QM: Potential Energy & Box: U(x) & Schrodinger EQ: \psi (x)

[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?

 

[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.

 

[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?

 

[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.

 

[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.

 

[dextercioby]

In short,non mathematized () 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.

 

[jtbell]

hmmm sorry, but i really haven't have a thorough background in CM yet.

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.

 

[masudr]

I consider myself to be (rather big-headedly) very knowledgeable 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 learned QM: http://users.ox.ac.uk/~quee1685/main.pdf . I would appreciate if a few people (in particular dextercioby and zapperz) could provide criticism on it.

 

[dextercioby]

I consider myself to be (rather big-headedly) very knowledgeable 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?

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

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).

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

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.

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...

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).

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.

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.

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

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

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.

 

[masudr]

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

I know Weyl ordering; it makes sense now.

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

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?

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...

OK. I can accept that.

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

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.

 

[dextercioby]

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?

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

By unnecessary mathematics, I assume you mean groups/fields? This was the way I was introduced to vector spaces;

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

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

Masud.

If you have the time & the will...

Daniel.

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

 

[selfAdjoint]

U may search for "Weyl ordering".

I tried googling on Weyl ordering, and I found this excellent tutorial on Deformation Quantizing: http://idefix.physik.uni-freiburg.de/~stefan/Skripte/intro/node1.html