12
$\begingroup$

To find out the stationary states of Hamiltonian, we will be finding the eigenvalues and eigenstates. Is there any condition that form of the Hamiltonian should be like, $$\hat{H}=\hat{T}(\hat{p})+\hat{V}(\hat{x}).$$ I mean the sum of Kinetic operator and potential operator term. Can I have Hamiltonian without a kinetic term, $$\hat{H}=\hat{x}^{2}$$ or in other words? Can I have a Hamiltonian with just potential term alone? $$\hat{H}=\hat{V}(\hat{x}).$$

Does quantum mechanics allow that?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ Sure! Have you heard Heisenberg Hamiltonian? Or the Hamiltonian of a spin system in a magnetic field? $\endgroup$
    – rnels12
    Commented Aug 1, 2019 at 9:11

3 Answers 3

Reset to default
12
$\begingroup$

In principle, the Hamiltonian represents the energy of a system. Whether or not you want to model your system to have kinetic energy is up to you and what you need. For example: consider an atom with an electron that can be approximated as a two level system (i.e. it as only its ground state and an excited state).

The ground state $|g\rangle$ has some energy $E_0$, and the excited state $|e\rangle$ has some energy $E_0+\Delta E$, you are always free to chose the ground state energy as you like, so chose $E_0=-\frac{\Delta E}{2}$ and you have the Hamiltonian of your system

$$H=\frac{\Delta E}{2}\left(|e\rangle\langle e|-|g\rangle\langle g|\right)=\frac{\Delta E}{2}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=-\frac{\Delta E}{2}\sigma_z $$

with $\sigma_z$ the Pauli matrix. This is a perfectly valid (albeit a bit boring) Hamiltonian that has no kinetic term, basically, you decided that you don't really care about the kinetic energy of the atom, you're interested in the state of the electron. More interesting Hamiltonians that don't model kinetic degrees of freedom are given in Paradoxy's answer.

And as a note, a Hamiltonian is just a hermitian operator bounded below. There is, in principle, no further requirement. You can take any hermitian operator bounded below and start solving the Schrödinger equation. Of course, whether or not this has any physical meaning is another story.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I completely agree, Can I have hamiltonian which is independent of $p$ and just depending on $x$, like $H=x^{2}$. What is the physical meaning of this? $\endgroup$
    – user135580
    Commented Aug 1, 2019 at 11:28
  • 3
    $\begingroup$ As I said, the only requirement is that it's Hermitian and bounded from below, so $H=A\hat{x}$, where $A$ is some constant to adjust the units, doesn't work because it has eigenvalues $Ax, x\in \mathbb{R}$, so not bounded (you must have a ground state energy!), in principle something like $H=A\hat{x}^2$ is allowed, but I don't know if it has a physical meaning, and I don't know if there are Hamiltonians with only the position operator that have physical meaning, sorry! $\endgroup$
    – user2723984
    Commented Aug 1, 2019 at 12:42
  • $\begingroup$ There is a reason for asking this question. If I am solving $H_{2}$ molecule (without electron-electron repulsion term), First I will solve the electronic part of the Hamiltonian, The electronic energy eigenvalue is $E=E(R)$ will be a function of the internuclear distance $R$, One has to solve the Schrodinger equation for the nuclear coordinates, $E(R)$, however, they will ignore the kinetic term of the nucleus as it is heavier than electron, and simply find out the minimum energy of $E(R)$ classically and mention that energy. Please clarify $\endgroup$
    – user135580
    Commented Aug 1, 2019 at 19:22
  • $\begingroup$ By the way why $H=Ax^{2}$ if I plot the energy spectrum $E_{x}=Ax^{2}$ the lowest energy value is just 0, the energy function is just continuous and degenerate. Am I correct? If I am wrong please explain to me why this is unbounded $\endgroup$
    – user135580
    Commented Aug 1, 2019 at 19:38
  • 1
    $\begingroup$ @user135580 $H=Ax^2$ is indeed bounded, as I said in my comment. I said that $Ax$ is not bounded. As for the specific question about the $H_2$ molecule, I think you should post a new question with that specific problem, as it is not clear to me what you're asking from your comment $\endgroup$
    – user2723984
    Commented Aug 1, 2019 at 22:07
9
$\begingroup$

