# What does $[X,P]=ci I$ imply for the dimension of the underlying space?

Let $X$ and $P$ be linear operators on a $\mathbb C$ vector space $V$ and $I$ the identity operator. Suppose the commutator

$$[X,P] = XP - PX = ciI$$

for some real, positive constant $c$.

What can we say about the dimension of $V$?

The motivation for this problem was a throw away line by a theoretical physicist in a lecture I saw where he said this necessarily implied $V$ is infinite dimensional. $X$ and $P$ will be recognized by anyone with training in quantum mechanics as labels for two popular operators. However their particularities shouldn't matter as I want to see what we can say about $V$ with just this information.

Any ideas?

• Do you define $[X,P] = i(XP - PX)$? (This is standard for physics, but not for math). Commented Feb 24, 2015 at 21:33
• Let me write it out Commented Feb 24, 2015 at 21:34
• Great, thanks ${}$ Commented Feb 24, 2015 at 21:34

In particular, we can show that this is impossible to have $[X,P] = cI$ with $c \neq 0$ in a finite-dimensional vector space with relative ease.
Assume $V$ is finite dimensional. Fix a basis so that $X$ and $P$ are now matrices over the vectors spaces. We note that \begin{align} \operatorname{trace}([X,P]) &= \operatorname{trace}(XP - PX) = \operatorname{trace}(XP) - \operatorname{trace}(PX) \\ & = \operatorname{trace}(XP) - \operatorname{trace}(XP) = 0 \end{align} So: if $[X,P] = cI$, then $c = 0$.
• The secret is to make it look easy :P. In general, "check the determinant" and "check the trace" are good tricks when $V$ is finite dimensional (or when your operators are sufficiently nice, i.e. compact or of finite rank). Commented Feb 24, 2015 at 21:42
• I somehow thought that the fact that the field was $\mathbb C$ was important. But no. Now I see this result is quite general. Commented Feb 24, 2015 at 21:45
• So how does having infinite-dimensional $V$ fix this? Commented Feb 17, 2019 at 11:20
• @spraff if $V$ is infinite dimensional, we can no longer define a trace function that will work for every linear operator over $V$, so the argument no longer applies. Commented Feb 17, 2019 at 12:54