# Quantum Hamiltonian

1. Jan 15, 2015

### DiracPool

I'm a bit puzzled over the structure of the Hamiltonian in the Schrodinger equation (SE). First we take the famous expression, Eψ=Hψ.

From what I'm aware of, the classical Hamiltonian is H=Kinetic energy (KE) + Potential energy (PE). However, in the SE, there appears to be a negative sign before the KE term in the rhs representing the Hamiltonian, i.e., E = -KE + PE, which would seem to imply that H=PE-KE. Can someone help reconcile this apparent discrepancy for me?

2. Jan 15, 2015

### Staff: Mentor

There is no minus sign. The kinetic energy for a particle of mass $m$ is $\hat{P}^2 / 2m$, which is what enters in the Hamiltonian. The minus sign only appears in the position representation because then
$$\hat{P} = -i \hbar \frac{\partial}{\partial x},$$
therefore
$$\hat{P}^2 = -\hbar^2 \frac{\partial^2}{\partial x^2}.$$

3. Jan 15, 2015

### Clear Mind

Mmmmmh ...
Suppose to have a classic hamiltonian (in 1 dimension): $H=\frac{p^2}{2m}+V(q)$ (where $KE=\frac{p^2}{2m}$ and $PE=V(q)$) in classical physics q and p are coordinate of your system, wich means that they are numbers. In QM q and p are represented by operators and their "form" depends on which representation you choose, for example in Schrodinger representation you have that:
$p \rightarrow i \hbar \frac{\partial}{\partial x}$
$q \rightarrow x$
So your QM Hamiltonian become:
$H=- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+V(x)$
The minus on the KE comes from the representation of p, but it doesn't mean that the KE is negative

4. Jan 15, 2015

### Staff: Mentor

That's not called the Schrödinger representation. And you are missing a minus sign there.

5. Jan 15, 2015

### Clear Mind

My apologies, i've missed a sign in the p

$p \rightarrow -i \hbar \frac{\partial}{\partial x}$

By the way, isn't that the Schrödinger representation? The one where de q are diagonal operators?

6. Jan 15, 2015

### stevendaryl

Staff Emeritus
The Schrodinger picture (if that's what you mean) is one in which the operators--position, momentum, energy, etc.--are constants, while the wave function evolves in time. In contrast, the Heisenberg picture is one in which the wave function is a constant, and the operators evolve in time.

So Schrodinger versus Heisenberg has nothing to do with which operators are diagonal.

7. Jan 15, 2015

### Clear Mind

Ok, now i'm starting to understand the confusion ...
Don't ask me why in italian the representation where the $q$ act like a diagonal operator is called Schrodinger rapresentation (or more rarely used, coordinate picture)
So what is the name in english for the coordinate picture?

8. Jan 15, 2015

### Staff: Mentor

I'm used to "position representation," but they may be other common ways of saying it.

9. Jan 15, 2015

### Clear Mind

Thank you, i hope to not make any other nomenclature mistakes :s

10. Jan 16, 2015

### vanhees71

That's an important point. So let's clarify it once and forever:

Picture: In the abstract formulation of quantum theory in terms of selfadjoint operators, representing observables on a Hilbert space and the Hilbert-space vectors, representing states you can shuffle the time dependence between these operators and states pretty arbitrary. The specific way you do this "split of the time dependence" is called the choice of the picture of time evolution. Which one to use is just a matter of convenience. The physical outcome of your calculation, i.e., probabilities, expectation values of operators for a given state are independent of the choice of the picture.

The Schrödinger picture is the one, where the entire time dependence is put on the state vectors and thus the operators representing observables are time independent. In the Heisenberg picture it's the opposite. The general idea of the picture was discovered by Dirac, and usually applied as the "interaction picture" in time-dependent perturbation theory.

