# Why should not $\hat X$ and $\hat P$ commute in quantum mechanics?

I have heard if we want to obtain classical results from quantum mechanics, we have to choose the commutator of $\hat X$ and $\hat P$ to be $\left[\hat X,\hat P\right]=i\hbar$. Is there any reason support this statement?

EDIT: I want to understand how inventors of QM deduced the $\hat X$ and $\hat P$ should not commute. I really wonder why numerous lecturers postulate it.

• The classical analog of the commutator is the Poisson bracket. Apr 2, 2017 at 17:20
• I gave rise how you can derive the commutation relations by the quantum action principle: physics.stackexchange.com/q/315825 Apr 2, 2017 at 17:45
• @Alpha001 I read that. thanks but don't you think there is a similar argument using wave mechanics? I mean Heisenberg or Schrodinger didn't know the Hilbert space language so how they deduced the commutator of position and momentum should not be zero?! I guess it's really a revolutionary result and I want to make it more clarified. Apr 2, 2017 at 18:18
• Apr 2, 2017 at 18:40

The idea goes back to Heisenberg. He believed that physics could only describe quantities that could be measured experimentally, and sought to develop a mathematical theory that would reflect this and correctly predict the relative intensities of spectral lines.

In classical physics, the radiated intensities depend (in the first approximation) on electric dipoles, which are dependent on the position of the electrons. To account for the fact that the position of the electrons cannot be measured during a transition, he introduced a number $x_{nm}$ to characterize the position in the transition from $n\to m$. He also introduced $v_{nm}$ for the velocities of the electrons during the transitions and a related acceleration $a_{nm}$.

Heisenberg was eventually able to reproduce the energy levels $E_n$ (actually their differences $E_n-E_m$) using these quantities but only if the quantities satisfied the "unusual" combination properties $$\sum_{m} x_{nm} v_{mk} = A_{nk} \ne A_{kn} = \sum_m v_{km} x_{mn}\, .$$ In particular, using his "tables" of $x_{nm}$ and $v_{nm}$ he was able to work out what we now write as $[x,p]$.

The story goes that Pascual Jordan met Heisenberg in a train at the time Heisenberg was working on this. Jordan, who had mathematical training, recognized the combination rule as matrix multiplication. Jordan, together with Max Born and Paul Dirac, realized that the use of non-commutative quantities was essential to Heisenberg's description.

Dirac in particular postulated that the multiplication rules had to follow from dynamical considerations; inspired by the correspondence principle, he was able to relate, up to an overall factor, the classical Poisson bracket to a quantum bracket to find the now-famous $$[q_i,q_j]=0\, ,\qquad [p_i,p_j]=0\, ,\qquad [q_j,p_k]=i\hbar \delta_{jk}\, .$$

There are several accounts of this discovery. The most historical is by Max Jammer, who was able to interview firsthand some of actors in the story. There is also an interesting and more recent text by Roland Omnes but it doesn't focus so much on the history. I'm sure there are others.

Edit: After reading @hyportnex account, I found Jammer online and checked. hyportnex account is accurate when it comes to Born recognizing the matrix form of Heisenberg's expression. As to the story of the train: it is Born that met Jordan in a train. Quoting from Jammer, page 109:

Now it happened that Born, while traveling by train to Hanover, told a colleague of his from Gottingen about the fast progress of his work but also mentioned the peculiar difficulties involved in the calculations with matrices. It was fortunate and almost an act of providence that Jordan, who shared the same compartment in the train, overheard thispiece of conversation. At the station in Hanover Jordan then introduced himself to Born, told him of his experience in handling matrices, and expressed his readiness to assist Born in his work.

• That's the answer I was looking for. nice answer and thanks Apr 2, 2017 at 18:51
• Nice bit of history. The account I read had Born saying, "But Heisenberg, those are matrices", but Pascaul Jordan as the catalyst for matrix mechanics seems more plausible. I'm pretty sure the account I read was fairly light and not historically deeply researched. Apr 2, 2017 at 23:44
• @WetSavannaAnimalakaRodVance From what I recall Heisenberg was coming back from Copenhagen and met Jordan perchance on a train platform. I read Jammer a while ago so my memory could be rusty. Apr 2, 2017 at 23:46
• Actually Hyportex's answer cites Bartel van der Waerden's personal correspondence with Born and Jordan on the matter; this would seem to imply that it was Born who first noticed the matrix operations. Apr 3, 2017 at 4:40
• @WetSavannaAnimalakaRodVance Interesting. I will get Jammer from the library and double check. Apr 3, 2017 at 5:58

To study the history of the subject I recommend van der Waerden: Sources of Quantum Mechanics (Dover Publications). This book only discusses the history of matrix mechanics, so the development of wave mechanics (de Broglie, Schrodinger, etc.) is left for a never finished/published 2nd volume. To write this book Waerden contacted the main actors directly: Pauli, Heisenberg, Born, Jordan, etc. Let me quote a letter that Waerden received from Born, see pages 36-37.

