonical commutation relation, from QM to QFT.

[annihilatorM]

Homework Statement

This is a system of n coupled harmonic oscillators in 1 dimension.
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Since the distance between neighboring oscillators is ## \Delta x ## one can characterize the oscillators equally well by ## q(x,t) ## instead of ## q_j(t) ##. Then ## q_{j \pm 1} ## should be replaced by ## q(x \pm \Delta x , t) ##. Show that the quantization condition

$$ [\dot{q}_j(t), q_k (t)] = - \frac{i}{\mu} \delta _{jk} $$

can be expressed as$$ [ \dot{q} (x,t), q (x',t)] = - \frac{i}{\mu} \int_{ \frac{- \Delta x}{2}}^{\frac{ \Delta x}{2}} dy \delta (y-x'+x) $$

Hint: Distinguish the two cases ## x=x' ## and ## | x-x' | = j \Delta x ##

Homework Equations

The Attempt at a Solution

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Making the substitution of ## q(x,t) ## in place of ## q_j (t) ## I obtain

$$ [ \dot{q} (x,t), q (x',t)] = \delta (x-x') $$

which is the expected commutation relation I've seen from other sources, however, I don't understand why the integration with respect to y is introduced or where the integration bounds come from. What is wrong with simply writing down what I have above?

 

[_coyote_]

A:You have an expression for the commutator between two operators at the same point, $x = x'$. This is$$[\dot q(x,t),q(x,t)] = \delta(x-x').$$But you also need an expression for the commutator between two operators at different points, $|x-x'| = j\Delta x$. To get this you need to consider how $q(x,t)$ and $\dot q(x,t)$ vary over a small region around $x$, say from $x-\frac12\Delta x$ to $x+\frac12\Delta x$. If you expand $\dot q(x,t)$ and $q(x',t)$ into a Taylor series around $x$ up to second order, then you should find that$$[\dot q(x,t),q(x',t)] = \int_{-\frac12\Delta x}^{\frac12\Delta x} dy\delta(y+x-x') + O(\Delta x^2).$$The integral with the delta function is what you need, and the remainder is a higher order term which vanishes in the limit $\Delta x\to 0$.