# B Why does waveform exp[iwt] have negative kinetic energy?

1. May 10, 2017

### Happiness

Why so?

Quoted from Quantum Physics 3rd ed. by Stephen Gasiorowicz, p. 26.

It was explaining why we ignore the terms with exp[iwt] when adding plane waves to form a wave packet:

2. May 10, 2017

### Staff: Mentor

Please give the reference that you are quoting from.

3. May 12, 2017

### blue_leaf77

Try applying the energy operator $\hat H = i\hbar \frac{\partial }{\partial t}$ to the wavefunction with $\exp(i\omega t)$. In fact the time-exponential factor arises from the application of time evolution operator $\exp(-i\hat H t/\hbar)$, thus it makes perfect sense that one should use $\exp(-i\omega t)$ since the kinetic energy is positive.

4. May 13, 2017

### vanhees71

The Hamilton operator is NOT $\mathrm{i} \partial_t$ but a function of operators that represent observables. For a particle in a potential in non-relativistic physics, e.g., it's
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{x}}).$$
The Schrödinger equation in the position representation reads (I set $\hbar=1$ for convenience)
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$
The cited passage in #1 is as enigmatic to me as it is to Happiness. The sign in question is completely determined by the Schrödinger equation. Obviously the author is treating energy eigenstates, i.e., solutions of the Schrödinger equation of the form
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \omega t) u_{E}(\vec{x})$$
which obey
$$\hat{H} \psi_E(t,\vec{x})=E \psi_E(t,\vec{x}).$$
For a time-independent Hamiltonian for the above ansatz we have
$$\mathrm{i} \partial_t \psi_E(t,\vec{x})=\omega \exp(-\mathrm{i} \omega t) u_{E}(\vec{x})=\omega \psi_E(t,\vec{x}),$$
i.e., we have (uniquely
$$\omega=+E.$$
The time-independent eigenfunction of the Hamiltonian has to be determined by
$$\hat{H} u_E(\vec{x})=E u_E(\vec{x}),$$
which in general is a pretty complicated partial differential equation (usually called "the time-independent Schrödinger equation"). Only for simple potentials can we solve it analytically. The most simple case is the free particle, i.e., $$V=0$$. Then we have
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}=-\frac{\Delta}{2m}.$$
$$-\frac{\Delta}{2m} u_E(\vec{x})=E u_E(\vec{x}).$$
It is easy to solve by a separation ansatz
$$u_E(\vec{x})=u_1(x_1) u_2(x_2) u_3(x_3).$$
Plugging this ansatz into the equation leads to the solutions
$$u_1(x_1)=N_1 \exp(\mathrm{i} p_1 x_1), \quad \ldots$$
or
$$u_E(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x}).$$
Plugging this into the eigenvalue equation for the Hamiltonian yields
$$\hat{H} u_E(\vec{x})=\frac{\vec{p}^2}{2m} u_E(\vec{x}) \; \Rightarrow \; E=\frac{\vec{p}^2}{2m},$$
i.e., a free particle's energy eigenvalues are related to the momentum of the particle as in classical physics.

The reason that this works is, of course, that the Hamiltonian commutes with all momentum operators, i.e.,
$$[\hat{H},\hat{\vec{p}}]=0,$$
and thus the momentum eigenstates are also energy eigenstates. It's more convenient to use this particular set of energy eigenstates, because the three momentum components build a complete set of compatible observables, so that specifying the three momenta determine the eigenfunction uniquely. The Hamiltonian eigenvalues themselves are degenerated, since for a given eigenvalue $E$ all the momentum eigenstates with $\vec{p}^2/(2m)=E$ (i.e., an entire sphere in momentum space) are eigenvectors.

For the free particle the time-dependent energy eigensolution finally reads
$$\psi_E(t,\vec{x})=N \exp[-\mathrm{i} (E t-\vec{p} \cdot \vec{x})],$$
and the signs here are completely determined by the defining equations. There's no doubt as suggested by the passage cited in #1.

There is an issue in relativistic quantum mechanics concerning the sign of energy eigenvalues of free particles, finally leading to the conclusion that it is more adequate to formulate relativistic QT as a QFT right from the beginning, but that's another issue.