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In quantum mechanics the momentum and energy operators appear in Schroedinger's equation. In fact in the derivation of Schroedinger's equation from the classical wave equation the momentum operator comes from the space derivative, and the energy operator comes from the time derivative, which has nice links to Noether's theorem.

My question is where does the position operator come from? What does it mean for a wave to have a position anyway? What I mean is what were the reason for choosing the position operator?

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  • $\begingroup$ Suppose you Fourier transform to the momentum representation, the FT of ψ(x) being φ(p). The position operator will now be a gradient with respect to momentum, and, the momentum operator simply multiplication by the variable p. Noether's theorem would give you conservation of x if only your hamiltonian did not depend on p. This is, of course, hardly unusual for conventional hamiltonians with stock kinetic terms, but you can contrive highly bizarre mirror systems where such things happen, if only for formal mathematical convenience. $\endgroup$ – Cosmas Zachos Jun 6 '16 at 21:46
  • $\begingroup$ What do you mean "where does it come from"? The very definition of canonical quantization is that position and momentum become operators. $\endgroup$ – ACuriousMind Jun 7 '16 at 13:18
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Purely formal considerations suggest that the Lie bracket of quantum observables should be proportional to the Poisson bracket of their classical counterparts. (In particular, the classical Poisson bracket satisfies$ \lbrace u,wx\rbrace v +u \lbrace v,wx\rbrace =\lbrace uv,w\rbrace x + w \lbrace uv,x\rbrace$, and the proportionality follows from this and a little algebraic manipulation (after dropping the commutativity assumption.))

Classically, the Poisson bracket of position and momentum is a constant (namely $ 1$), so in the quantum setting we want the Lie bracket of position and momentum to be a constant (in this case purely imaginary so that $p$, $q$ and $[p,q]$ can all be Hermitian). That is, we want to identify position and momentum with Hermitian operators satisfying $[p,q]=i\hbar$ for some real constant $\hbar$.

The usual choices satisfy this condition, and are particularly simple, so they get used. Other choices would work just as well in principle but might be a little messier to manipulate. This is all explained quite clearly in Dirac's book on Quantum Mechanics.

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The position operator also appears in the Hamiltonian, as soon as you have a potential. However, when using the position representation (the Schroedinger's "wave equation") it then appears as a multiplication of the wave-function by a real number.

Don't put too much in the wave part of "wave equation". This is just one of the multiple representation of the Schroedinger's equation, which can be described as a wave equation, or a more standard matrix equation (think of the harmonic oscillator, that can be solved either using the wave equation, or the algebra of the creation-annihilation operators).

Finally, the position operator is also useful for symmetry reasons, as it is the momentum shift operator (when exponentiated), much in the same way that momentum is the position shift operator. This is useful to perform Galilean boosts.

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