# A few QM questions.

1. Mar 2, 2005

### misogynisticfeminist

1. Why is the potential function U(x) infinite outside the 1-d rigid box and 0 inside? Is this proven by the schrodinger equation when $$\psi (x)$$ is 0 outside the box?

2. Why is it that in QM, $$potential energy= \frac {p^2}{2m}$$.

I heard that it is a consequence of the debroglie relations but how can it be if the relations have no mass involved?

Last edited: Mar 2, 2005
2. Mar 2, 2005

### dextercioby

1.That's the model.The potential is assumeed a constant value inside the box and infite outside.The infinite value of the potential automatically implies the wavefunction to be zero outside the box.

2.You mean KE=p^2/2m...That's CM and the quantization postulate (assuming you mean operators,else it's plain simple CM).

Daniel.

3. Mar 2, 2005

### misogynisticfeminist

^ CM?? uhhhh i dun really understand the acronym, can you explain that? thanks alot....And why is the relation a result of the quantization postulate?

4. Mar 2, 2005

### dextercioby

Classical Mechanics...This $$\hat{H}=\frac{\hat{p}^{2}}{2m}$$ is a result of the 2-nd postulate.

What does that postulate say...?

Daniel.

5. Mar 2, 2005

### misogynisticfeminist

hmmm sorry, but i really haven't have a thorough background in CM yet. Is it alright if you explain it here? thanks alot. I'm actually self-taught in QM.

6. Mar 2, 2005

### dextercioby

In short,non mathematized (:yuck:) version,you quantize every classical observable by making the function become a linear operator on the Hilbert space of states...

The momentum & the Hamiltonian are 2 examples...

Daniel.

7. Mar 2, 2005

### Staff: Mentor

Have you at least seen the classical-mechanics formulas for momentum and kinetic energy?

$$p = mv$$ and $$K = \frac {1}{2} mv^2$$

Solve the first equation for v and substitute into the second one.

8. Mar 2, 2005

### masudr

I consider myself to be (rather big-headedly) very knowledgable in QM. But I have to ask, just 'cos $T=\frac{p^2}{2m}$ in CM doesn't mean it should translate directly over into QM, or does it? I know that the observables can be associated with any Hermitian operator, and also that the operators associated with observables should satisfy Heisenberg's Uncertainty relations (namely $\mathbf{xp-px}=i\hbar$).

So once we have chosen the operator for position to be pre-multiply by the position, the moment operator follows, since

$$(\mathbf{x})(-i\hbar\nabla)\psi-(-i\hbar\nabla)(\mathbf{x})\psi=i\hbar\psi$$

as required. So I'm happy with the operators for position and momentum. But I question the logic in choosing the operators for energy and angular momentum based on their classical definitions. (I'm not saying they're wrong because I know they work remarkably well).

My misunderstanding may well arise because (as we were discussing earlier) I only know the Lagrangian formalism in detail, and also QM in detail, but do not yet know the Hamiltonian formalism in detail.

In fact, this document shows how I learnt QM: http://users.ox.ac.uk/~quee1685/main.pdf [Broken]. I would appreciate if a few people (in particular dextercioby and zapperz) could provide criticism on it.

Last edited by a moderator: May 1, 2017
9. Mar 2, 2005

### dextercioby

There are rules.The postulate of quantization explains them very well.U may search for "Weyl ordering".

No,that's a particular case of a far more general statement

There's much more to it.The proof for $$\langle \vec{r}|\hat{\vec{P}}|\psi\rangle =-i\hbar\nabla_{\vec{r}}\psi(\vec{r})$$ is rather tedious...

Then u should read either Roger P.Newton's latest book on QM ("Quantum Theory:A Text for Graduate Students",Springer Verlag,2002) (i'm sure you've heard of him,he's an Englishman),or J.J.Sakurai's masterpiece.They take nothing for granted and they don't use the "traditional" axiomatic approach to (non-relativistic) QM in Dirac's formulation.

As i said,u should... If you're really interested in an overview of this theory...

Rushed,missed a few key points in the axioms (which is very bad),introduced (for the purpose of the article) unnecessary mathematical details...

Useless...Better put a hand (actually both) on David J.Griffiths' (another Englishman) book.

Daniel.

Last edited by a moderator: May 1, 2017
10. Mar 2, 2005

### masudr

I know Weyl ordering; it makes sense now.

I am aware that HUP applies to any pair of observables whose operators don't commute. Is there a more general statement I'm missing?

OK. I can accept that.

Yes it was rushed! I have, funnily enough, better things to do than summarise my knowledge on QM! Although I admit that's no excuse for sloppiness. Care to elaborate on which axioms I missed out? By unnecessary mathematics, I assume you mean groups/fields? This was the way I was introduced to vector spaces; besides it just delegates listing all the axioms of a Hilbert space to previous sections; and gives the reader glimpses into other fields of mathematics.

You have recommended many books to me; I will certainly take a look.

Masud.

11. Mar 2, 2005

### dextercioby

Yes,any QM book has a jusitification for the general relation which applies to all possible pairs of QM observables.

At least that should have been the prerequisite in maths one might have...

If you have the time & the will...

Daniel.

EDIT:And one more thing,Masud,HAPPY BIRTHDAY!!

Last edited: Mar 2, 2005
12. Mar 2, 2005