# Lie algebra vs. position and momentum commutators

Most theoretical texts on high energy physics make statements like below:

$$[A_i , A_j] = i C^k_{i,j} A_k$$

(I suppose $$\hbar$$ may or may not be needed) and of course they describe this as being the Lie algebra over some group. I think in this case, what I have just describes is the general lie algebra.

Meanwhile, there is the standard relation found in quantum mechanics texts:

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

My question is two-fold. First, I am confused about the second equation. The way I see it:

$$[\hat x, \hat p ] = \hat x \hat p - \hat p \hat x$$ which in turn equals: $$=\hat x \frac{\partial}{\partial x} - \frac{\partial}{\partial x} \hat x$$ $$=\hat x \frac{\partial}{\partial x} - 1$$ I may be a bit unclear on that last line above but the way I see it, it sure doesn't equal $$i \hbar$$.

I guess my question is what gives?

• For starters, you may represent $\hat p$ by hermitean $-i\hbar\partial_x$, not your option. Your also mishandled the operators badly in you penultimate line, which the answers alert you to. – Cosmas Zachos Apr 14 at 1:08

$$\hat x$$ and $$\hat p$$ are operators, they act on states. In coordinate representation, $$\hat x$$ becomes the operator that multiplies with $$x$$, and $$\hat p$$ becomes the operator that multiplies with $$\hbar /i$$ and derives with respect to $$x$$. The meaning of $$\left[ {\hat x,\hat p} \right] = \hat x\hat p - \hat p\hat x$$ can be expressed on a function: \begin{aligned} \left( {x \cdot \frac{\hbar }{i}\frac{d}{{dx}}\left( \bullet \right) - \frac{\hbar }{i}\frac{d}{{dx}}\left( {x \cdot \bullet } \right)} \right)f\left( x \right) & = x \cdot \frac{\hbar }{i}\frac{d}{{dx}}f\left( x \right) - \frac{\hbar }{i}\frac{d}{{dx}}\left( {x \cdot f\left( x \right)} \right) \\ & = x \cdot \frac{\hbar }{i}\frac{d}{{dx}}f\left( x \right) - \frac{\hbar }{i}x\frac{d}{{dx}}f\left( x \right) - \frac{\hbar }{i}f\left( x \right) \\ & = i\hbar f\left( x \right) \\ \end{aligned}
With other words, the muted product symbol between the operators are the $$\circ$$ composition symbol, just like when it comes to matrices, $$ABx = \left( {A \circ B} \right)x = A\left( {B\left( x \right)} \right).$$
What you missed is that in coordinate representation, there is a division with $$i$$ in the definition of $$\hat p$$ and $$\frac{d}{{dx}}x$$ means $$\frac{d}{{dx}}\left( {x \cdot \bullet } \right):f\left( x \right) \mapsto \frac{d}{{dx}}\left( {x \cdot f\left( x \right)} \right).$$
• I agree that your answer is technically correct as stated, but I don't see any way that your answer can reconcile the dilemma that $[\hat x, \hat p]$ is suppose to somehow equal $i \hbar$, or even $i$ if we assume that $\hbar = 1$. – the_photon Apr 14 at 0:45
• @the_photon Do you agree that in the first equation he computed $[\hat{x},\hat{p}] f(x) = i \hbar\,f(x)$ for arbitrary $f$? – MannyC Apr 14 at 3:23
The set $$\{x, p/i, x^2, (p/i)^2, N = (xp+px)/i\}$$ forms the osp(1/2) super-Lie algebra, where $$\{x, p/i\} = (xp+px)/i = N, \\ \{x, x\} = 2x^2, \\ \{p/i, p/i\} = 2(p/i)^2,$$ obeys the anti-commuting relationship, and the rest observe the commuting Lie relationships. For example $$[x, N] = xN - Nx = x, \\ [p/i, N] = -p/i.$$ I encourage you to work out the other Lie relationships.