Born's conjecture on pq — qp

On July 19, Born took the train to Hannover to attend the meeting of the Deutsche Physikalische Gesellschaft. His own account, confirmed by Jordan's testimony, runs thus:

After having sent Heisenberg's paper to the Zeitschrift fur Physik for publication, I began to ponder about his symbolic multiplication, and was soon so involved in it that I thought the whole day and could hardly sleep at night. For I felt there was something fundamental behind it ... And one morning ... I suddenly saw light: Heisenberg's symbolic multiplication was nothing but the matrix calculus, well known to me since my student days from the lectures of Rosanes in Breslau. I found this by just simplifying the notation a little: instead of $q(n, n+\tau)$ I wrote $q(n, m)$, and re-writing Heisenberg's form of Bohr's quantum conditions I recognised at once its formal significance. It meant that the two matrix products $\mathbf{pq}$ and $\mathbf{qp}$ are not identical. I was familiar with the fact that matrix multiplication is not commutative; therefore I was not too much puzzled by this result. Closer inspection showed that Heisenberg's formula gave only the value of the diagonal elements (m=n) of the matrix pq — qp: it said that they were all equal and had the value $h/i2\pi$. But what were the other elements $m \ne n$ ? Here my own constructive work began. Repeating Heisenberg's calculation in matrix notation, I soon convinced myself that the only reasonable value of the non-diagonal elements should be zero, and I wrote down the strange equation

$$\mathbf {pq-qp} = \frac{h}{2\pi i} \mathbf{I}$$ where $\mathbf{I}$ is the unit matrix. But this was only a guess, and my attempts to prove it failed.

• Ah yes, I'd quite forgotten van der Waerden's foray into physics; he was quite a polymath. I first came to "know" him through this amazing paper and there's a wonderful interview with him and his wife Camilla, especially about his time at Göttingenhere Apr 3, 2017 at 4:37
• I could not read the paper, not in English. But the interview was indeed wonderful. Quite a history, and so wonderful that he researched in many multiple areas that interested him, from the history of math and astronomy, to math itself, and some physics. And many areas In math including topology, geometry and algebra. I did not know of him and am happy to have read of him. Apr 3, 2017 at 5:20
• updated my post to correct an error in my recollection. The part you give is more or less also in Jammer. Apr 3, 2017 at 6:48
• @BobBee Yes, he was quite a guy. The paper I first read proves that if one has a compact, semisimple Lie group, and then imagines trying to strip away the topology and putting another in its place to equip the same algebraic structure with a different Lie group structure, then you must wind up with the same structure as you began with if the topology must be compact and connected. What this says is that, for compact, semisimple Lie groups the very topology is encoded in the group's algebraic structure alone. Of course, one can always take any Lie group and give it the .... Apr 3, 2017 at 7:33
• ..... discrete topology making it a totally disconnected Lie group of dimension 0, but, if you want to keep the group compact in the topology you assign, the theorem says that the topology of a compact, semisimple Lie group is unique. One can broaden this to any semisimple group, but then you have uniqueness modulo fairly restricted discontinuous automorphisms.\ Apr 3, 2017 at 7:35

The commutator arises in canonical quantisation which is a procedure to quantise a classical theory, ubiquitous in most quantum mechanics and quantum field theory texts. One imposes the condition that,

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

which is in accordance with the rule of thumb (as there are many subtleties as to its applicability),

$$\{x,p\} \to \frac{1}{i\hbar}[x,p]$$

attributed to Dirac. The quantisation is a means of going from a theory specified by either an action or a phase space with a particular symplectic structure. So, your question boils down in essence to the validity of quantisation in this manner.

In order to rigorously justify this, I would recommending reading the nLab article; it has a habit of over-complicating certain matters especially with jargon but I think will address your concerns.

Another way to see it is by trying to define $x$ and $p$ in a quantum theory sensibly. For example, the momentum operator has an intuitive meaning and an expected behaviour on states.

Furthermore, in for example quantum field theory, one can apply Noether's theorem in the case of translational symmetry to derive the conjugate momentum, and then upon quantising the fields of the theory, modulo potential ordering ambiguities, one arrives at an operator $p$.

Explicitly, one may write out $\phi(x)$ and $\pi(x)$ say, and check that,

$$[\phi(x),\pi(x)] = i\hbar \delta^{(3)}(x-y)$$

which is the analogue of $[x_i,p_j] = i\hbar \delta_{ij}$ though this derivation would depend on the fact that $[a,a^\dagger]= 1$ for the ladder operators and so the two commutators' definitions are intertwined in a sense. Asking for a justification of the canonical relations is linked to the relation for ladder operators.