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I have two question about variational method of quantum mechanics.

  1. Why we always find the ground state energy by this approach. Why not the other excited states?

  2. When we find the ground state energy of helium by this method, at one stage we consider $Z$ (charge of nucleus) is a variable parameter in stead of using "2". But we have done it because our analytical value was not matching with experimental value. My question is, if experimental value always guide us to modify the Hamiltonian, I mean if we use experimental value of ground state energy of helium at first place, then what is the benefit of this method? it sounds quite silly to me. If anybody can not understand me then I will edit my question. No problem. But give some response at least.

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1. Why can you only get the ground state?

According to the wikipedia article linked to, the variational approach works by taking a wave function with some parameters (for example, a gaussian with mean $\mu$ and standard devation $\sigma$) and minimizing the energy with respect to these parameters. The minimum energy parameters give you a wave function that approximates the ground state (it won't be perfect because the ground state wave function probably won't be a gaussian or whatever). You can't get the other eigenstates this way, because the other eigenstates are typically not extrema of the hamiltonian.

2. Is the variational method useless if you already know the ground state energy?

No. The variational method winds up giving you a wave function that is supposed to approximate the ground state wave function. This wave function contains a lot more information than just the ground state energy. For example, you could get a particle's expected position.

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  • $\begingroup$ "You can't get the other eigenstates this way, because the other eigenstates are typically not extrema of the hamiltonian". Can you make me understand this line elaborately? $\endgroup$ – Hani10 Jun 25 '15 at 21:19
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    $\begingroup$ The ground state is the state with the lowest energy. That's what it means to be the ground state. The other eigenstates have higher energies, because if they had lower energies, they would be the ground state. So the other eigenstates will have a range of values all higher than the ground state. Therefore, you can't say that one of these states will be an extermum of the energy. Since these other eigenstates are not extrema of the energies, you cannot find them by extremizing the energy. $\endgroup$ – Brian Moths Jun 25 '15 at 21:24
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As another side note, if you know the ground state, it IS possible to use the variational method to get the next-lowest energy. You simply restrict your trial wavefunctions to be orthogonal to the ground state, and then minimize as usual. If you know the two lowest ground states, you can use the variational method to find the third-lowest energy, etc.

Also, if you believe the energy of your system is bounded from above, you can use the variational method to find the HIGHEST energy. You just maximize instead of minimize!

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This is just an additional remark – the other answers are fine.

However, you can get all states from the variational method, just not with the trial function method (unless your trial function space contains the exact excited states).

While the excited states are usually not extrema of the variation functional, they are stationary points! That is, by varying the functional you get the excited states as solutions, by requiring the variation vanish. Consider the variation functional with a Lagrange multiplier to enforce the norm of the states: $$E\big[\left|\psi\right> , \left<\psi\right|\big] = \left< \psi \middle| H \middle| \psi \right> - \lambda \big( \left< \psi \middle| \psi \right> -1 \big).$$ Computing the variation one gets: $$\delta E = \left< \delta \psi \middle| H \middle| \psi \right> + \left< \psi \middle| H \middle| \delta \psi \right> - \lambda \left< \delta \psi \middle| \psi \right> - \lambda \left< \psi \middle| \delta \psi \right> $$ (Note, that we consider $\left| \psi \right>$ and $\left< \psi \right|$ to be independently varied.)

From this we can see, that the functional is stationary for: $$H \left| \psi \right> = \lambda \left| \psi \right>.$$ Which is the time independent Schrödinger equation, which gives all the stationary states. The Langrange multiplier turns out to be the eigenenergy $E$ of the state.

In other words, when viewed from this angle, the variational principle is equivalent to the time independent Schrödinger equation.

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