I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is the proof of this canonical commutation relation?

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    $\begingroup$ This is a key axiom of matrix mechanics. It can be demonstrated in wave mechanics, but then you ask "why is p a differential operator?". The derivation is heuristic, it only works within the old quantum theory, and it is reproduced on Wikipedia's "matrix mechanics" page. $\endgroup$
    – Ron Maimon
    Sep 29 '12 at 16:55
  • $\begingroup$ Interesting point of view on the canonical commutation relation: johncostella.com/physics/… $\endgroup$ Jul 28 '16 at 1:16

As Lubos has mentioned


is one of the basic requirements of quantum mechanics. Classically observables are functions of variables $q$, and $p$ and Poisson bracket relation read

$\{q,p\}=1$ (note that $\{q,p\}$ is unitless quantity )

In QM observables are required to be hermitian operators (so that they can have real eigenvalues). In particular for position we have an operator $Q$, and for momentum we have an operator $P$. Poisson bracket is replaced by commutator and we require


In analogy with classical Poisson brackets we would have required


But this is not possible since

i) We have already required that $Q,P$ be hermitian. So $[Q,P]^\dagger=(QP-PQ)^\dagger=(QP)^\dagger-(PQ)^\dagger=PQ-QP=-[Q,P]$. So if we require $[Q,P]$ to be a constant (i.e. constant multiple of Identity matrix) it should be purely imaginary.

ii) $[Q,P]$ has units of $ML^2T^{-1}$. You can see this by noting that $Q$ is a position operator, so it has units of $L$, and that $P$ is a momentum operator, so it has units of $MLT^{-1}$.

Two natural choices are $[Q,P]=i\hbar$ and $[Q,P]=-i\hbar$. They are both equivalent and choice of $[Q,P]=i\hbar$ is just a convention.

No two finite dimensional matrices can satisfy $[Q,P]=i\hbar$. This can be seen by taking trace on both sides. However this relation can be satisfied by infinite dimensional matrices. More explicitly take vector space to be space of functions of $q$. Define $Q$ as $Qf=qf$, and $P$ as $Pf=-i\hbar\partial f/\partial q$. Then it can be seen that these operators satisfy required commutation relation. Moreover if we define our inner product as

$(f,g)=\int f^* g\: dq$

then $Q$ and $P$ defined as above will also be hermitian.

  • $\begingroup$ Just to check: how do we get that $[Q,P]$ has units of $ML^2T^{-1}$? $\endgroup$
    – War
    Sep 29 '12 at 10:58
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    $\begingroup$ @War $Q$ is position operator so it is required to have units of length $L$. Similarly $P$ is required to have units of momentum $MLT^{-1}$. So $QP-PQ$ should have dimensions of $ML^2T^{-1}$. $\endgroup$
    – user10001
    Sep 29 '12 at 11:06
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    $\begingroup$ @War There is a correction to be made in last para of above answer. Definition of $P$ should be $Pf=-i\hbar\partial f/\partial q$. $\endgroup$
    – user10001
    Sep 29 '12 at 11:44
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    $\begingroup$ @dushya, I incorporated the content of your comments into the answer. I hope that this is okay. $\endgroup$ Sep 29 '12 at 16:01

In a matrix representation, $Q$ and $P$ are infinite-dimensional matrices. (Finite-dimensional matrices wouldn't do; the trace of the two sides of the commutation relation gives a contradition.)

Many possible pairs of matrices qualify; the nicest ones are obtained when expressing position and momentum in a basis of eigenstates of the harmonic oscillator. See
Indeed, this is Heisenberg's original representation for ''matrix mechanics''.)

The more frequently used position representation (or momentum representation) takes $Q$ (resp. $P$) as a multiplication operator on wave functions depending on position (or momentum), and $P$ (resp. $Q$) as a first order differential operator chosen to match the commutation relation.

To prove the relation when $Q$ and $P$ are given, simply apply both sides to an arbitrary state vector and check that one gets the same result.

  • $\begingroup$ For $\varphi$ a smooth Schwartz function $ \in S(\Bbb{R})$ the simplest case that the CCR is generalizing is $(Q \varphi)(x) =x \varphi(x), (P\varphi)(x) =-i\hbar\varphi'(x),\hbar >0$ both self-adjoint operators we have $(PQ \varphi)(x) = -i\hbar\varphi(x)-i\hbar x\varphi'(x), (QP\varphi)(x) = -i\hbar x\varphi'(x)$ so $[Q,P]\varphi(x) = (QP-PQ)\varphi(x) = i\hbar \varphi(x)$. $\endgroup$
    – reuns
    May 5 '19 at 0:09

One may formulate more general rules that determine the commutators of operators in a quantum theory obtained from a Lagrangian or a Hamiltonian. However, there always have to be some axioms. In the simplest models of quantum mechanics, it's most legitimate to say that $xp-px=i\hbar$ is simply a key axiom of quantum mechanics, so it can't be derived from anything "deeper".

The fact that they don't commute – that their product depends on the order – implies that $x,p$ cannot be ordinary numbers. Instead, they are operators: operators don't have to commute. An $\hat L$ operator is something that assigns to each vector $|\psi\rangle$ of a space – in the case of quantum mechanics, Hilbert space – the result $\hat L |\psi \rangle$.

If one chooses a basis of the space with finitely or countably many elements, all the information about a linear operator may be expressed in terms of matrix elements $\langle i|\hat L |j\rangle = b(j,L(i))$ and the set of these inner products i.e. matrix elements is a matrix.

This picture of physics is consistent but the consistency has many aspects so it's a broad question. In essence, you want to explain why everything in quantum mechanics works. Well, it does but it isn't quite a 1-line proof. Quantum mechanics is the right theory of everything so it would be unwise to expect 1-line proofs of its consistency or validity. If you have some more particular concern or hypothetical consistency, please feel to update your question.

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    $\begingroup$ More can be said, namely, at least one of $ x $ and $ p $ has to be an unbounded operator on the state space. Helmut Wielandt first proved that no pair of bounded operators on a Hilbert space can ever satisfy the Heisenberg Commutation Relation. $\endgroup$ Dec 28 '17 at 21:18

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