Solutions to a quantum harmonic oscillator - desperate for help

[Bowenwww]

I need to find the value σ for which:

ψ₀(x) = (2πσ)^-1/4 exp(-x²/4σ)

is a solution for the Schrodinger equation

I know the equation for the QHO is:

Eψ = (P²/2m)ψ + 1/2*mw²x²ψ

I've tried normalizing the wavefunction but I end up with a σ/σ term :(

Any help would be greatly appreciated :)

 

[Clever-Name]

You aren't using the right form of the Schrodinger Equation. Try plugging it into the form with the derivatives:

[tex]
E\Psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\Psi + \frac{1}{2}m\omega^{2}x^{2}\Psi[/tex]

Solve the resulting equation

edit - wrote the wrong equation.

 

[Bowenwww]

that is the equation I was using I just put P as the momentum operator :( I'm confused! Do I enter the wavefuntion into schrodinger and then solve for what?

<3 xoxo

 

[Clever-Name]

Are you given the general expression for [itex] E_n [/itex] ?? You are given the expression for [itex] \psi_n [/itex] for [itex] n = 0 [/itex]. So take the necessary derivatives, plug in the expression for E_0 and recall that this solution must be valid for all x. Do you know a specific value for x that might make this much easier?

edit - the general expression for E isn't really necessary now that I think about it. You'll still be able to solve for [itex] \sigma [/itex] in terms of E

 

[Bowenwww]

I can find σ in terms of E but the question says give the value for it that makes ψ₀ (the one I originally posted) a solution to the schrodinger

Then part II asks me to find the energy associated by putting the wavefunction ψ₀ into schrodinger like you said, so I'm guessing there must be an actual value or σ?

I'm really confuzzled! :(

All I'm given is the function ψ₀ like I originally posted, nothing else.

<3 xoxo

 

[Clever-Name]

Huh, I'm confused then as well, typically when a question asks you what value for a constant makes it a solution to an equation the thing to do is plug it into said equation. Could you possibly write out the entire problem? Or is it in a certain textbook? If so, what text and what's the problem number.

 

[Bowenwww]

Part I

Find the value σ₀ for which:

ψ₀(x) = (2πσ)^-1/4 exp(-x²/4σ)

is a solution to the Schrodinger equation

Part II

What is the associated energy in part I? [hint: insert the given wave function into the schrodinger equation]

That's it word for word.

Thank you so much for helping I really appreciate it! <3 xoxoxo

 

[qbert]

when you substitute ψ into schrodingers equation
Hψ=Eψ, you know E is a constant (independent of x)
but when you calculate Hψ you'll get terms proportional
to x^2. you need to pick σ so that those terms go away.

 

[Bowenwww]

I put the wavefunction in and took the 2nd derivative and all that I canceled a load of common terms and ended up with

E = (-h(bar)²/2m)*(x²)/(4σ²) + 1/2*(mω²x²)

Is this right? I don't get how to pick a value so all the x's disappear?

<3 Cheers guys xoxox

 

[Clever-Name]

qbert, that would mean [itex] \sigma [/itex] is a function of x^2. This is definitely not the case.

 

[Clever-Name]

Bowenwww, You should end up with the following:

[tex] E = \frac{\hbar^{2}}{4m\sigma} - \frac{\hbar^{2}x^{2}}{8m\sigma} + \frac{1}{2}m\omega^{2}x^{2} [/tex]

 

[qbert]

more like:
[tex]E = \frac{1}{4}\frac{\hbar^2}{m \sigma}-\frac{1}{8}\frac{\hbar^2x^2}{m\sigma^2}+\frac{1}{2} m \omega^2 x^2[/tex]

Clever-Name, picking σ to make the x^2 terms 0 does not
mean that σ is a function of x. in fact, if you just do it
you'll see precisely what σ has to be!

 

[Bowenwww]

So that's how I find out the energy levels, right? But I still don't know how to get the sigma?

I really appreciate this guys I just need pointing in the right direction ^ _ ^<3 xoxoxox

 

[qbert]

i told you how to get σ.

 

[Bowenwww]

But there's still two unknowns E and sigma - sorry I know I'm being really stupid here; I know how to get to the form you've given I just don't know how to get from there to a value for sigma?

:( xoxox

 

[qbert]

pick σ to make the x^2 terms go away.

 

[gabbagabbahey]

If [itex]C_1 = (a-2)x^2 +C_2[/itex] where [itex]a[/itex], [itex]C_1[/itex] and [itex]C_2[/itex] are all constants, then [itex]a[/itex] must equal 2 (otherwise [itex]C_1[/itex] would be a function of [itex]x[/itex], not a constant) and [itex]C_1[/itex] must equal [itex]C_2[/itex]

 

[Clever-Name]

qbert you're right, my mistake, but your attitude is unnecessary. Bowen is obviously struggling to figure this out and your bitter responses aren't very helpful. I also made a typo in the equation you corrected. The appropriate thing to do is assume I made that typo instead of being so pompous in response.

Bowen, The last two terms (the ones with x^2) should cancel out based on what [itex] \sigma [/itex] is. In other words, their sum should be zero! Work that out and tell me what your answer for sigma is! i.e. solve this equation:

[tex]
-\frac{\hbar^{2}x^{2}}{8m\sigma^{2}} + \frac{1}{2}m\omega^{2}x^{2} = 0 [/tex]

 

[qbert]

qbert you're right, my mistake, but your attitude is unnecessary. Bowen is obviously struggling to figure this out and your bitter responses aren't very helpful. I also made a typo in the equation you corrected. The appropriate thing to do is assume I made that typo instead of being so pompous in response.

Bowen, The last two terms (the ones with x^2) should cancel out based on what [itex] \sigma [/itex] is. In other words, their sum should be zero! Work that out and tell me what your answer for sigma is! i.e. solve this equation:

[tex]
-\frac{\hbar^{2}x^{2}}{8m\sigma^{2}} + \frac{1}{2}m\omega^{2}x^{2} = 0 [/tex]

wow. bitter and pompous and here i thought i was being helpful.
i'm sorry if i offended anyone, that wasn't my intention.