# How to understand momentum is the generator of translation ?

• HxWang
In summary, the concept of momentum as a generator of translations comes from classical mechanics and is related to the idea of canonical transformations. In quantum mechanics, momentum is represented by -id/dx, which can be seen as the generator of translations in the context of wavefunctions. However, in classical mechanics, momentum is seen as the generator of passive transformations, while in group theory, the gradient operator is seen as the generator of active transformations. This concept is further explained by Noether's theorem, which states that for every continuous symmetry of a Lagrangian, there is a conserved quantity that acts as the generator of the symmetry.
HxWang
How to understand "momentum is the generator of translation" ?

"As it is known from classical mechanics, the momentum is the generator of translation", it is from WIKI, (http://en.wikipedia.org/wiki/Momentum_operator)
I don't understand its meaning.
$T(ε)$ is translation operator, then $T(ε)f(x)=f(x-ε)=f(x)-ε\frac{\partial f}{\partial x}=(1-ε \frac{\partial }{\partial x}) f(x)$ ,
so $T(ε)=1-ε \frac{\partial }{\partial x} = 1 + i ε (i \frac{\partial }{\partial x})$,
we can say $i \frac{\partial }{\partial x}$ is the generator of translation, where is momentum ??

This is an interesting question.
Of course you will have to think about the correspondance CM-QM, and thats' why it will be interesting.

http://en.wikipedia.org/wiki/Canonical_transformation

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HxWang said:
"As it is known from classical mechanics, the momentum is the generator of translation", it is from WIKI, (http://en.wikipedia.org/wiki/Momentum_operator)
I don't understand its meaning.
$T(ε)$ is translation operator, then $T(ε)f(x)=f(x-ε)=f(x)-ε\frac{\partial f}{\partial x}=(1-ε \frac{\partial }{\partial x}) f(x)$ ,
so $T(ε)=1-ε \frac{\partial }{\partial x} = 1 + i ε (i \frac{\partial }{\partial x})$,
we can say $i \frac{\partial }{\partial x}$ is the generator of translation, where is momentum ??
You're asking about a generator in classical mechanics, but your calculation involves a translation operator acting on a wavefunction. If CM has taught us to think of momentum as a generator of translations, then your result looks like a good reason to guess that momentum can be represented by -id/dx in QM. (Or +id/dx...I'm a bit puzzled by the sign, but I don't have time to think it through right now).

If we want to argue that momentum is the generator of translations in CM, then we must do a calculation in CM, no wavefunctions involved. I don't remember the reason why momentum is said to be a generator of translations in CB (not sure I ever knew it), so I checked the mechanics books I have at home. They don't explain this, so I decided to try to figure it out. I still don't know what the answer is, but this looks like an answer:

If
$$S[q]=\int_0^t L(q(t),\dot q(t)) \mathrm dt,$$ where L is a Lagrangian and S is an action functional, then ##S[q+\varepsilon]=S[q]+\varepsilon p(t)## to first order in ε. This is perhaps a reason to call p a "generator of translations".

But maybe this isn't it at all. I have a feeling that we should be looking at Noether's theorem. Maybe there's a reason to call the conserved quantities "generators". If someone has a better answer, I would like to hear it.

Go to Chapter 2.8 in Shankar's QM text.

Edit: Section attached

He then goes on to apply this theorem to translations, but for some reason I can't clip that part without my computer freezing up.

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Thanks a lot, maajdl, Fredrik and HomogenousCow.

This is my argument about the question:
The word "generator" appears in "Group Theory" and "CM", this is the reason why we confused.
In Group Theory, ∇ (or $\pm i ∇$) is the generator of translation, the manner we obtain it is "translating object $f(x)$". Generally, we call this “active transformation”.
In CM, canonical transformations is "coordinate transformation", that is "passive transformation". In passive infinitesimal canonical transformations,
$q'=\hat{T}(ε)(q,p)=q+ε\frac{\partial g}{\partial p}$
$p'=\hat{T}(ε)(q,p)=p-ε \frac{\partial g}{\partial q}$​
where $\hat{T}(ε)$ is "transforming operator", which act on (q,p) simultaneously. we call $g=g(q,p)$ generator of transformation. when $g(q,p)=p$, corresponding transformation is spatial translation.

Summary:
∇ is the generator of active transformation, g(q,p) is the generator of passive transformation.

What do you think of my argument?

PS：I'm from china, so I'm very grateful for someone pointing out my mistake in Physics and language. :)
Thanks to maajdl, Fredrik and HomogenousCow again.

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Fredrik said:
I have a feeling that we should be looking at Noether's theorem. Maybe there's a reason to call the conserved quantities "generators".
Yes, it's all about Noether's theorem (or, deeper, Lie transformation theory). For every (continuous) symmetry of the Lagrangian there's a conserved quantity which generates the symmetry.

E.g., for a variation of a canonical coordinate ##q_k \to q_k + \delta q_k##, the conserved quantity turns out to be $$\frac{\partial L}{\partial \dot q_k}$$which is the canonical momentum.

Note the adjective "canonical". The correspondence with familiar everyday quantities gets trickier when dealing with non-free or relativistic situations.

BTW, this can also be regarded as a form of the general principle that for every continuous symmetry of a (set of) dynamical equation(s) of motion (i.e., a symmetry which takes solutions to solutions), there's necessarily a generator. This is the essence of Lie transformation theory.

## 1. What is momentum?

Momentum is a physical quantity that describes the amount of motion an object has. It is determined by an object's mass and velocity, and is a vector quantity with both magnitude and direction.

## 2. How is momentum related to translation?

Momentum is the generator of translation, meaning it is responsible for the motion or translation of an object. The greater the momentum of an object, the more difficult it is to change its state of motion.

## 3. Can momentum be conserved?

According to the law of conservation of momentum, the total momentum of a system remains constant unless acted upon by external forces. This means that momentum can be conserved in a closed system, where no external forces are present.

## 4. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The equation for momentum is p = m * v, where p is momentum, m is mass, and v is velocity.

## 5. Does momentum have units?

Yes, momentum has units of kilogram-meters per second (kg*m/s) in the SI system. This reflects the fact that momentum is the product of mass (kg) and velocity (m/s).

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