How Does the Quantum Harmonic Oscillator Transition from U(x) to E?

[6Stang7]

Can someone explain to me how they went from U(x)=(1/2)kx^2 to E=(n+1/2)(h/2pi)w?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html

 

[malawi_glenn]

when you solve the shrödinger equation for that kind of potentail, the soloutins are eigenfunctions with eigenvalues (n+1/2)h_bar * omega

do you have a course book in QM ?

http://www.oru.se/oru-upload/Institutioner/Naturvetenskap/Dokument/Fysik/PJ/Kvantmekanik/stegop.pdf

 

[malawi_glenn]

and yeah, the thing i post is an alternate way to solve it, by using ladder operators. But you can find the solutions for the differential equations needed for solving the Shrödinger equation for this potential in almost any basic QM book.

 

[malawi_glenn]

LOL look at this:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc2.html#c1

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

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html#c1

a more stringent solutions can be found in a QM book as i said.

 

[6Stang7]

Ahhhhh, now this make since.

I have a problem I am working on where there a 2N electrons (of mass m) that are free to move along the x-axis. The potential energy for each electron is U(x)=(1/2)kx^2, where k is a positive constant. I need to find the total energy of the system for a) integer angular momentum particls, and b) half-interger. (all magnetic and electric forces can be ignored.

so for a), the particles would act like bosons and not be restricted by the exclusion princple, i.e. they would all sit in the same quantum state. That would give a total energy of E=(2N)(1/2)k. (the x^2 can be dropped because all particles will be in the lowest and same state), giving E=Nk. omega=(k/m(r))^(1/2), and I can solve this in terms of k. However, would m(r), the reduced mass, be [2Nm(e)m(n)]/[(2Nm(e)+m(n))]? My thinking is that all electrons would be in one state and can be viewed as a single point mass of 2Nm(e). Can I do that?

As for part b with half-integer angular momentum, it would be just be E=(n+1/2)(h/2pi)w, which would become E=(n+1/2)(h/2pi)(k/m(r))^(1/2), correct?However, this does make a lot more sense now.

 

[malawi_glenn]

Is a statistical mechanics problem? Were you want to calculate <E> ? Expactation value of energy (as usual in Quantum physics). And for that we get a geometric serie..

If this is the problem you want to do, there is good info in this:
http://www.oru.se/oru-upload/Institutioner/Naturvetenskap/Dokument/Fysik/PJ/Kursplaner/instud05.pdf

 

[6Stang7]

this is it word for word:"there are 2N electrons (of mass m) that are free to move along the x-axis. The potential energy for each electron is U(x)=(1/2)kx^2, where k is a positive constant. I need to find the total energy of the system for a) integer angular momentum particls, and b) half-interger. (all magnetic and electric forces can be ignored."

I know that for part (b) i treat it like a quantum harmonic oscillator. However, I am unsure of what the reduced mass would be (although as I understand it electrons are still though of as point masses; therefore i can treate a group of 2N particles as a single mass of 2Nm(e).)

 

[malawi_glenn]

I do not think I can help you any further =(

do you by "angular momentum" mean intristinc angular momentum, spin ?

 

[6Stang7]

ya. one is to think of the electrons has having there normal 1/2 spin, and the other is to look at them as whole integers, i.e. think of the elctrons as bosons and then as fermions.

 

[malawi_glenn]

Well for the bosons, you just add them up 2N times, all will be in the ground state. 2N(0+1/2)h_bar*omega.

For the fermions, you get this m_s quantum number (spin "up" or "down"), so there can only be two fermions for each n.

So you get this sum

E = 2 * sum{n= 0 to r}((n+1/2)*h_bar*omega)
were r is N/2 - 1

this should be right =)