# On a constructive method of quantum state preparation

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)?

1. I don't have Ballentine with me, but this seems to be directly answered in response to your earlier question. As you quoted there, Ballentine earlier explained the "strategy of waiting." Your quote says "it is possible to prepare the lowest energy state of a system simply by waiting for the system to decay to its ground state." This was explained as the system being open (i.e., interacting with an external environment), there being a nonzero probability of the system transitioning to the ground state, and there being irreversibility conditions such that the system will remain in its ground state forever (or for a long enough time. So here Ballentine is saying that if you start in any initial state, set the potential $$W_1$$, allow the system to interact with an environment like a thermal bath at zero temperature that can whisk away all energy and not give it back for a long time, then the system will go to the ground state of $$W_1$$, regardless of how you chose the potential and regardless of the actual value of the ground-state energy. We actually don't care about the value of the ground-state energy at all, we just construct a potential $$W_1$$ with a variety of eigenenergies and make sure that the lowest energy corresponds to the wavefunction $$\Psi_1$$.
2. Again, none of this depends on $$E$$. Increasing the potential by a constant value will simply increase all of the eigenenergies by that constant value. By setting the potential $$W_1$$ with a given value of $$E$$, all of the eigenstates will have their energies be something offset from $$E$$ by the same amount regardless of our choice of $$E$$.

Maybe you meant to ask, through these two questions: if we set $$W_1$$ in such a way, we can observe that $$\Psi_1=R$$ is an eigenstate with energy $$E$$; how do we know that there is no other eigenstate with lower energy? Luckily, this is directly answered by the third question!

1. For any one dimensional quantum potential, the ground state has zero nodes, and each subsequent excited state has one more node, exluding possible nodes at the boundaries of the wavefunction. This has been asked multiple times on this website, with a great answer here. Since $$\Psi_1=R$$ has no nodes and is an eigenstate of the Schrödinger equation with potential $$W_1$$, it must be the ground state! Any other eigenstate will have a nonzero number of modes. So this is indeed all one question, answered by the statement that the ground state of the one-dimensional time-independent Schrödinger equation has no nodes, the first excited state has one node, and so on. Then, by construction, since we know that $$\Psi_1=R$$ is an eigenstate that happens to have no nodes, it must be the ground state. Conversely, if we wanted to construct a wavefunction for which $$R$$ had a node, then it must not be the ground state of any Hamiltonian, because there would be another wavefunction with no nodes that would have less energy.

Your question does not ask how to prove that the ground state has no nodes, and indeed this has been asked other times (Number of Nodes in energy eigenstates) (Simple proof of oscillation/nodal theorem in quantum mechanics) (Physical intepretation of nodes in quantum mechanics) in addition to the linked answer. The answer also has a comment with a video explaining the "node theorem" with a nice intuitive strategy: for any potential, if you just kept the area near the minimum of the potential and sent the rest to infinity then the eigenfunctions would be similar to that of the infinite square well. If you slowly move the boundaries to encompass the rest of the true potential, you can't develop a node because then the wavefunction and its derivative would both have to simultaneously take the value zero somewhere at the same position, which is impossible.

Now, it looks like Ballentine is actually talking about 3D instead of 1D. Then one is no longer guaranteed that the node theorem will hold, as far as I'm aware, but it's probably still a good intuition. Some exceptions in multiple dimensions, from Googling, seem to be this and this.

• This is a beautiful answer, thank you so much! "Maybe you meant to ask, through these two questions: if we set $W_1$ in such a way, we can observe that $\Psi_1=R$ is an eigenstate with energy $E$; how do we know that there is no other eigenstate with lower energy?" Yes, this is precisely what I am asking with my question #1. To confirm, your answer to #2 is "no restrictions". And for #3, I am indeed asking about the 3D case, but you say I should just use this node-rule as "intuition"/a suggestion in that case?
– EE18
Commented Aug 14, 2023 at 19:13
• @EE18 yes, I can confirm. I am confident in the node rule for 1D and I am fairly certain it is not always true in 3D. However, the intuition behind the rule in 1D still holds for 3D (more nodes means the function is changing more, so it has a higher second derivative and thus more kinetic energy), so I get the feeling that Ballentine is trying to rely on that. Or else being sloppy and using a rule from 1D, but who knows Commented Aug 14, 2023 at 19:15
• I see. Thank you!
– EE18
Commented Aug 14, 2023 at 19:36