# Harmonic oscillator Hamiltonian.

1. Feb 22, 2014

### Catflap

I know I've seen this point about roots discussed somewhere but I can't for the life of me remember where. I'm hoping someone can point me the right direction.

Here's the situation:-

The standard derivation of the quantum HO starts with the classic Hamiltonian in the form H = p2 + q2 (ignoring constants)

Then the raising and lowering operators are introduced with complex values such that the QM hamiltonian becomes something of the form (p + iq)(p-iq), which expands into p2 + q2 plus the commutator [q.p] as an imaginary part and the rest follows.

I found myself questioning this. Beyond the fact that 'it works' (which is fair enough). When you consider that the commutator in question evaluates to ih/2, what we have actually done is to simply add h/2 to the original classical Hamiltonian without any clear justification, disguising the fact by a subtle ordering of the steps.

I know this can be justified mathematically in other ways. I just need to know where I can find it.

2. Feb 22, 2014

### stevendaryl

Staff Emeritus
I'm not sure what you mean. If you choose units so that
$H = p^2 + x^2$, that means in terms of raising and lowering operators:

$H = 2 \hbar (A^\dagger A + \frac{1}{2})$

where $A = \frac{1}{\sqrt{2 \hbar}} (x + ip)$
$A^\dagger = \frac{1}{\sqrt{2 \hbar}} (x - ip)$

You haven't changed the value of the Hamiltonian. (In these units, the natural frequency $\omega = 2$ and the mass is $\frac{1}{2}$ and the spring constant $k$ is $m \omega^2 = 2$)

3. Feb 22, 2014

### Catflap

A*A = 1/2h(x-ip)(x+ip) = x2 + p2 + i(xp-px)

Classically the imaginary term is zero so you have A*A = x2 + p2 = H
This gives the result you would expect classically and the lowest energy possible is zero.

In QM, the commutator has the value ih/2. so that the imaginary term now has a value of h/2 -
i.e the non-zero ground state energy.

You have simply placed a value of 1/2 in the QM Hamiltonian you give - that's often the case in textbooks and I seek to justify that addition.

Last edited: Feb 22, 2014
4. Feb 23, 2014

### Bill_K

If it bothers you, use instead H = ½(a*a + aa*), which gives you back the classical Hamiltonian exactly.

5. Feb 24, 2014

### Catflap

That's just H + h/2 plus H - h/2. :)

My point is that the classical H is an approximation to the quantum version, which is actually H_classical + h. You can't take an approximation and use it to derive the correct result.
The textbook argument somehow suggests that the QH is derived from the classical when in truth, it's merely a fudge to get the 'right answer' according to experiment. That's fair enough as far as it goes but the 'derivation' is not really valid.

When p2 + q2 is expanded in terms of real and imaginary parts, it can be done in more than one way. Half of these lead to a correct result. The others do not. We 'fix' that by throwing away those results where the energy turns out negative. Obviously they must be wrong so we can ignore them.
Even the steps leading to the 'correct' result are dubious because they involve introducing imaginary quantities into supposedly real dynamic quantities, which conveniently turn out to have just the right value as a commutator.

The 'proof' slides from being an equation among dynamic quantities to being an operator expression without any clear point where the transition takes place.

I wrote my original post hoping to find a better form of the proof but having thought about it, I realise there can't be one - as I said, you can't argue from an approximation to an accurate result. The approximation contains nothing that could act as a seed for it's own correction. All you can really do is show the equations are not totally inconsistent.

6. Feb 24, 2014

### ChrisVer

the classical limit is taken by send h to zero.
Of course you couldn't wait for energy=0 state in QM, since due to the Heisenberg's uncertainty principle, you'd still have some energy even in the ground state something you can expect in the classical case. Of course again, taking the approximation of h going to zero, you'll still get the zero-energy ground state result you'd get from the classical HO as well...

I don't really understand your question.
The addition to classical hamiltonian is just the result of the fact that the momentum and the position operators don't commute in QM.

7. Feb 24, 2014

### Bill_K

No it isn't. You've missed the point entirely.

The classical Hamiltonian is not uniquely determined - one can add to it an arbitrary constant. The quantum Hamiltonian shares this ambiguity. One can choose to write H = c a*a + d aa* where c + d = 1 and get the same result.

Nature does not "derive" quantum theory from classical - it goes the other way. The world is quantum to begin with, and reduces to classical in a certain limit. We try to reverse this. Starting with a known classical theory, we try to guess the original quantum theory from it. Unsurprisingly, this process is not a rigorous derivation.

