Lie algebra vs. position and momentum commutators

Question

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?

Answer 1

$\hat x$ 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 $$\frac{d}{{dx}}\left( {x \cdot \bullet } \right):f\left( x \right) \mapsto \frac{d}{{dx}}\left( {x \cdot f\left( x \right)} \right).$$

Answer 2

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.