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How do I find the average kinetic energy and average potential energy of a hydrogen electron in the ground state?

In my modern physics class, we are wrapping up the 3D Schrödinger equation, and I am more than a little lost. A few chapters ago, we learned about operators, and I have an equation for both these things in 1D. It looks like $$ \left<K\right> = \int \psi K \psi \, \mathrm{d}x,$$ where $$K =-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}.$$

So,

  • How do I make that work for 3D, and is that even what I want to do here?
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    $\begingroup$ Yes you need to do that integral, but the ke operator in 3d involves $\nabla^2$ rather than just an x derivative $\endgroup$ – danimal Mar 19 '15 at 14:35
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    $\begingroup$ You are on the right track. Why not just launch into the calculation building your operators from their classical analogues? Sometime flailing and failing a few times in the process of learning is helpful. $\endgroup$ – dmckee Mar 19 '15 at 14:57
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    $\begingroup$ @Ashley I think that has to be an $\hbar$ in the definition of $K$ unless you are working with funny units. You can change $h$ to $\hbar$ by typing \hbar instead of h. $\endgroup$ – Gonenc Mar 19 '15 at 16:51
  • $\begingroup$ Besides $\left<K\right> = \int \psi K \psi \, \mathrm{d}x$ should have been $\left<K\right> = \int \psi^* K \psi \, \mathrm{d}x$ you can also change that by changing \int \psi K \psi to \int \psi^* K \psi $\endgroup$ – Gonenc Mar 19 '15 at 16:59
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One method is to use/prove the virial theorem: for a potential of the form $V\left(r\right) \propto {r^p}$ (for the hydrogen atom, $p = -1$), $$ \left<T\right>_n = \frac{p}{2} \left<V\right>_n $$ for the $n$-th energy eigenstate.

Use this, together with $$ E_n = \left<E\right>_n = \left<T\right>_n + \left<V\right>_n $$ and $$ E_n = \frac{E_1}{n^2}, \ \ \ \ \ E_1 = - \frac{1}{2} \alpha^2 m_e c^2 = - 13.6 \ eV. $$

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I don't want to spoil the beauty of 3D QM for you especially if you want to explore it by your own. However I thought I would give you some advise on how to approach to the transition from 1D to 3D.

Considering 1D QM first of all think about what it means to be on the x-axis. Who is to decide which axis is x and which axis is the y or z axis for that matter. This means that there has to be a symmetry between these axes ie. the operator $K$ cannot be an operator like $K=\hat{x}+\hat{y}^2$ if there is that kind of a symmetry between the axis. Furthermore the 3D operator $K$ must reduce to the 1D operator that you have given above if you consider the 3D operator to be 1D. Note also that your equation has to be dimensionally consistent.

The expectation value of this operator is also very straight forward. You just have to adjust something in the integral and that is all.

If you want more detailed explanation, indicate this in the comments and I'll edit my answer or answer your question in the comment section accordingly.

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You can also use that the shift in the energy eigenvalue due to a perturbation to first order is the expectation value of the perturbation in the unperturbed state. If you multiply the kinetic energy term by $\lambda$ and the potential energy term by $\mu$, you have essentially the same Hamiltonian, so you can write down the ground state energy without much effort. The derivatives w.r.t. $\lambda$ and $\mu$ will then give you the two desired expectation values.

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