The easiest one which comes to mind is $$H=j\sum_{<ij>} \sigma_{i}\sigma_j$$ The Hamiltonian of ising model. (They use it in Schrödinger equation too sometimes with numerical calculations) https://en.m.wikipedia.org/wiki/Ising_model

Or as it's said in comments by @rnels12 Hamiltonian of Heisenberg model $$H=j\sum_{<ij>} \sigma_{i}.\sigma_j-h\sum_i \sigma_i$$

https://en.m.wikipedia.org/wiki/Heisenberg_model_(quantum)

Edit: I am not sure why I got downvoted, but I was trying to say QM does allow this. In fact, whenever you don't care about motion of your particles or their motion are relatively small you can just drop kinetic energy term to make Hamiltonian even easier. And if you are worried about uncertainty principle or something, there won't be any problem with it if you calculate wave function and then variances. Because after all neglecting kinetic term is not equal to assuming zero momentum.

Edit 2: Thanks to @d_b I understood that the first term in hofstadter butterfly Hamiltonian is lattice version of a kinetic term, although it's not clearly $\frac{\hat{p}^2}{2m}$, in continuum limit it becomes one.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ You mean to say that, if the mass of the particle is heavy (this is the case in quantum chemistry, where, nuclear masses are heavier than electrons), we can neglect the kinetic term and just consider the potential term Am I correct here? $\endgroup$
    – user135580
    Commented Aug 1, 2019 at 19:07
  • 1
    $\begingroup$ @user135580 exactly. Actually I have done some numerical calculations for ising model myself, where we assumed a fullerene like structure (en.m.wikipedia.org/wiki/Fullerene) and safely neglected the motion of atoms, because carbon is heavy enough. Note that if you want to ignore kinetic energy term, you need one thing, and that's $T<<V$ in Hamiltonian. $\endgroup$
    – Paradoxy
    Commented Aug 1, 2019 at 19:43
  • 1
    $\begingroup$ The Hofstadter hamiltonian definitely has a kinetic energy term. In fact it's only got a kinetic energy term. $\endgroup$
    – d_b
    Commented Aug 2, 2019 at 3:05
  • 1
    $\begingroup$ @d_b please check this out:(journals.aps.org/prb/abstract/10.1103/PhysRevB.60.10054) the Hamiltonian of Hofstadter is $H=\sum_{i,j} c_jc_i^\dagger e^{i2\pi\phi_{ij}}+\sum_i v_i c_ic_i^\dagger$. Which is defined on square lattice.$i$ and $j$ denotes different sites $c_i$ and $c_i^\dagger$ are creation and annihilation operators $v_i$ is on the site potential and $\phi$ is magnetic fIeld. I can't see any term related to kinetic energy of particle which indicats its motion, that's $p^2/2m$, or am i missing something here? $\endgroup$
    – Paradoxy
    Commented Aug 2, 2019 at 9:48
  • 2
    $\begingroup$ Thanks, I'm familiar with the Hofstadter model. I would call that hamiltonian the Hofstadter hamiltonian with a mass term. The first term is a hopping term, which describes particles moving from site to site. It's the lattice version of a kinetic term. If you take the continuum limit you will see that that term becomes $(p - eA)/2m_{\ast}$ with effective mass $m_{\ast}$. $\endgroup$
    – d_b
    Commented Aug 2, 2019 at 16:49
3
$\begingroup$

In studying the quantum Hall effect, it is common to assume that all particles have the same kinetic energy (sometimes people describe this by saying the kinetic energy is "quenched"). The assumption is reasonable when all of the particles occupy a single Landau level and the energy spacing between Landau levels (the cyclotron energy) is large relative to other energy scales in the problem. In this case, the kinetic energy just gives a constant shift in the energy spectrum, so we can ignore it and consider a hamiltonian with only an interaction term (projected to a Landau level), $H = PVP$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.