# How to determine whether a state is a ground state

• Haorong Wu
In summary: I think it's worth trying to solve the Schrodinger equation, though.In summary, the F-H theorem would not apply here. So do the Virial theorem.
Haorong Wu
Homework Statement
Given a potential field, ##V \left ( x \right ) =
\begin{cases}
0 & x \leq 0 \\
- \frac {\alpha \hbar ^2} {m x} & x \gt 0
\end{cases}
##, where ##\alpha \gt 0## is a parameter, and a state, ## \psi _0 \left ( x \right ) ##, determine whether ##\psi_0=
\begin {cases}
0 & x \leq 0 \\
Nxe^{- \alpha x} & x \gt 0 \end{cases}
## is the ground state or not.
Relevant Equations
None
I guess the hard way is to solve the Schrödinger equation, but that would be exhausting.

I think the F-H theorem would not apply here. So do the Virial theorem.

Are there other theorems I forget?

Is your Hamiltonian bounded from below?

Antarres said:
Is your Hamiltonian bounded from below?
Hi, Antarres. I am sorry I do not get your point. The Hamiltonian is infinite at ##x =0##.

Also, I have made a mistake, that the Hamiltonian should be##
V \left ( x \right ) =
\begin{cases}
\infty & x \leq 0 \\
- \frac {\alpha \hbar ^2} {m x} & x \gt 0
\end{cases}##.

I do not know why I can not edit my post now.

Well the point was, that in order to have a ground state in the first place, the Hamiltonian must be bounded from below. Your potential was going to negative infinity as ##x## goes to zero from the positive side, which means that such a Hamiltonian doesn't have a ground state. The corrected potential you have given isn't much different, as it goes to negative infinity as ##x## goes to zero from positive side, and then jumps up to positive infinity. So it is still unbounded.

So I would say from inspection that your potential shouldn't allow a ground state, you can check it by solving the Schrodinger equation. That's what I would say at first look.

Antarres said:
Well the point was, that in order to have a ground state in the first place, the Hamiltonian must be bounded from below. Your potential was going to negative infinity as ##x## goes to zero from the positive side, which means that such a Hamiltonian doesn't have a ground state. The corrected potential you have given isn't much different, as it goes to negative infinity as ##x## goes to zero from positive side, and then jumps up to positive infinity. So it is still unbounded.

So I would say from inspection that your potential shouldn't allow a ground state, you can check it by solving the Schrodinger equation. That's what I would say at first look.
Thanks, Antarres. But how about a delta potential, which is zero everywhere except at ##x=0## where the potential is negtive infinity. This potential allows only one bound state with ##E \lt 0##. Then the potential is not bounded but with a ground state.

That is an interesting remark, I forgot about the delta potential. Just so I don't sound like I'm pulling things out of back of my brain, I'll refer you to a discussion on bound states in Cohen Tannoudji vol 1. complement ##M_{III}##. I'll sum up the discussion here, just listed it in case you want more details that what could be put in one post.

Basically, we look at time-independent Schrodinger equation in one dimension for some solution ##\phi(x)##, and arbitrary potential ##V(x)##. In the first part of the discussion it is proven that bound states always happen at negative energies and the energy spectrum is discrete. I will focus on the discussion on ground state. Acting with the conjugate of the solution and integrating, we get the following form:
$$\int_{-\infty}^{\infty} dx\left( \phi^{*}(x)\frac{d^2}{dx^2}\phi(x) + V(x)\vert \phi(x) \vert^2 \right)= E$$
Using partial integration and property of state functions that they are defined to be square-integrable, you find that the first term on the left, which we can call mean kinetic energy is nonnegative, while the second term is mean potential energy. So we have:
$$E = \langle T \rangle + \langle V \rangle$$
Now we're looking for the lower boundary of energy in order to find the ground state. If potential has a minimum, then we can make the expectation value of potential strictly bigger than that minimal value, using normalization of the states. However, if it has no minima, then we can't find a minimal energy here.

So we conclude, bound states exist in attractive potentials, and they have discrete energy levels, but they don't have to have a ground state, in a sense, such a system would be unstable since under external perturbation it could fall deeper and deeper into the potential well.

Now about the Dirac delta, that function isn't treated in the reference I gave you, since they only treat bound potentials extensively. The reason why Dirac delta allows only one bound state, is because of normalization condition on the state. That is, the zero width of this distribution amounts to such a conclusion, because normalization directly correlates the energy of the bound state to the coefficient in front of delta potential. Potentials of non-zero width wouldn't amount to such an effect, since the states could be properly normalized at different energies, not just one. If you look at Dirac delta as a limit of some sequence of distributions, for example Gaussian or Lorentzian, or even box distributions, you can track how this effect comes about as you push the width of the distribution to zero.

So I would still say I'm certain that in your case you don't have a bound state of minimum energy, aka ground state in this kind of system. When I get time(unless you figure it out yourself), I'll look into that Schrodinger equation to try to create something like annihilation and creation operators for it. Action of annihilation operator on that state would also show if it is a ground state, though finding such operators in this case might be a troublesome thing to do.

