Hamiltonian problem concerning the simple harmonic oscillator

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noblegas
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Homework Statement



use the hamiltonian equation H=H_x+H_y+H_z to show that wave functions of the form
[tex]\varphi[/tex](r)=[tex]\phi[/tex]i(x)[tex]\phi[/tex]j(y)[tex]\phi[/tex]k(z)

where the functions phi_i(x) are the energy eigenfunctions for a 1-d SHM , satisfy H*phi=E*phi , and find the followed values of E for the 3-d oscillator

Homework Equations

H=-h-bar^2/2m*del^2+1/2*K*r^2, k is a constant

r^2=x^2+y^2+z^2

del^2=d^2/dx^2+d^2/dy^2+d^2/dz^2

The Attempt at a Solution



H=d^2/dx^2(-h-bar^2/2m)+1/2*k*x^2+

Homework Statement



use the hamiltonian equation H=H_x+H_y+H_z to show that wave functions of the form
[tex]\varphi[/tex](r)=[tex]\phi[/tex][SUBi[/SUB](x)[tex]\phi[/tex]j(y)[tex]\phi[/tex]k(z)

where the functions phi_i(x) are the energy eigenfunctions for a 1-d SHM , satisfy H*phi=E*phi , and find the followed values of E for the 3-d oscillator

Homework Equations

H=-h-bar^2/2m*del^2+1/2*K*r^2, k is a constant

r^2=x^2+y^2+z^2

del^2=d^2/dx^2+d^2/dy^2+d^2/dz^2

The Attempt at a Solution



H=d^2/dx^2(-h-bar^2/2m)+1/2*k*x^2+

Homework Statement



use the hamiltonian equation H=H_x+H_y+H_z to show that wave functions of the form
[tex]\varphi[/tex](r)=[tex]\phi[/tex][SUBi[/SUB](x)[tex]\phi[/tex]j(y)[tex]\phi[/tex]k(z)

where the functions phi_i(x) are the energy eigenfunctions for a 1-d SHM , satisfy H*phi=E*phi , and find the followed values of E for the 3-d oscillator

Homework Equations

H=-h-bar^2/2m*del^2+1/2*K*r^2, k is a constant

r^2=x^2+y^2+z^2

del^2=d^2/dx^2+d^2/dy^2+d^2/dz^2

The Attempt at a Solution



H=d^2/dx^2(-h-bar^2/2m)+1/2*k*x^2+d^2/dy^2(-h-bar^2/2m)+1/2*k*y^2+d^2/dz^2*(h-bar^2/2m)+1/2*k*z^2

not sure how to proceed with my solution but I am sure the equation i*h-bar*dphi/dt=H*ohi will play a role in helping me form my final solution
 
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kuruman said:
The problem is asking you to solve the 3-D Schrödinger equation by using the method of separation of variables. Are you familiar with this method?

Somewhat, the second derivative are already explicitly given in the equation for the hamiltonian. you would not take the second derivative of phi with respect to x, y, and z and then plug them into the hamiltonian expression would you?
 
kuruman said:
Yes I would. Then I would divide by φ(r) and see what I get.

yeah, but phi is not given, only my hamiltononian equation is given and I have to show that the equation [tex]\varphi[/tex](r)=[tex]\phi[/tex]i(x)[tex]\phi[/tex]j(y)[tex]\phi[/tex]k(z)
 
kuruman said:
The problem is asking you to solve the 3-D Schrödinger equation by using the method of separation of variables. Are you familiar with this method?

I don't think the problem requires him to actually solve the 3D-Schrödinger equation, only show that as long as [itex]\phi_i(x)[/itex], [itex]\phi_j(y)[/itex] and [itex]\phi_k(z)[/itex] satisfy the corresponding one-dimensional Schrödinger equations, [itex]\varphi(\textbf{r})=\phi_i(x)\phi_j(y)\phi_k(z)[/itex] will satisfy the 3D equation.

@noblegas Just substitute [itex]\varphi(\textbf{r})[/itex] into the 3D Schrödinger equation and carry out the derivatives...what do you get?
 
kuruman said:
φ(x,y,z) is given. You are told it is the product of harmonic oscillator solutions to the one-dimensional Schrödinger equation. Each factor in the product is a function of a single independent variable, x, y or z.

you are right.

I got my final solution to be:

-h-bar^2/2m*((phi_i(x)''/phi_i(x))+(phj_j(y)''/phj_j(y))+(phk_k(z)''/phk_k(z))+1/2*k*(x^2+y^2+z^2)=E
 
Correct. Now look at the piece of this equation that depends only on x. This is

[tex]-\frac{\hbar ^{2}}{2m}\left(\frac{1}{\phi _{i}(x)}\frac{\partial ^{2} \phi_{i}(x)}{\partial x^{2}}\right )+\frac{1}{2}k x^{2}[/tex]

Then look at the one-dimensional Schrödinger equation in x. What can the expression above be replaced with? Do the same for y and z.
 
kuruman said:
Correct. Now look at the piece of this equation that depends only on x. This is

[tex]-\frac{\hbar ^{2}}{2m}\left(\frac{1}{\phi _{i}(x)}\frac{\partial ^{2} \phi_{i}(x)}{\partial x^{2}}\right )+\frac{1}{2}k x^{2}[/tex]

Then look at the one-dimensional Schrödinger equation in x. What can the expression above be replaced with? Do the same for y and z.
[tex]-\frac{\hbar ^{2}}{2m}\left(\frac{1}{\phi _{i}(x)}\frac{\partial ^{2} \phi_{i}(x)}{\partial x^{2}}\right )+\frac{1}{2}k x^{2}[/tex]=[tex]E[/tex]? Sorry about my latex
 
kuruman said:
Which E? It can't be the same as the E in the original equation because that is the sum of three such terms. Better call it E1. Now do the same for y and z and put it together.

I think it would be appropriate to call this particular expression E_x since the derivatives of phi are taking with respect to x.
 
Correct. You know from having solved the one-dimensional problem what the allowed values for Ex are. You should write them in terms of quantum number nx. Do the same for the other two directions, put it back in the 3-D equation and you should end up with an expression for the total energy E for the 3-D oscillator.
 
kuruman said:
Correct. You know from having solved the one-dimensional problem what the allowed values for Ex are. You should write them in terms of quantum number nx. Do the same for the other two directions, put it back in the 3-D equation and you should end up with an expression for the total energy E for the 3-D oscillator.

Should I apply the Energy equation for the simple harmonic oscillator E=(1/2+n)*[tex]\hbar[/tex]*[tex]\varpi[/tex]