# Why isn't the constant force problem in QM classes?

1. Jul 12, 2005

### pellman

I have not been able to find the quantum constant force problem in the various QM texts that I have checked. I am certain it was not covered in any of the classes I took. Why is that? I would think it would be next thing covered after free particles.

2. Jul 12, 2005

### dextercioby

There's no such thing as "force" in QM.

Daniel.

P.S. A linearly coordinate depending potential energy operator (in the Schrödinger picture, Dirac fomulation) (let's say coming from a constant and homogenous gravitational field of a massive pointlike particle) has a discrete spectrum of eigenvalues and the Airy functions as eigenfunctions.

3. Jul 12, 2005

### QMrocks

You mean solving of Schrodinger equation with a constant force?

4. Jul 12, 2005

### dextercioby

There's no force in the SE, get over it.

Daniel.

5. Jul 12, 2005

### QMrocks

Physically, u just need to have a position dependent potential.

6. Jul 12, 2005

### Gokul43201

Staff Emeritus
Have you never come across a triangular well ? I think that's what you are looking for.

7. Jul 12, 2005

### dextercioby

Not necessarily. Think of a free particle acted upon by a time dependent perturbation to the Hamiltonian.

Daniel.

8. Jul 12, 2005

### James R

But $F = -\nabla U$

and I believe U appears in the Schrodinger equation.

9. Jul 12, 2005

### MalleusScientiarum

This is true; however as you know the non-relativistic Schrodinger equation is a scalar-valued equation. To be analogous to Newton's laws directly, you would have to have a vector-valued function in a differential equation that becomes the Schrodinger equation with some mathematical trickery. If you want to follow the classical prescription.

10. Jul 13, 2005

### vanesch

Staff Emeritus
It is the triangular potential, and it is used, because the earth gravity field (near its surface) does this. I have a collegue here at the ILL which plays with bouncing cold neutrons in the gravity field and he studies their quantum properties, V. Nesvizhevsky. Do a search on his name, you'll find some papers about it.

cheers,
Patrick.

11. Jul 13, 2005

### QMrocks

My lecturer worked out the problem of Schrodinger equation with constant force term (i.e FX, where F is a the constant value of the force and X is the postition operator) in his notes, just by using the Heisenberg equation of motion.

Also in electron transport problems in devices, one have to deal with electron transport with a triagular potential, but with open boundary.

12. Jul 13, 2005

### QMrocks

Sorry, i do not understand this concept. Does using the concept of 'force' (as in the gradient of potential V) in QM leads to serious contradictions in theory? Maybe when one try to deal with vector potential A in EM?

13. Jul 13, 2005

### dextercioby

There is no force, there are only observables which are mathematical objects (1). "Force" is the key concept in the newtonian formulation of classical mechanics. However, to apply the quantization scheme proposed by Dirac (2), on has to use the Hamiltonian formulation of classical dynamics. J.Schwinger (3) proposed a construction which starts from the Lagangian action. However, classically, before quantization, such concepts as "momentum, force, acceleration, velocity" are basically useless when applying the quantization postulate.

Daniel

Notes
(1) densly defined selfadjoint linear operators acting on the (rigged, if unbounded) separable Hilbert space of states.
(2) Graded Dirac bracket on the reduced phase space goes to $$\frac{1}{i\hbar}$$ times graded Lie gracket.
(3) See R. Newton "Quantum Physics for Graduate Students", Springer Verlag.

14. Jul 13, 2005

### Observable

convert it into potential

Constant force can be treated as linear potential. Then you can solve it sectionally. A similar problem has been investigated: V(x)=ABS(x). The solution to this potential is just Airy functions. But it doesn't have as much practical value as the ones show up in QM books(like oscillator potential and coulomb potential) and usully linear effect could be easily dealt with perturbation theory.You can check it in some mathematical handbook for Airy functions.

15. Jul 13, 2005

### ZapperZ

Staff Emeritus
Surprisingly enough, the linear potential IS quite widely used, especially in many tunneling barrier model. One of the most common application is in the Fowler-Nordheim model of field emission (see, for example, http://ece-www.colorado.edu/~bart/book/msfield.htm). The simplest approximation to such a phenomenon is a triangular barrier. This is the model most people use to describe field emission, and if you know what that is, you'll know that your plasma screen and many other devices make use of this effect.

Zz.

16. Jul 14, 2005

### Observable

I think those applications are actually treated by WKB approximation, so comes the linear potential

17. Jul 14, 2005

### ZapperZ

Staff Emeritus
Yes, but it is STILL a linear potential barrier.

The WKB approximation simply allows one to already "pre-determine" the wavefunction, rather than solving for one explicitly. In many cases, this is a detail that isn't necessary since all one cares about is the transmission probability that determine the current density. Thus, the WKB wavefunction is good enough.

Zz.

18. Jul 14, 2005

### Gokul43201

Staff Emeritus
Besides FETs, there's a whole host of other places where you see approximations to triangular wells or linear potentials - in modulation doped heterostructures, metal-semiconductor interfaces, biased junctions, etc.

And in most of these places, the linearity of the potential does not come out of the approximation used to model the system (in any case, such an approximation is chosen for a reason) but is the expected/designed profile.

19. Jul 20, 2005

### aav

See J. J. Sakurai, Modern Quantum Mechanics, revised ed. Problem 24, Chapter 2.

20. Jul 22, 2005

### Gokul43201

Staff Emeritus
And the bit about gravity-induced interference between neutron beams (also in Ch. 2)