Last edited:
Haorong Wu said:
Are there other theorems I forget?
There are theorems regarding the number of nodes in the bound-state energy eigenfunctions. For example, see this paper or this video .

Last edited:
Looking around, I found a treatment of this type of potential in an old paper:
Loudon, R. (1959). One-Dimensional Hydrogen Atom. American Journal of Physics, 27(9), 649–655. doi:10.1119/1.1934950
Your potential has very similar form to 1d hydrogen atom potential, yours is just defined for positive ##x##. So you can check how this problem is solved there, with help of some special functions. It's not a long paper, so it might be useful for this.

Antarres said:
Looking around, I found a treatment of this type of potential in an old paper:
Loudon, R. (1959). One-Dimensional Hydrogen Atom. American Journal of Physics, 27(9), 649–655. doi:10.1119/1.1934950
Your potential has very similar form to 1d hydrogen atom potential, yours is just defined for positive ##x##. So you can check how this problem is solved there, with help of some special functions. It's not a long paper, so it might be useful for this.
Hi, Antarres. I figured it out.

It turns out the Hamiltanian is just like the one of a hydrogen atom. Then the remaining job is just replace the ##e^2## with ##\frac {\alpha \hbar ^2} {m }## in the ground state energy of the hydrogen atom.

PeroK
This is just to complete the "node theorem" approach to the problem. It might not be an acceptable approach for your course if you have not covered this theorem.

It is easy to check that the given function ##\psi_0(x)## is an energy eigenstate. You can also see that ##\psi_0(x)## has no nodes (other than the left end point x = 0 of the well). The node theorem implies that the only energy eigenstate that can have zero nodes is the the ground state. So, we can conclude that ##\psi_0(x)## is the ground state.

PeroK
Haorong Wu said:
Hi, Antarres. I figured it out.

It turns out the Hamiltanian is just like the one of a hydrogen atom. Then the remaining job is just replace the ##e^2## with ##\frac {\alpha \hbar ^2} {m }## in the ground state energy of the hydrogen atom.
Just to make it clear, it is like 1D hydrogen atom. Normal hydrogen atom is a 3D system, and the consequence of this is that equation along radial coordinate has a bounded potential, since there is an extra term appearing from the angular part that makes the effective potential bounded. So here the potential is definitely singular and unbounded, while normal hydrogen potential isn't. When I say bounded here, I mean obviously bounded from below.

As far as I've seen in that treatment I linked you, they've approximated the potential with a potential that isn't singular and calculated the wave function in terms of confluent hypergeometric functions. Properties of those functions are not really important for understanding the solution, as the series expansion is given. When they perform a limit towards the correct singular potential, the ground state drops down into a delta function.

I've seen another treatment where analysis wasn't as deep, but the solution was searched for in a similar way as the one for radial equation in hydrogen. That equation yields associated Laguerre polynomials as solution, and so ground state appears exactly in the form that has been presented to you. To me, this kind of treatment is a bit approximate, but oh well, the paper I linked you seems more thorough. So it depends on the treatment that is expected of you, maybe it is put just as an exercise reminding of standard hydrogen. Thing with 1d hydrogen system is that it appears that it's been worked on in various ways so far, and there isn't an agreement on the solution to this problem. The problem is much less trivial than the 3D case(curiously). That said, I'm glad I've been able to help, if at all.

## 1. How do I determine if a state is a ground state?

The ground state of a system is the state with the lowest energy. To determine if a state is a ground state, you need to calculate the energy of the state and compare it to the energies of other possible states. If the energy of the state is the lowest, then it is the ground state.

## 2. What factors affect the determination of a ground state?

The determination of a ground state depends on several factors, including the physical properties of the system, the interactions between particles, and the external conditions such as temperature and pressure. These factors can influence the energy levels and thus affect which state is the ground state.

## 3. Can the ground state of a system change?

Yes, the ground state of a system can change. This can happen when there is a change in the external conditions or when there is an interaction with another system. When the ground state changes, the energy of the system also changes.

## 4. How is the ground state determined in quantum mechanics?

In quantum mechanics, the ground state is determined by solving the Schrödinger equation for the system. The solution to this equation gives the possible energy levels of the system, and the ground state is the state with the lowest energy. This approach is used to determine the ground state of atoms, molecules, and other quantum systems.

## 5. How can knowing the ground state of a system be useful?

Knowing the ground state of a system can provide important information about the system's stability, reactivity, and other properties. It can also help in predicting the behavior of the system under different conditions and in designing experiments or technologies that utilize the system's properties.

Replies
3
Views
1K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
1
Views
2K
Replies
4
Views
1K
Replies
4
Views
4K
Replies
1
Views
2K
Replies
1
Views
873
Replies
5
Views
2K
Replies
2
Views
674