# Momentum and Kinetic Energy

1. Sep 25, 2005

### LeonhardEuler

Hello, I'm studying P-chem and I'm having a hard time with a concept. In quantum mechanics the momentum operator is $\hat{P_x}=-i\hbar\frac{\partial}{\partial x}$. The trouble I have is seeing how this reduces to the classical concept of momentum. I tried to follow the reasoning that leads to this operator back as far as I could, and what I found was that the idea was somewhat based on or inspired by the DeBroglie equation $\lambda=\frac{h}{p}$. This equation implies that if the momentum of a particle is zero, then its wavelength is infinite. According to the classical concept of momentum, this would apply to any body at rest. I think the difficulty with applying this equation to a body at rest is that it is in fact made up of many particles which are not at rest and therefore do not have infinite wavelengths. So you would add thier momentums to get the toal momentum and this would imply that the wavelength is not infinite (if it even makes sense to talk about the wavelength of a group of particles). The kinetic energy operator is kind of based on the momentum operator (or the other way around) so it also does not make much sense to me. What I am looking for is an explaination of how it is that momentum and kinetic energy increase when an object starts to move according to quantum mechnaics, and also a clarification of the meaning of the DeBroglie equation. Thanks you all in advance for your replies!

2. Sep 25, 2005

### MalleusScientiarum

I'll give you a good hint: the concept of a particle "moving" in quantum mechanics does not make a whole lot of sense in the classical meaning of the word.

3. Sep 26, 2005

### Crosson

You need some Hamilton-Jacobi theory:

http://philsci-archive.pitt.edu/archive/00001193/

Hope you have at least studied Lagrangians or variational calculus. Good luck.

4. Sep 27, 2005

### Gokul43201

Staff Emeritus
Perhaps someone can do a more elegant job, but here goes my attempt, nevertheless :

(Warning : watch for violent skipping)

PART I : The classical story ...

Recall the classical canonical transformations

$$Q_i = Q_i (q,p)$$
$$P_i = P_i (q,p)$$

transforming the positions and momenta $q_i,p_i$ to the new set $Q_i,P_i$.

Next recall the suitably chosen generating functions of the type F(q,P) which give the transformations through

$$p_i = \frac{\partial F}{\partial q_i}$$

$$Q_i = \frac{\partial F}{\partial P_i}$$

Now, specifically consider the generating function $F(q,P) = q_iP_i$. From above, the transformations yield

$$p_i = \frac{\partial F}{\partial q_i} = P_i$$

$$Q_i = \frac{\partial F}{\partial P_i} = q_i$$

The new and old co-ordinates are the same, so this F generates the identity transformation.

Next, we skip ahead to the idea of infinitesimal canonical transforms (ICTs), given by

$$Q_i = q_i + \delta q_i$$
$$P_i = p_i + \delta p_i$$

Notice that to first order, these are nothing but the identity transformation - or the ICTs differ only infinitesimally from the identity transformation. Naturally, we expect the generating function for an ICT to differ infinitesimally from the generator for the identity transform (covered above). So, in general we have

$$F = q_iP_i + \epsilon G(q,P)$$

where $\epsilon$ is an infinitesimal parameter and G is any differentiable function. Now, if we specifically choose F to be

$$F(q,P) = q_iP_i + p_i \delta q_i$$

then, the transformations generated are

$$Q_i = q_i + \delta q_i$$
$$P_i = p_i$$

Notice that the above generating function, which generates infinitesimal spatial translations, differs from the identity transformation by the extra term $p \delta q [/tex]. PART II : The quantum story ... In QM, infinitesimal translations are generated by a unitary operator [itex] {\cal J} (dx)$ defined as

$${\cal J} (dx) |x \rangle = |x + dx \rangle$$

where $|x \rangle$ is a position eigenstate of the system. It is found that the required properties (unitarity, composition, inversion, etc.) of this translation operator can be satisfied by chosing it to be of the form

$${\cal J}(dx) = \mathbf{1} - iK \cdot dx$$

where $\mathbf{1}$ is the identity operator. So, we see that in the quantum case, the generator of infinitesimal translations differs from the identity operator by a term that looks like Kdx, where K is some undetermined (but Hermitian) operator. So, noticing the striking similarity with the classical result, it makes sense to suggest that K be an operator that is proportional to the momentum operator $\hat{p}$. To get the dimensions right (and borrowing de Broglie's idea), we set

$${\cal J} (dx) = \mathbf{1} - \frac {ip \cdot dx}{ \hbar}$$

If you operate this infinitesimal translation operator on a general state and play around with the math, (too lazy to do this now) you arrive at the position representation of the momentum operator : $\hat{p} = -i \hbar \partial /\partial x$

5. Sep 27, 2005

### Galileo

I`ll try to give an elementary approach as to why $\hat P_x=\frac{\hbar}{i}\frac{\partial}{\partial x}$ is the right operator to represent momentum, with the help of deBroglie: $p=\lambda/h$. (I apologize if it's too basic, or not what you're looking for)

First, we'll look at the eigenstates of $P_x$ (in the position representation -> wavefunctions)
$$P_x \psi(x)=\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)=p \psi(x)$$
The eigenvalue is ofcourse the momentum that we measure. This is an easy differential equation, the answer is a plane wave $A\exp(ikx)$, with $p=\hbar k$. (k is the wavenumber and is related to the wavelength by $k=2\pi/\lambda$.) So $p=\hbar k = h/\lambda$, in accordance with deBroglie.
So you see the meaning of deBroglie's relation. A plane wave has only one wavenumber or wavelength and according to deBroglie corresponds to a state with a well defined momentum. P_x has exactly that plane-wave (one wavelength) as an eigenstate.

6. Sep 27, 2005

### reilly

It's not that hard -- just look at the behavior of wave packets, with small sigma. This is discussed in many texts, particularly in older ones like Kemble, for example. A pure plane wave solution of Maxwell's equations carries a distinct momentum, demonstrated by Poynting's Thrm.

Regards,
Reilly Atkinson

7. Sep 28, 2005

### LeonhardEuler

Thank you everyone for your responses. I've been really busy the past few days and will be for a while, so it'll take me a while to make sense of all this, but just glancing over it, it seems to be what I was looking for. Thank you.