Question about Operators in Quantum Mechanics

[Another]

I study on quantum mechanics and I have question about operator.
In one dimension. How do we know ## \hat{x} = x## and ## \hat{p}_{x} = -i \bar{h} \frac{d}{dx} ##

When schrodinger was creating an equation, which later called "the schrodinger equation".
How does he know momentum operator equal to ## -i \bar{h} \frac{d}{dx} ## and position operator ## \hat{x} ## equal to ## x ## ?

 

[A. Neumaier]

When schrodinger was creating an equation, which later called "the schrodinger equation".
How does he know momentum operator equal to ## -i \bar{h} \frac{d}{dx} ## and position operator ## \hat{x} ## equal to ## x ## ?

It was part of Schrödinger's insight. It earned him a Nobel prize!

 

[jedishrfu]

Here's a derivation of the momentum operator that may help:

 

[bhobba]

It was part of Schrödinger's insight. It earned him a Nobel prize!

For some of the gory detail see here:
https://arxiv.org/pdf/1204.0653.pdf

It basically was inspired guess work - he even made errors that canceled themselves out.

In modern times the correct derivation is from symmetry as you will find in Ballentine. At least in part Wigner got a Nobel for that.

Thanks
Bill

 

[chmodfree]

You can regard the momentum operator as generator of spatial translation group.

 

[vanhees71]

I study on quantum mechanics and I have question about operator.
In one dimension. How do we know ## \hat{x} = x## and ## \hat{p}_{x} = -i \bar{h} \frac{d}{dx} ##

When schrodinger was creating an equation, which later called "the schrodinger equation".
How does he know momentum operator equal to ## -i \bar{h} \frac{d}{dx} ## and position operator ## \hat{x} ## equal to ## x ## ?

The question is, from what you start. A modern approach to QT is to start in the "representation free" way with bras and kets (i.e., Dirac's approach to QT, which is the most simple and lucid of the three historical approaches; the other two are wave mechanics (Schrödinger 1926) and matrix mechanics (Heisenberg, Born, and Jordan 1925)).

If you start with the representation free approach for a particle moving along a line (to keep the notation simple; there's no difficulty to extend this approach to 3D motion of a particle), the only thing you have is the Heisenberg algebra (which should be more justly called Born algebra for that matter), i.e., the commutation relation
$$[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1}.$$
This commutation relation can be guessed from the fact that in classical Hamiltonian mechanics momentum is associated with spatial translations thanks to Noether's theorem, i.e., momentum is the generator for spatial translation symmetry transformations.

This also hints at how to construct the Hilbert space in the position representation which leads to the formulation of non-relativistic QT of a single (scalar) particle in terms of Schrödinger's wave mechanics. This comes about in the following way: Suppose ##|x=0 \rangle##, i.e., the eigenvector of the position operator with eigenvalue 0 exists. Then it's easy to prove via the commutation relation above that
$$|x \rangle=\exp(-\mathrm{i} x \hat{p}/\hbar) |x=0 \rangle.$$
is an eigenvector of ##\hat{x}## with the arbitrary real eigenvalue ##x##.

The sketch of the proof goes as follows. First you can easily show by induction that for any power of ##\hat{p}## you have
$$[\hat{x},\hat{p}^k]=\mathrm{i} k \hbar \hat{p}^{k-1},$$
and applying this to the power series of the exponential, you get
$$[\hat{x},\exp(-\mathrm{i} x \hat{p}/\hbar)]=\mathrm{i} \hbar \partial_p \exp(-\mathrm{i} x \hat{p}/\hbar)=x \exp(-\mathrm{i} x \hat{p}/\hbar).$$
This now implies the proof of the above claim, because
$$\hat{x} \exp(-\mathrm{i} x \hat{p}/\hbar) |x=0 \rangle = \left \{ [\hat{x},\exp(-\mathrm{i} x \hat{p}/\hbar)]+\exp(-\mathrm{i} x \hat{p}/\hbar) \hat{x} \right \} |x=0 \rangle= x \exp(-\mathrm{i} x \hat{p}/\hbar) |x=0 \rangle$$.
We thus can define the position eigenvectors by the unitary translation operators
$$|x \rangle=\exp(-\mathrm{i} x \hat{p}/\hbar) |x=0 \rangle,
\quad \langle x |=\langle x=0| \exp(+\mathrm{i} x \hat{p}/\hbar).$$
Then it's also clear how the momentum operator acts on the position representation of the state, i.e., the wave function
$$\hat{p} \psi(x) = \langle x|\hat{p} \psi \rangle=\langle x=0|\exp(\mathrm{i} x \hat{p}) \hat{p} |\psi \rangle =
-\mathrm{i} \hbar \partial_x \langle x=0|\exp(\mathrm{i} x \hat{p}) |\psi \rangle
=-\mathrm{i} \hbar \partial_x \langle x|\psi \rangle = -\mathrm{i} \hbar \partial_x \psi(x).$$
What also immediately follows is the momentum eigenfunction in position representation
$$u_p(x)=\langle x|p \rangle=\langle x=0|\exp(\mathrm{i} x \hat{p}/\hbar) p \rangle = \langle x=0|p \rangle \exp(\mathrm{i} x p/\hbar).$$
The undetermined factor can be fixed by the usual normalization condition
$$\langle p|p' \rangle=\delta(p-p')$$
which, up to an unimportant phase factor leads to
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} x p/\hbar).$$