# Help Needed: Dirac's Book

1. Jul 25, 2014

### anorlunda

I understand from Susskind's lectures that if we need a constant, that we can assign any label to it, and that i$\hbar$ is merely a prescient choice. But I fail to understand Dirac's logic of why "$\hbar$ must simply be a number."

Also, am I right in feeling that HUP just appeared out of nowhere in this mysterious algebraic step? If so, Dirac didn't call attention to it in this chapter.

p.s. I apologize for a previously fumbled attempt to post this question.

2. Jul 25, 2014

### bolbteppa

What else could it be? You can't have it as an operator because you wont be able to cancel it from both sides when you plug

u₁v₁-v₁u₁ = iℏ[u₁,v₁]
u₂v₂-v₂u₂ = iℏ[u₂,v₂]

into

[u₁,v₁](u₂v₂-v₂u₂)=[u₂,v₂](u₁v₁-v₁u₁)

As far as I can see, Dirac is saying that because of [u₁,v₁](u₂v₂-v₂u₂)=[u₂,v₂](u₁v₁-v₁u₁) we need something like
[u₁,v₁](u₂v₂-v₂u₂)=[u₂,v₂](u₁v₁-v₁u₁) ---> (u₁v₁-v₁u₁)(u₂v₂-v₂u₂)=(u₂v₂-v₂u₂)(u₁v₁-v₁u₁)
to hold, however there's no reason why we can't include a multiplicative constant on both sides
λ(u₁v₁-v₁u₁)(u₂v₂-v₂u₂)=(u₂v₂-v₂u₂)λ(u₁v₁-v₁u₁)=λ(u₂v₂-v₂u₂)(u₁v₁-v₁u₁) (last equality not possible in general if λ is an operator)
Now, on page 28 Dirac points out that even though u₁ and v₁ are real, we might have that u₁v₁ is complex, however i(u₁v₁ - v₁u₁) is always real, so if [,] is to copy the Poisson bracket (which gives real results) we should have that λ = i, though λ = μi is entirely possible as well (though something like λ = 3 + μi isn't). Experiments show that μ = h/2pi so we say μ = h/2pi = ℏ.

Dirac does Heisenberg uncertainty a few pages later. How have you found the first 3 chapters of Dirac?

3. Jul 25, 2014

### anorlunda

Thanks for the help. It will take a while to absorb your answer. I'm not a grad student, I'm 69, retired, and studying QM for the first time, so it takes me more time.

I've been thinking of this approach:

[u₁,v₁](u₂v₂-v₂u₂) = [u₂,v₂](u₁v₁-v₁u₁)​
rearranging
[u₁,v₁]/(u₁v₁-v₁u₁) = [u₂,v₂]/(u₂v₂-v₂u₂)​

thus separating u1 v1 on the left and u2 v2 on the right. Then if the equality of those ratios holds true even if u1 v1 are unrelated to u2 v2, then it must hold for any pairs of variables. As you say, "what else could it be" besides a constant?

I'm finding it progressively more difficult, the more I go. I frequently need to backtrack and consult earlier sections; as in your explanation of page 86 referring back to page 28. But learning this stuff the "real" way with math rather than with metaphors and analogies, is rewarding.

I am surprised to learn how much of QM Dirac deduces from the calculus alone; such as the profound step that is the subject of this thread.

4. Jul 25, 2014

### bolbteppa

Okay well first off, apologies - in your OP you quoted Dirac as saying

[u₁,v₁](u₂v₂-v₂u₂)=[u₂,v₂](u₁v₁-v₁u₁)

but he actually says

[u₁,v₁](u₂v₂-v₂u₂)=(u₁v₁-v₁u₁)[u₂,v₂]

and we have to be careful about that - you can't write it the first way because these are operators we're multiplying (i.e. matrices), and these do not commute (AB =/= BA, in general).

Second, your proof is more or less the real jist of what's going on. But notice you divided - you're not dividing numbers you're dividing operators, so we have to be more careful and realize division is really multiplying by an inverse [i.e. 3/4 = 3.(4)⁻¹]. For operators, inverses do not always exist (if the determinant is zero it's not invertible), so even going from
[u₁,v₁](u₂v₂-v₂u₂)=(u₁v₁-v₁u₁)[u₂,v₂]
to
(u₁v₁-v₁u₁)⁻¹[u₁,v₁](u₂v₂-v₂u₂) =[u₂,v₂]
may not be possible in general, and we need generality.

Intuitively we know your method should basically work, i.e. it should work in the nice case of invertibility anyway, so we just come up with some other reason why we should get the answer you got, and if it's right it's right...

Removing the subscripts
[u,v](uv-vu)=(uv-vu)[u,v]
and seeing that by setting [u,v] = λ(uv-vu) we can make this identity hold (where λ is a constant), voila. Note that by defining [u,v] = 0 everything would also work, and the axioms would hold, so [u,v] = λ(uv-vu) is not the only possibility, we just found one that works nicely. This definition has the same form as the classical Poisson bracket Dirac mentioned, and also - you get this kind of definition arising naturally if you take one of the other approaches to QM (e.g. Landau).

Don't be shy with questions about Dirac, I'm hoping to finally read it properly over the next two months (having constantly skipped over, backtracked, closed, given away etc... etc... Dirac this year :tongue:).

5. Jul 25, 2014

### bhobba

I learnt QM from Von-Neumann and Dirac's book.

I came across that one and mucked around with that equation and by dint of effort managed to prove it but cant recall the key idea - your approach looks vaguely familiar.

That said if your goal is to learn QM I would NOT recommend Dirac. I did it but from that experience would not put others through it.

https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465036678

Then move onto Ballentine:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

Ballentine will be slower going - but persevere. Chapter 3 is tough mathematical going - but even if the detail defeats you, you will get the gist.

Ballentine explains carefully what h bar really is. You can prove that for the momentum and velocity operator P = MV where M is a constant. 1/h bar is defined to be M/m where m is the particles mass - its basically a constant to convert between whatever units that M is in to the units of mass. One also finds similar equations where the same constant crops up - and again it simply converts units.

Then you can come back to Dirac.

That's the funny thing about math and physics in general.

It reaches its ultimate expression in Noethers Theorem IMHO:
http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html

Thanks
Bill

Last edited by a moderator: May 6, 2017
6. Jul 25, 2014

### marcusl

Dirac is pretty intense, you should use an undergrad book instead. There are tons of them (Griffiths is popular), and numerous PF threads for you to search on their pros and cons.