0
$\begingroup$

If we have a free particle with a wave function consisting of a sine wave/exponential wave moving in the negative-x direction, its wave function will be: $$|\psi\rangle=Ae^{i(kx+\omega t)}$$ The energy eigenvalue for this wavefunction will be $-\hbar\omega$: $$\hat E|\psi\rangle=i\hbar\dfrac{\partial}{\partial t}(Ae^{i(kx+\omega t)})=-\hbar\omega Ae^{i(kx+\omega t)}=-\hbar\omega|\psi\rangle$$ What is the significance of this and how is it different from a wave going in the other direction?

$\endgroup$
  • $\begingroup$ What you've proven is that that isn't actually the wavefunction of a particle moving to the left, it's $e^{-i(kx+\omega t)}$. You can instead think of what you have as a particle moving to the right backwards in time, sort of. $\endgroup$ – knzhou Dec 11 '17 at 19:43
2
$\begingroup$

$i\hbar \frac{\partial}{\partial t}$ is not the energy operator. For a free particle of mass $m$, the energy operator (in position space) is $$ \hat E = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$$

Applying that to your wave function gives $$\hat E \psi = -\frac{\hbar^2}{2m}(-k^2)\psi = \frac{\hbar^2k^2}{2m}\psi$$ so the eigenvalues of the free particle energy operator are non-negative.


Separately, Schrodinger's equation tells us that $$i\hbar \frac{\partial \psi}{\partial t} = \hat E \psi$$

If we plug in your chosen wave function, then we find that $$-\hbar \omega \psi = \frac{\hbar^2 k^2}{2m} \psi$$

Because $\frac{\hbar^2 k^2}{2m}$ is nonnegative, it follows that $\omega$ must be negative. In other words, you have written down a negative frequency plane wave.

Conventionally we avoid this issue by working only with positive frequency solutions. If $\psi$ is an energy eigenstate, then $$\psi \propto e^{-i\omega t}$$ where $\omega$ is taken to be strictly positive. In such a convention, it follows that $E = \hbar \omega$. If you choose not to follow this convention, then you find that $E = \pm \hbar \omega$ depending on the sign in the exponent $e^{\mp i \omega t}$.

$\endgroup$
1
$\begingroup$

This is one of those questions whose answer depends a lot on the depth you want.

Facetious answer, that will look less stupid in a few paragraphs: you've affected $t\to -t$, so the particle is moving backwards in time.

We can shift energies by an arbitrary additive constant $E_0$, which is achieved in QM with the unitary transformation of multiplying wavefunctions by $\exp -iE_0 t/\hbar$. So there isn't much physical significance in the positive-negative energy distinction; we can only say how much energy a state has relative to another one.

If you take relativity into account, it's easy to run into cousins of the Schrödinger equation such as the Klein-Gordon equation, which unlike Schrödinger have a solution symmetry $\psi\to\psi^\ast$. This has the effect of changing the sign of energy, and understanding these negative-energy states was a big headache for a while. This is because many systems have no highest energy, and if you allow a sign change in eigenenergies there won't be a ground state either.

Eventually the Dirac equation gave us a slightly better understanding of "negative-energy" states by associating them with antimatter, whose energy measure requires a sign change to make sense of energy conservation in such events as annihilation or pair creation. This means the energies end up positive even for antimatter, but you wouldn't know that from the $\exp i\omega t$ form of its part of the Dirac spinor. In view of CPT invariance, one can say an antiparticle is like a particle moving backwards in time. So the shortest answer I could have given to your question is "it's antimatter".

But that would leave out one further detail.

The above illustrated why we ended up with quantum field theory. It's possible in an expanding universe for an initial $\exp -i\omega t$ dependence to become of the form $\alpha \exp -i\omega \eta + \beta \exp i\omega \eta$, $\eta$ the conformal time. (This is an example of a Bogoliubov transformation.) The $\beta$ term yields particle creation, because the wave's evolution hasn't "stayed in sync" with that of the vacuum definition. As a result, we get a particle density proportional to $|\beta|^2$.

$\endgroup$
0
$\begingroup$

its wave function will be: $|ψ\rangle=Ae^i(kx+ωt)$

For a stationary state (energy eigenstate), the time dependence is $|\psi(t)\rangle = |\psi\rangle e^{-i\frac{E}{\hbar}t}$ so that

$$H|\psi(t)\rangle = i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle = i\hbar\frac{\partial}{\partial t}|\psi\rangle e^{-i\frac{E}{\hbar}t} = E|\psi(t)\rangle$$

It follows that the wavefunction of a free particle with definite momentum in the negative $x$ direction is

$$\psi(x,t) = e^{-i(kx + \omega t)}$$

where $\omega \equiv \frac{\hbar k^2}{2m}$. Then

$$P_x\psi(x,t) = -i\hbar \frac{\partial}{\partial x}\psi(x,t) = -\hbar k\, \psi(x,t)$$

$$H\psi(x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x,t) = \frac{\hbar^2 k^2}{2m}\psi(x,t) = \hbar\omega\,\psi(x,t)$$

$\endgroup$

protected by Qmechanic Dec 11 '17 at 20:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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