Finding the parameters for Harmonic Oscillator solutions

AI Thread Summary
The discussion focuses on determining the parameter α for the Harmonic Oscillator solution Ψ(x)=Axe^{-\alpha x^2} using the Schrödinger equation. The derived equation shows that the relationship must hold for all x, leading to the conclusion that specific values of α and E are necessary. By evaluating the equation at x=0, α is found to be mω/2ħ, corresponding to the energy level of 3ħω/2, which represents the first excited state of the oscillator. The importance of grouping terms by powers of x is emphasized to ensure the polynomial equality holds across all x values. The conversation concludes with clarity on the necessity of these conditions for a valid solution.
gabu
Messages
5
Reaction score
0

Homework Statement



Using the Schrödinger equation find the parameter \alpha of the Harmonic Oscillator solution \Psi(x)=A x e^{-\alpha x^2}

Homework Equations



-\frac{\hbar^2}{2m}\,\frac{\partial^2 \Psi(x)}{\partial x^2} + \frac{m \omega^2 x^2}{2}\Psi(x)=E\Psi(x)

E=\hbar\omega(n+\frac{1}{2})

The Attempt at a Solution



Using the Schrödinger equation we have arrive at

-2\alpha (2x^2\alpha-3)+\frac{m^2\omega^2x^2}{\hbar^2} = \frac{2m}{\hbar^2}E

If I make x=0 I obtain

\alpha = \frac{m\omega}{2\hbar}

using the information that the energy level of the oscillator is the same as the highest power in the solution, meaning E=3\hbar\omega/2.

Now, my problem with this solution is the need to make x=0 to arrive at it. I know that the equation holds for every x, so it is justifiable to consider the origin. The thing is, however, that shouldn't I be able to solve the equation without this assumption? Shouldn't it be independent of x?

Thank you very much.
 
Physics news on Phys.org
Hello.
gabu said:
Using the Schrödinger equation we have arrive at

-2\alpha (2x^2\alpha-3)+\frac{m^2\omega^2x^2}{\hbar^2} = \frac{2m}{\hbar^2}E
This equation must be satisfied for all ##x##. This can be true only for one specific value of ##\alpha## and one specific value of ##E##.

Group together terms of like powers in ##x## to form a polynomial in ##x##. Use the fact that if a polynomial in ##x## is equal to zero for all values of ##x## in some finite domain, then each coefficient in the polynomial must be zero.
 
  • Like
Likes gabu
gabu said:
Using the Schrödinger equation find the parameter αα\alpha of the Harmonic Oscillator solution Ψ(x)=Axe−αx2Ψ(x)=Axe−αx2\Psi(x)=A x e^{-\alpha x^2}
This is actually the first excited state of a 1-D quantum harmonic oscillator having energy 3/2ħω
Where α is mω/2ħ
 
TSny said:
Hello.

This equation must be satisfied for all ##x##. This can be true only for one specific value of ##\alpha## and one specific value of ##E##.

Group together terms of like powers in ##x## to form a polynomial in ##x##. Use the fact that if a polynomial in ##x## is equal to zero for all values of ##x## in some finite domain, then each coefficient in the polynomial must be zero.

Oh, I can see now. Thank you very much.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top