# On a constructive method of quantum state preparation (Ballentine)

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• EE18
EE18
Ballentine, in his Chapter 8.1, appears to give the attached recipe for *in principle* preparing an (almost) arbitrary (pure) state (of a particle with no internal degrees of freedom) by the method of "waiting for decay to the energy ground state". My questions are fourfold:

1) From (8.1), we are clearly constructing the potential ##W_1## so that ##R(\mathbf{x})## is an eigenstate of the corresponding Hamiltonian. However, why on Earth should we expect it to be the ground state energy? Do we need to fix ##E## in a certain way so that this is true? Is there some theorem which guarantees that, if I pick ##E## a certain way, then that will be the lowest eigenstate for the corresponding potential?2) Following this question we are led naturally to ask what restrictions, if any, must be imposed on the assumed ground state energy ##E##, in order that the potentials be physically reasonable? I can't seem to think of any since ##E## is not in a denominator or something like that.

3) Ballentine writes "We must restrict ##R(\mathbf{x})## to be a nodeless function, otherwise it will not be the ground state." Why is this true? I can see from (8.1) that nodes in ##R## may cause trouble (although we seem to divide by functions which take on the value 0 all the time when using the method of separation of variables, so perhaps this isn't the issue). Is there a theorem which says ground states can't have nodes? And how, if at all, does this tie into question 1 (i.e. if it doesn't have a node, how can I guarantee this will be the ground state)?

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EE18 said:
1) From (8.1), we are clearly constructing the potential so that is an eigenstate of the corresponding Hamiltonian. However, why on Earth should we expect it to be the ground state energy? Do we need to fix in a certain way so that this is true? Is there some theorem which guarantees that, if I pick a certain way, then that will be the lowest eigenstate for the corresponding potential?
In 1D, it follows from Sturm–Liouville theory, because the Eigenfunctions are orthogonal with respect to the associated scalar product. But because Ballentine probably doesn't intent to restrict himself to 1D, this does not answer your question. In n-dimensions, we still have an associated scalar product, which actually measures the energy. I don't see immediately why the positive real valued Eigenfunction (I see why there is at most one such Eigenfunction) is known to have the lowest energy, but of course I believe this to be true.

EE18 said:
3) Ballentine writes "We must restrict to be a nodeless function, otherwise it will not be the ground state." Why is this true?
If it would get negative, then the above mentioned theory and thoughts should explain why. But of course, a negative sign could have been absorbed in ##S(x)##. But the resulting formulas would still have singularities at the nodes, and ##\nabla^2 R(x)## would make trouble at such singularities.

vanhees71
gentzen said:
In 1D, it follows from Sturm–Liouville theory, because the Eigenfunctions are orthogonal with respect to the associated scalar product. But because Ballentine probably doesn't intent to restrict himself to 1D, this does not answer your question. In n-dimensions, we still have an associated scalar product, which actually measures the energy. I don't see immediately why the positive real valued Eigenfunction (I see why there is at most one such Eigenfunction) is known to have the lowest energy, but of course I believe this to be true.
How come you believe this to be true? Even a slight hint would get me towards understanding how this construction actually means that $R$ as described is the ground state wavefunction.
gentzen said:
If it would get negative, then the above mentioned theory and thoughts should explain why. But of course, a negative sign could have been absorbed in ##S(x)##. But the resulting formulas would still have singularities at the nodes, and ##\nabla^2 R(x)## would make trouble at such singularities.
Can you explain why being negative should be a problem?

Also, would you be able to make a comment about 2) at all if you get the chance? Thank you!

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