# Variational principle in quantum mechanics

• tom.stoer
In summary: For example, the CI method does a very good job of converging the energy of a system, but often fails to converge the correct wavefunction. This can be due to a number of reasons, but it is a problem that occurs pretty frequently.In summary, the variational principle guarantees that, given a good reference Hamiltonian, the wavefunction and energy will converge to the exact values. However, there is no guarantee that your chosen method will converge to the correct wavefunction and energy for the physical system being studied.
tom.stoer
I have a question regarding the variational principle in quantum mechanics.

Usually we have a Hamiltonian H and we construct a state |ψ> using some trial states. Then we minimize E = <ψ|H|ψ> and get an upper bound for the ground state energy. In many cases the state |ψ> is then used to calculate other quantum mechanical observables O. E.g. in the non-relativistic quark model one is interested in the magnetic moment of the nucleon and things like that.

The interesting fact is that in order to do that one must assume that we have not only weak convergence of <ψ|H|ψ> to the true ground state energy E° but instead strong convergence of |ψ> to the true ground state |ψ°> .

And this is my question: under which conditions does this strong convergence follow from the variational principle?

tom.stoer said:
[...] one must assume that we have not only weak convergence of <ψ|H|ψ> to the true ground state energy E° but instead strong convergence of |ψ> to the true ground state |ψ°> .

And this is my question: under which conditions does this strong convergence follow from the variational principle?

Since no one else has answered yet, I'll ask a clarification question...

What meanings do you understand by "weak" and "strong" convergence?
If I understand your question, convergence of <ψ|H|ψ> to a real value E° is convergence
using the standard topology in C (or R). But |ψ> could converge to |ψ°> either in
Hilbert space norm topology, or Hilbert space weak topology.

I'm guessing that you meant "when does a sequence |ψ_n> defined via
$$\langle \phi | H | ψ_n\rangle ~\rightarrow~ \langle \phi | H | ψ°\rangle ~,~~~~ [\mbox{for all}~ \phi],$$
converge to |ψ°> in Hilbert space weak topology, if the sequence (E_n) defined by
$$E_n ~:=~ \langle ψ_n | H | ψ_n\rangle$$
converges to E° in the standard topology of C (or R)? "

OTOH, maybe you want
$$| ψ_n\rangle ~\rightarrow |ψ°\rangle$$
in norm topology?

(Apologies that this doesn't answer your question -- I'm just trying to get it clear...)

Weak convergence means that a sequence of points (in a Hilbert space) (xn;x) converges to some (xn;x°); then one says that xn converges weakly to x.

Strong and norm convergence are the same and means that ||xn-x|| converges to 0; then one says that xn converges strongly to x.

So I am asking for a condition to derive strong or norm convergence from weak convergence.

tom.stoer said:
So I am asking for a condition to derive strong or norm convergence from weak convergence.

The standard results are [cf. Kreyzsig or Lax] :

(a) strong convergence implies weak convergence with the same limit.

(b) The converse of (a) is not generally true.

(c) In the finite-dimensional case, weak convergence implies strong convergence.

But there are apparently examples of inf-dim spaces where weak convergence
implies strong convergences. The oft-quoted example is l^1.

However, that's not very relevant to a Hilbert space where square-integrability
is more important. E.g., if {x_n} is an orthonormal sequence in a Hilbert space H,
then {x_n} tends to 0 weakly but not strongly.

But maybe you knew this much already? It's a bit difficult to say more without
extra details of your specific context. Is your space of trial states finite-dimensional,
(as is typically the case in elementary examples of variational perturbation theory) ?

Last edited:
Infinite-dimensional, typically L², i.e. square integrable bound states

tom.stoer said:
Weak convergence means that a sequence of points (in a Hilbert space) (xn;x) converges to some (xn;x); then one says that xn converges weakly to x.

Strong and norm convergence are the same and means that ||xn-x|| converges to 0; then one says that xn converges strongly to x.

So I am asking for a condition to derive strong or norm convergence from weak convergence.

Sorry, but looking again at the variational principle the question seems to be incorrect.

I am looking for an additional condition to derive (xn; Txn) → (x; Tx) and (xn; y) → (x; y) provided that (xn; Hxn) → E° holds. In physical terms H is the Hamiltonian, T is an observable and x,xn are normed state vectors, typically bound states in L².

tom.stoer said:
I have a question regarding the variational principle in quantum mechanics.

Usually we have a Hamiltonian H and we construct a state |ψ> using some trial states. Then we minimize E = <ψ|H|ψ> and get an upper bound for the ground state energy. In many cases the state |ψ> is then used to calculate other quantum mechanical observables O. E.g. in the non-relativistic quark model one is interested in the magnetic moment of the nucleon and things like that.

The interesting fact is that in order to do that one must assume that we have not only weak convergence of <ψ|H|ψ> to the true ground state energy E° but instead strong convergence of |ψ> to the true ground state |ψ°> .

And this is my question: under which conditions does this strong convergence follow from the variational principle?

I think it always follows, at least for systems with a discrete spectrum of eigenstates. That is, for a method like the Rayleigh-Ritz method, if you increase the basis to infinite size, the resulting wavefunction and energy will converge on the "true" ground state wavefunction and energy. The speed of convergence will depend on the choice of the reference Hamiltonian used to define the basis functions. if it is a good approximation to the full Hamiltonian, then convergence will be faster, but it should always converge. As I understand it, this is because the eigenbasis of the reference Hamiltonian is a complete set.

However, there can be another problem ... namely, if your full Hamiltonian is not exact, then there is no guarantee that the ground state energy and wavefunction that you are converging on are representative of the true energy and wavefunction of the physical system being studied. We run into this problem in electronic structure calculations rather frequently. We almost always care about properties of the system other than just the energy, so we need the correct wavefunctions as well. It can be the case that, while a particular method does a good job of reproducing the correct ground state energy, when other physical observables (i.e. dipole moment, polarizability, vibrational frequencies) are calculated from the corresponding wavefunction, they show much poorer agreement with experiment than the energies.

Perhaps that last part is obvious to you already, but it seemed relevant to this thread. Particle physics isn't my field, so it may be the case that for quarks you can always work with the exact Hamiltonian ... we always have to use approximate Hamiltonians in electronic structure theory of molecules.

It's the same for elementary particle physics; the QCD Hamiltonian is much too complicated to be solved via simple QM methods - and the no-relativistic quark model is not really a realistic model - only in rare cases. It was only an example.

OK, so you say that for a discrete spectrum based on the correct H weak convergence implies strong convergence. Fine, thanks.

But if you read my last post carefully you see that I was wrong and we do NOT have weak convergence (xn; y) → (x; y) but only (xn; Hxn) → E°.

Is this still sufficient for strong convergence?

tom.stoer said:
OK, so [...] for a discrete spectrum based on the correct H weak
convergence implies strong convergence. [...]

For the record, that's only true if the discrete spectrum is finite.
For the infinite case, I already mentioned a counter-example in post #4.

[...] we do NOT have weak convergence (xn; y) → (x; y) but only (xn; Hxn) → E°.

Is this still sufficient for strong convergence?

Since this condition is "weaker" than the usual weak convergence,
and since weak convergence does not guarantee strong convergence in
the case of an infinite discrete spectrum, this seems unlikely in general.

Last edited:
yes, thanks, seems reasonable

## 1. What is the variational principle in quantum mechanics?

The variational principle in quantum mechanics is a fundamental principle that states that the actual energy of a quantum system is always greater than or equal to the lowest possible energy of that system. It is used to determine the ground state energy of a quantum system by finding the minimum value of an energy functional.

## 2. How is the variational principle used in quantum mechanics?

The variational principle is used in quantum mechanics to find the ground state energy of a system by minimizing the energy functional. This involves varying a trial wavefunction and calculating the expectation value of the energy operator. The trial wavefunction that results in the lowest energy is considered to be the closest approximation to the true ground state wavefunction.

## 3. What is a trial wavefunction in the context of the variational principle?

A trial wavefunction is an initial guess at the wavefunction of a quantum system. It is used in the variational principle to find the ground state energy by varying the parameters within the trial wavefunction. The closer the trial wavefunction is to the true ground state wavefunction, the lower the energy will be.

## 4. What are the benefits of using the variational principle in quantum mechanics?

The variational principle allows for a systematic approach to finding the ground state energy of a quantum system, even in cases where the exact solution is unknown. It also provides a way to improve upon approximate solutions by using more complex trial wavefunctions. Additionally, the variational principle can be extended to other physical quantities besides energy, such as magnetic moments and electric dipole moments.

## 5. Are there any limitations to the variational principle in quantum mechanics?

While the variational principle is a powerful tool, it does have some limitations. It can only provide an upper bound on the ground state energy, meaning that it may not give the exact solution. It also relies on having a good initial guess for the trial wavefunction, which may be difficult to obtain in some cases. Additionally, the variational principle may not work well for systems with degenerate energy levels.

Replies
11
Views
2K
Replies
3
Views
1K
Replies
14
Views
2K
Replies
3
Views
1K
Replies
5
Views
937
Replies
7
Views
2K
Replies
7
Views
1K
Replies
1
Views
988
Replies
0
Views
619
Replies
4
Views
959