Consider [position](Position%20Operator.md) and [momentum](Momentum%20Operator.md) operators $\hat{q}$ and $\hat{p}$ respectively. Assume that they both belong on a [finite dimensional Hilbert space](Hilbert%20Space%20dimension%20in%20quantum%20mechanics#Finite%20dimensional%20Hilbert%20spaces%20in%20quantum%20mechanics) and thus can be represented as [$n \times n$ matrices.](Matrices.md#Square%20matrices) We could then take the [traces](Trace.md) $\mbox{tr}(\hat{q}\hat{p})$ and $\mbox{tr}(\hat{p}\hat{q}).$ By the [cyclic property of the trace](trace#^c768aa), $\mbox{tr}(\hat{q}\hat{p})-\mbox{tr}(\hat{p}\hat{q}) = 0$ However, for the [commutation relation to hold,](Position-Momentum%20Commutators.md) where [$[\hat{q},\hat{p}]=i\hbar\hat{\mathbb{1}}$](position-momentum%20commutators#^a96259) it must be the case that $\mbox{tr}([\hat{q},\hat{p}])=\mbox{tr}(\hat{q}\hat{p})-\mbox{tr}(\hat{p}\hat{q})=i\hbar\mbox{tr}(\mathbb{1})=0,$ which would imply that these operators belong in a [0 dimensional vector space](Finite%20dimensional%20vector%20spaces.md#0%20dimensional%20vector%20space) containing only the vector $\mathbf{0}=\mathbb{1}$ such that $\mbox{tr}(\mathbb{1})=0.$ ^b72ffb Thus, by contradiction, $\hat{q}$ and $\hat{p}$ cannot belong in a [finite dimensional vector space,](Finite%20dimensional%20vector%20spaces.md) implying that they must instead be in an [infinite dimensional](Infinite%20dimensional%20vector%20spaces.md) space and infinite dimensional. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^44ac54 This commutation relation is the basis for the [Heisenberg algebra](Heisenberg%20group.md#Heisenberg%20algebra) and thus this proof also implies that all [Representations of the Heisenberg algebra](Heisenberg%20group.md#Representations%20of%20the%20Heisenberg%20algebra) must also be infinite dimensional. %%This could follow very closely from a general proof that representation of the Heisenberg algebra are infinite dimensional. Does this also necessarily imply that these operators must also be in Hilbert Spaces? How do we know this as well.%% %%This form sheet 2 of the 2018/2019 qm semester and may also be found in earlier homework.%% #QuantumMechanics/QuantumMeasurement/QuantumObservables #Proofs