- #1

LeonhardEuler

Gold Member

- 859

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter LeonhardEuler
- Start date

- #1

LeonhardEuler

Gold Member

- 859

- 1

- #2

MalleusScientiarum

- #3

- 1,256

- 4

The trouble I have is seeing how this reduces to the classical concept of momentum.

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

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

- 7,082

- 21

Perhaps someone can do a more elegant job, but here goes my attempt, nevertheless :LeonhardEuler said:Hello, I'm studying P-chem and I'm having a hard time with a concept. In quantum mechanics the momentum operator is [itex]\hat{P_x}=-i\hbar\frac{\partial}{\partial x}[/itex]. The trouble I have is seeing how this reduces to the classical concept of momentum.

(Warning : watch for violent skipping)

Recall the classical canonical transformations

[tex]Q_i = Q_i (q,p) [/tex]

[tex]P_i = P_i (q,p) [/tex]

transforming the positions and momenta [itex]q_i,p_i[/itex] to the new set [itex]Q_i,P_i[/itex].

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

[tex]p_i = \frac{\partial F}{\partial q_i} [/tex]

[tex]Q_i = \frac{\partial F}{\partial P_i} [/tex]

Now, specifically consider the generating function [itex]F(q,P) = q_iP_i [/itex]. From above, the transformations yield

[tex]p_i = \frac{\partial F}{\partial q_i} = P_i [/tex]

[tex]Q_i = \frac{\partial F}{\partial P_i} = q_i [/tex]

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

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

[tex]Q_i = q_i + \delta q_i [/tex]

[tex]P_i = p_i + \delta p_i [/tex]

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

[tex]F = q_iP_i + \epsilon G(q,P) [/tex]

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

[tex]F(q,P) = q_iP_i + p_i \delta q_i [/tex]

then, the transformations generated are

[tex]Q_i = q_i + \delta q_i [/tex]

[tex]P_i = p_i [/tex]

Notice that the above generating function, which generates infinitesimal spatial translations, differs from the identity transformation by the extra term [itex]p \delta q [/tex].

In QM, infinitesimal translations are generated by a unitary operator [itex] {\cal J} (dx) [/itex] defined as

[tex] {\cal J} (dx) |x \rangle = |x + dx \rangle [/tex]

where [itex] |x \rangle [/itex] 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

[tex] {\cal J}(dx) = \mathbf{1} - iK \cdot dx [/tex]

where [itex]\mathbf{1} [/itex] 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 [itex] \hat{p} [/itex]. To get the dimensions right (and borrowing de Broglie's idea), we set

[tex] {\cal J} (dx) = \mathbf{1} - \frac {ip \cdot dx}{ \hbar} [/tex]

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 : [itex] \hat{p} = -i \hbar \partial /\partial x [/itex]

- #5

Galileo

Science Advisor

Homework Helper

- 1,991

- 6

First, we'll look at the eigenstates of [itex]P_x[/itex] (in the position representation -> wavefunctions)

[tex]P_x \psi(x)=\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)=p \psi(x)[/tex]

The eigenvalue is ofcourse the momentum that we measure. This is an easy differential equation, the answer is a plane wave [itex]A\exp(ikx)[/itex], with [itex]p=\hbar k[/itex]. (k is the wavenumber and is related to the wavelength by [itex]k=2\pi/\lambda[/itex].) So [itex]p=\hbar k = h/\lambda[/itex], 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

reilly

Science Advisor

- 1,077

- 1

Regards,

Reilly Atkinson

- #7

LeonhardEuler

Gold Member

- 859

- 1

Share: