# What is a positive definite Hamiltonian?

1. Aug 20, 2008

### arroy_0205

Can anybody explain what is meant by positive definite Hamiltonian? All I know is that if a Hamiltonian can be factorized as
$$H={Q}^{\dagger}Q$$
then that Hamiltonian is one such example. But I am not sure if that is the definition of a positive definite Hamiltonian. In the quantum mechanical context what is meant by a positive definite operator, that also is not clear to me.
After Google search I found there is something called positive definite operator in mathematics, defined as: the operator A is a positive definite operator if for all non-null vector x,
$$xAx>0$$
But clearly the definition/concept used in quantum mechanics is a bit different.

2. Aug 20, 2008

### CompuChip

It's not different at all.

Since you are talking about quantum mechanics let me use my favorite, bra-ket notation. I assume you are familiar with it (if not just shout and I'll rewrite my post in matrix notation).
The example of the Hamiltonian you gave is
$$H = Q^\dagger Q$$.
Now if x is any Hilbert space vector, then
$$\langle x | H | x \rangle = \langle x | Q^\dagger Q | x \rangle = \langle Q x | Q x \rangle > 0,$$
where in the last equality, I had the conjugated operator work to the left ($\langle x | A^\dagger = \langle A x |$) and then used that for Qx nonzero, the inner product is strictly positive.

I hope this makes it look more familiar and you see the connection.

3. Aug 20, 2008

### clem

Positive definite means that all the eigenvalues are real and positive.
A consequence of this is that any expectation value <x|H|x> is positive.
The Q*Q example you give is a special case that is positive definite, but not true for most positive definite H.

4. Aug 20, 2008

### arroy_0205

Thanks for the responses.
I have a related question. Suppose I know the Hamiltonian corresponding to a Schroedinger equation. Is there any theorem/result which can be used to tell if the eigenvalues will be positive definite or not? I assume that the exact solution of the equation is not possible.

5. Aug 20, 2008

### maverick280857

First of all, the Hamiltonian H is a Hermitian operator (which implies that all its eigenvalues are real). You can then look at http://en.wikipedia.org/wiki/Positive-definite_matrix and read the set of equivalent conditions under the 'characterizations' sub-topic. Perhaps the simplest 'theorem' that you can use is that when diagonalized, the diagonal entries of H are its own eigenvalues. But if you want to determine whether H is positive definite without going through this exercise, then you will have to use the other equivalent conditions given on that page.

6. Aug 20, 2008

### Count Iblis

"But clearly the definition/concept used in quantum mechanics is a bit different"

In QM we require that the spectrum is bounded from below.

7. Aug 21, 2008

### ismaili

In order that $$\langle{\alpha}H{\alpha}\rangle$$, to be positive definite, i.e. the expectation value of Hamiltonian to be positive definite, we have to write all the annihilation operators to the right, and all the creation operators to the left. In this way, the Hamiltonian can be factorized as what you wrote down, and the energy expectation value is positive definite.

In usual QFT, we often use the so-called normal ordering to ensure this. Actually, what we did is throw away the cosmological constant term into the field which is in charged of quantum gravity. In string theory, besides imposing the normal ordering condition, we have to add a undetermined constant to account for the ordering ambiguity. It turns out the in the critical dimension, D = 26, the anomaly (constant from normal ordering) is zero.

By the way, the Hamiltonian must be written as the products of creation fields and annihilation fields because this ensures the cluster decomposition principle. This can be found in Weinberg's chap 4.

8. Aug 21, 2008

### maverick280857

Nice idea.

As for the rest, I think the OP's question was about nonrelativistic quantum mechanics