9 votes
Accepted

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

If you interpret the Klein-Gordon equation as describing a field theory (which is the modern way to look at it), then yes, the energy of the field is positive definite. The problem is if you try to ...
Andrew's user avatar
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8 votes

Why does $e^{-H}\partial_j e^{H} = \partial_j + \partial_jH$?

OP's identity$^1$ $$\begin{align} e^{-H}\partial e^H ~\equiv~&e^{-[H,\cdot]}\partial\cr ~\equiv~& \partial+[\partial, H] +\frac{1}{2}[[\partial, H],H] +\frac{1}{6}[[[\partial, H],H],H] +\ldots\...
Qmechanic's user avatar
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8 votes
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Why does $e^{-H}\partial_j e^{H} = \partial_j + \partial_jH$?

It's just Leibnitz rule. For any $\psi(x)$ we have $$ e^{-H(x)}\partial_x e^{H(x)} \psi(x) = e^{-H(x)} (\partial_x e^{H(x)})\psi(x) + e^{-H(x)} e^{H(x)}\partial_x \psi\\ = e^{-H(x)}e^{H(x)} \...
mike stone's user avatar
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5 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

You have a valid point. 'Negative energy solution' is a misnomer. Negative as well as positive frequency solutions have positive energy, as the Klein-Gordon hamiltonian correctly implies. As an ...
my2cts's user avatar
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5 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

Briefly speaking, there are at least two notions of energy $E$ and momentum ${\bf p}$: One notion stems from the Fourier-transform $({\bf x},t)\to ({\bf p},E)$ of spacetime, or equivalently, a plane-...
Qmechanic's user avatar
  • 200k
5 votes

Possible inconsistencies of the Hamiltonian in the two-body problem

This expression does not admit of an uncertainty in the position of the particle which generates the field, it implies that the particle generating the field is infinitely precisely localized at a ...
J. Murray's user avatar
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3 votes

Possible inconsistencies of the Hamiltonian in the two-body problem

When we solve the single coordinate Schrödinger equation... we imply the potential $V(x)$ is exactly defined everywhere on $x$ with no uncertainties... The potential is taken to be a well-defined ...
hft's user avatar
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3 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

I sometimes think that the whole negative energy thing is a bit overrated. For example one can solve the classical Klein-Gordon equation and find via the appropriate ansatz that the solutions will be ...
Albertus Magnus's user avatar
2 votes

Connection of a concrete Hamiltonian to the generator of time-translations

Of course not - it is one property of the Hamiltonian, but it does not determine it. It leaves many options possible, like Hamiltonian of free particle, Hamiltonian of harmonic oscillator, Hamiltonian ...
Ján Lalinský's user avatar
2 votes

How come time does not commute with $i\hbar\dfrac{\partial}{\partial t}$ but it does so with $H$?

The origin of this confusion stems from the fact that in (standard) quantum mechanics, time is actually not an operator. Rather, time is just a real variable. The way time enters in QM is, for example ...
lcv's user avatar
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1 vote
Accepted

Is there a name for a Heisenberg-like model, but instead of the ZZ operator, we have one that favor only spin-up-spin-up configurations?

As pointed out by @LPZ, the term you mention is \begin{equation} J_{\mathrm{new}}(4\sigma^z_i\sigma^z_{j}+2\sigma^z_i+2\sigma^z_j+1)/4, \end{equation} and therefore, if you plug this in to where you ...
Jun_Gitef17's user avatar
1 vote

How the supercharge operators act on superfields in quantum mechanics, and the adjoints of supercharges?

After applying an infinitesimal unbarred SUSY transformation generated by $\hat Q$, $$\delta_\epsilon X=[\epsilon\hat{Q},X]=\epsilon(\partial_\theta+i\bar\theta\partial_t)X(t,\theta,\bar\theta)$$ ...
Nihar Karve's user avatar
  • 8,405
1 vote
Accepted

Second quantization of hamiltonian of the Klein-Gordon field

Edited because previous answer was incorrect The integral in the Hamiltonian is over all the momentums $\vec{p}'$, not only the $\vec{p}$ associated to the one of the state. All the momentum should ...
Léo Vacher's user avatar
1 vote

$\mathbf k\cdot\mathbf p$ Hamiltonian

I am not sure what $P_{cv}(k)$ and $C(s)$ are without some more context, but we can work through the $k\cdot p$ model from the beginning and continue the dialogue if you're still unsure. $k\cdot p$ ...
intraband's user avatar

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