Representation: A representation is the choice of which basis you use to formulate quantum theory in terms of matrix or wave mechanics (or a mixed form, depending on the basis you choose). A basis is a (generalized) set of orthonormal common eigenvectors of a complete set of compatible operators. For a spinless non-relativistic particle common choices for wave mechanics is the position representation, where the complete set of compatible operators are the three components of the position vector operators or the momentum representation (three components of the momentum vector). A mixed representation in terms of a wave-matrix mechanics is given, e.g., by choosing $|\vec{x}|=r$, $\vec{L}^2$, and $L_z$ as a complete set of compatible observables. Then you have a mixed representation in terms of wave functions $\psi_{lm}(r)$ with $r \in \mathbb{R}_{>0}$, $l \in \{0,1,\ldots \}$, $m \in \{-l,-l+1,\ldots,l-1,l \}$. Finally an example for matrix mechanics is the choice of phonon eigenstates of a 3D harmonic oscillator, because there you have a basis given in terms of three occupation numbers for the modes of the oscillator.

The time evolution of the corresponding concrete realization of Hilbert space in terms of wave functions, infinitely-dimensional matrices, or a mixture there of is independent of the choice of the picture. Because the time dependence of the eigenvectors of observables, appearing in the left argument of the generalized scalar product that defines the "vector components" ("wave functions") of your representation, and the time dependence of the state ket, appearing in the right argument of this scalar product always combine to the unique Schrödinger equation, written in this representation, e.g., in the position representation
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x}) = \hat{H} \psi(t,\vec{x})=\left [-\frac{\hbar^2}{2m} \vec{\nabla}^2 + V(\vec{x}) \right ] \psi(t,\vec{x}),$$
where
$$\psi(t,\vec{x})=\langle \vec{x},t |\psi,t \rangle.$$
Of course, you must use the same picture of time evolution for both the eigenvectors of the operators representing the complete set of compatible operators and the state ket in this scalar product.

Here $\hat{H}$ is the Hamiltonian, which by definition is the generator of the physical time evolution. In the abstract formalism, it is very illuminating to introduce a covariant time derivative, which is however rarely done in textbooks (I only know a German one, where this is done: E. Fick, Einführung in die Grundlagen der Quantenmechanik). It consists of the postulate that if $\hat{O}$ is representing an arbitrary observable, then
$$\mathrm{D}_t \hat{O} = \frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}]$$
represents the time derivative of this observable. This leads to a picture-independent formulation of quantum theory in the abstract Hilbert-space formalism.

11. Sep 18, 2015

### Dirac62

I think after all, we would see a MINUS sign in the relation. Am I right? I mean we subtrahend KE and PE. Is it true?

12. Sep 18, 2015

### Staff: Mentor

What relation are you talking about?

13. Sep 18, 2015

### Dirac62

_h-2k2/2m+V(x)

14. Sep 18, 2015

### Staff: Mentor

I have no idea what that means. If you don't know LaTeX, at least use superscripts to make it more readable.

15. Sep 19, 2015

### lightarrow

Storically, the representation which uses eigenfunctions of position is called "Schrödinger representation", while the one which uses eigenvectors of di Hamiltonian is called "Heisenberg representation".
Of course this have not to be confused with the Schrödinger and Heisenberg "picture"! See Vanhees71 post.

--
lightarrow

16. Sep 19, 2015

### blue_leaf77

Seems like
$$\frac{\hbar^2k^2}{2m}+V(x)$$
The presence of minus sign in the kinetic energy term in Schroedinger equation does not harm the established physics because Hamiltonian is an operator, it's not a quantity. Moreover it can be shown that if the wavefunction vanishes at infinities the expectation value of kinetic energy is always positive.
$$\langle \hat{E}_k \rangle = -\frac{\hbar^2}{2m}\int_{-\infty}^{\infty} \psi^* \frac{d^2\psi}{dx^2} dx = -\frac{\hbar^2}{2m} \left( \psi^* \frac{d\psi}{dx} \bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \bigg| \frac{d \psi}{dx} \bigg|^2 dx \right)$$
$$= \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \bigg| \frac{d \psi}{dx} \bigg|^2 dx \geq 0$$
The same conclusion can also be found in momentum representation of the expectation value integral.

Last edited: Sep 19, 2015