I Question about the quantum harmonic oscillator

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Garlic

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Dear PF community, I am back with a question :)

The solutions for the quantum harmonic oscillator can be found by solving the Schrödinger's equation with:
Hψ = -hbar/2m d²/dx² ψ + ½mω²x² ψ = Eψ

Solving the differential equation with ψ=C exp(-αx²/2)
gives:

-hbar/2m (-α + α²x²)ψ + ½mω²x²ψ = Eψ

(α hbar²/2m - E)ψ + x²(½mω² - hbar²/2m α²)ψ = 0
And the solution in the internet says, that in order the Schrödinger equation to be solveable, the coefficients in the second term (with x²) has to be =0.

And we find the value of α that makes the term =0.

I don't understand: Why does the term with x² have to vanish, in order to make the Schrödinger eq. solveable?


And my second question:
we find α=mω/hbar

And with this relation: (α hbar²/2m - E ) =0 we find the Energy E0=½ω hbar.

But how do we find the other energy eigenvalues of the system? For E1=3/2 ω hbar the α term is 3 times larger.
The second term in the Schrödinger equation, (½mω² - hbar²/2m α²) won't be equal to zero if we take any other value for α that isn't =mω/hbar.

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc3.html#c1

Thank you for your time,
garlic
 
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BvU

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Hi,

You had a good chunk of ##\TeX## going already ! Only a few minor corrections to get something much more legible:

$$−\frac{\hbar}{ 2m} \frac {d^2} {dx^2}\Psi+\frac {1}{ 2} m \omega^2x^2\Psi=E\psi$$ gives

$$−\frac{\hbar}{ 2m} \frac {d^2} {dx^2}\Psi+\frac {1}{ 2} m \omega^2x^2\Psi=E\psi$$


Why does the term with x² have to vanish
Otherwise ##E## would not be a constant, but be dependent on ##x##

But how do we find the other energy eigenvalues of the system?
That's quite a bit of work.
By not setting ##C## = constant but instead taking ##C= C(x)## and looking at the equation for ##C(x)##. Solution is a power series that has to terminate and where it terminates determines ##E##
See e.g. here
It's worthwhile to invest the time to scrutinize this
 

Demystifier

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$$−\frac{\hbar}{ 2m} \frac {d^2} {dx^2}\Psi+\frac {1}{ 2} m \omega^2x^2\Psi=E\psi$$
$$−\frac{\hbar}{ 2m} \frac {d^2} {dx^2}\Psi+\frac {1}{ 2} m \omega^2x^2\Psi=E\psi$$
The two lines look exactly the same in the editor, why does the first line is not interpreted as a LaTeX formula?
 

DrClaude

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The two lines look exactly the same in the editor, why does the first line is not interpreted as a LaTeX formula?
The trick is to embed BB codes such that $$ is not interpreted as the equation delimiter. In this example, I put color tags around the second $ (choosing black, so the effect is invisible).
 

Garlic

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Thank you for your replies :)

You had a good chunk of ##\TeX## going already !
As I previewed my equations, they diddn't appear as Latex formulas, so I thought I got the syntax wrong and gave up :)

So today I've looked at the MIT text for a few hours and tried to understand every step of it. It was certainly very interesting.
I am surprised of the fact that the solution for this diff. equation
$$ \frac{d^2}{dx^2} \Psi = ( u^2 - \epsilon ) \Psi $$ took nine whole pages!
I don't think that I will be able to come up with this solution on my own.

I only have a small question: in the MIT text they used partial derivatives in the hamilton operator (which they called the energy operator) where my professors in Germany use total derivatives. This confuses me a little.
 
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BvU

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This confuses me a little
No need for confusion. For functions of one variable only, there is no need for partial derivatives. So your prof is correct.

Sometimes MIT still has an ##n## or an ##E## that has to be kept constant.
But my impression is they do everything writing partial derivatives, even for single variable functions (e.g. 0.1, 0.6). Matter of habit, I suppose.

nine whole pages!
Yes, but they hand you two important tools: asymptotic analysis and power series method.
These usually come back later in the curriculum: the hydrogen atom

On top of that, the harmonic oscillator comes back even further on in field theory -- but I don't know much about that :frown:
 

vanhees71

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Well, the brute-force treatment of the harmonic oscillator is a bit tedious, but a good exercise in calculus. The really elegant moder physicist's way is of course to use purely algebraic methods first as far as one can. In the case of the harmonic oscillator you get a complete solution by using the ladder operators to recursively define the energy eigenstates. On top this gives the very intuitive idea of quasiparticles (in this case phonons) and of the particle interpretation of free quantum fields (relativistic as well as non-relativistic).

Of course, finally you can use the algebraic result in the position representation to also derive the energy eigenfunctions. In this way you get an elegant derivation of the Hermite polynomials, including their generating function and all that.

The same holds also true for the hydrogen problem, where the key is the use of the large dynamical symmetry, providing an additional conserved quantity, the Runge-Lenz vector. This was by the way the first approach by Pauli to solve the hydrogen problem within matrix mechanics a la Born, Jordan, and Heisenberg, before Schrödinger presented his wave mechanics, which came as a relief to the physicists of this time who where not so familiar with abstract algebraic methods but rather with the analytical approach via partial differential equations.
 

BvU

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Later, later ....
First the hard way !
 

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