8. Feb 24, 2014

### stevendaryl

Staff Emeritus

$H = \dfrac{p^2}{2m} + \dfrac{m \omega^2 x^2}{2}$

and you make the substitutions:

$A = \sqrt{\dfrac{m \omega}{2 \hbar}} x + i \dfrac{1}{\sqrt{2 \hbar m \omega}} p$
$A = \sqrt{\dfrac{m \omega}{2 \hbar}} x - i \dfrac{1}{\sqrt{2 \hbar m \omega}} p$

then the same Hamiltonian in terms of $A$ and $A^\dagger$ is

$H = \hbar \omega (A^\dagger A + \frac{1}{2})$

not

$H = \hbar \omega A^\dagger A$

Of course, adding a constant to a Hamiltonian makes no difference at all, since energies are not defined in nonrelativistic physics, only energy differences. So it doesn't actually make any difference which Hamiltonian you use. In terms of $p$ and $x$:

$\hbar \omega (A^\dagger A + \frac{1}{2}) = \dfrac{p^2}{2m} + \dfrac{m \omega^2}{2}$

$\hbar \omega A^\dagger A = \dfrac{p^2}{2m} + \dfrac{m \omega^2}{2} - \dfrac{\hbar \omega}{2}$

9. Feb 24, 2014

### Catflap

I'm not talking about nature deriving anything from anything else. That's a bizarre idea.
I'm talking about a specific mathematical 'proof' offered in physics textbooks.

The problem given is to come up with a value for the Hamiltonian of the HO consistent with the concept of state vectors, operators and eigenvalues.
Given that it must reduce to the classical form in the limit as h->0, most (all?) books start with the classical Hamiltonian and 'expand' it in terms of imaginary numbers, show that the imaginary part has the form of a commutator and that replacing that commutator with the QM version of the Poisson bracket gives a form for the Hamiltonian that agrees with experiment.

I'm just saying that this is at best fortuitous and at worst a downright 'fiddle'.

It would be perfectly honest and acceptable to begin with 'let's guess what the Hamiltonian might be and see if we can get a result that agrees with experiment and reduces to the classical form when h->0' but that's not how it's presented.

_________________________________________________________________
Oh - and yes, it is. Being pedantic about it:-
hw(a a*) = (2m)-1(p - imwq)(p + imwq)
=(2m)-1(p2 + p2 + imw(qp-pq))
= H - 1/2 hw
whereas
hw(a* a) = (2m)-1(p + imwq)(p - imwq) merely changes the sign of the commutator giving H + 1/2 hw
Add the two together and you just get back the classic H.

10. Feb 24, 2014

### ChrisVer

So are you asking why the Hamiltonian of the HO is like the classical one with the momenta and position? that's because you just change the physical quantities of classical mechanics into operators in QM...
Then you can of course solve the problem.
One way to solve it is by introducing the ladder operators which help in simplifying the problem (and also contain information about symmetries). The reason you choose them imaginary is called complexification (of the algebra) and its how you make the root operators of your algebra. It's just an alternative way, and not the only one... for example you can still solve the HO without ladder operators, inserting the Hermit Polynomials etc...

11. Feb 24, 2014

### WannabeNewton

It's for the same reason we introduce the operators $J_{+} = J_x + iJ_y$ and $J_{-} = J_{+}^{\dagger}$ when solving for the eigenvalue spectrum of $J^2$ and $J_z$. As Chris mentioned, we're just performing a complexification of the Lie algebra.

As far as the eigenvalue problem of the QHO goes we don't need to use the extremely elegant method of complexification. We can just use Frobenius' method to solve the problem directly but that offers no insight.

12. Feb 24, 2014

### Catflap

I would just ask people to read what I've actually said - not what someone else may have said about what they think I might have been asking.

13. Feb 24, 2014

### ChrisVer

I didn't answer on anyone else- since I think the +/-constant in Hamiltonian has already been answered...I was just trying to understand what you asked for, and answered exactly that question of yours:
"The problem given is to come up with a value for the Hamiltonian of the HO consistent with the concept of state vectors, operators and eigenvalues.Given that it must reduce to the classical form in the limit as h->0, most (all?) books start with the classical Hamiltonian and 'expand' it in terms of imaginary numbers, show that the imaginary part has the form of a commutator and that replacing that commutator with the QM version of the Poisson bracket gives a form for the Hamiltonian that agrees with experiment."
through a mathematical approach (coming from group theory).

Last edited: Feb 24, 2014
14. Feb 24, 2014

### andrewkirk

Catflap you have said that you question the validity of the proof but have not said what the proof is whose validity you question. You've indicated that it has something to do with the Hamiltonian of the harmonic oscillator but that's not enough for others to understand exactly what proof you are questioning.

If you question the validity of some proof, there must be some theorem to which the proof relates, and you haven't said what that is. The raising and lowering operators will be valid tools for proving some theorems and not others. You need to tell us what theorem concerns you.

Is it the derivation of the eigenvalues of the Hamiltonian that you question? If so, I don't see the problem. My text (Shankar) uses $a^\dagger$ and $a$ to derive the eigenvalues of $H$, using the identity $\frac{H}{\hbar \omega}=(a^\dagger a + \frac{1}{2})$. Do you question the validity of that identity?

Once that identity is established, it is fairly straightforward to derive the eigenvalues. The only other input needed is that they cannot be negative.