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I've been told (in class, online) that the ground state of the 3D quantum harmonic oscillator, ie:

[tex]\hat H = -\frac{\hbar^2}{2m} \nabla^2 + \frac{1}{2} m \omega^2 r^2 [/tex]

is the state you get by separating variables and picking the ground state in each coordinate, ie:

[tex]\psi(x,y,z) = A e^{-\alpha(x^2+y^2+z^2)}[/tex]

where [itex]\alpha = m \omega/2\hbar[/itex], and this state has energy [itex]3\hbar \omega /2[/itex] (the sum of that from each coordinate).

On the other hand, going back to the schrodinger equation, assume a spherically symmetric solution (ie, l=0) and make the transformation [itex]\chi(r)= r \psi(r)[/itex]. Then the schrodinger equation becomes:

[tex]-\frac{\hbar^2}{2m} \frac{d^2 \chi}{dr^2} + \frac{1}{2} m \omega^2 r^2 \chi =E \chi[/tex]

which is the 1D SHO equation, and so we have the solution:

[tex]\chi(r) = B e^{-\alpha r^2} [/tex]

or:

[tex]\psi(r) = \frac{B}{r} e^{-\alpha r^2} [/tex]

with an energy [itex]\hbar \omega /2[/itex]. In fact, the 1st excited state of the 1D harmonic oscillator, with energy [itex]3\hbar \omega /2[/itex], is [itex]\chi(r)=A r e^{-\alpha r^2} [/itex], and this is just the solution found above by separating variables. So that wasn't the ground state, this new one (which isn't in separated variables form) is. Is this a correct analysis? Is there a general way to determine if a given state is the ground state of a given hamiltonian?

[tex]\hat H = -\frac{\hbar^2}{2m} \nabla^2 + \frac{1}{2} m \omega^2 r^2 [/tex]

is the state you get by separating variables and picking the ground state in each coordinate, ie:

[tex]\psi(x,y,z) = A e^{-\alpha(x^2+y^2+z^2)}[/tex]

where [itex]\alpha = m \omega/2\hbar[/itex], and this state has energy [itex]3\hbar \omega /2[/itex] (the sum of that from each coordinate).

On the other hand, going back to the schrodinger equation, assume a spherically symmetric solution (ie, l=0) and make the transformation [itex]\chi(r)= r \psi(r)[/itex]. Then the schrodinger equation becomes:

[tex]-\frac{\hbar^2}{2m} \frac{d^2 \chi}{dr^2} + \frac{1}{2} m \omega^2 r^2 \chi =E \chi[/tex]

which is the 1D SHO equation, and so we have the solution:

[tex]\chi(r) = B e^{-\alpha r^2} [/tex]

or:

[tex]\psi(r) = \frac{B}{r} e^{-\alpha r^2} [/tex]

with an energy [itex]\hbar \omega /2[/itex]. In fact, the 1st excited state of the 1D harmonic oscillator, with energy [itex]3\hbar \omega /2[/itex], is [itex]\chi(r)=A r e^{-\alpha r^2} [/itex], and this is just the solution found above by separating variables. So that wasn't the ground state, this new one (which isn't in separated variables form) is. Is this a correct analysis? Is there a general way to determine if a given state is the ground state of a given hamiltonian?

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