1
$\begingroup$

So a positive and a positive wave function create a bonding orbital where the probability of finding an electron is summed while a positive and a negative create an anti-bonding orbital with a lower electron probability in the region between them leading to a repulsion. My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?

$\endgroup$

4 Answers 4

4
$\begingroup$

The wavefunction contains all the information about the system, but it is not itself observable. Since our intuition tends to be linked to what we have observed in the past, it's not surprising that the wavefunction seems unintuitive. As you say it can be negative, but it can also be a complex number making it even less intuitive (if that's possible!).

Anything we can observe is obtained from the wavefunction using a function called an operator. In your example of calculating the probability of finding the electron in some infinitesimal volume $dV$, this probability is given by:

$$ P = \psi^*\psi dV $$

where $\psi^*$ is the complex conjugate of $\psi$. The product $\psi^*\psi$ is always real and positive, so the probability of finding an electron is always real and positive. We don't need to worry that the wavefunction can be negative or complex.

$\endgroup$
3
$\begingroup$

A wavefunction with negative sign works just like any other wave with negative sign. For example, water waves with negative height cancel out with waves of positive height. You can also make a 'negative' wave on a string by pulling the end down and back up, which will cancel with a positive wave.

$\endgroup$
7
  • $\begingroup$ I have no problem with a wave function for water where it's positive and negative WRT the waterline... But a 'probability amplitude' is a much stranger duck - what is negative probability? $\endgroup$
    – user263399
    Aug 19, 2015 at 4:23
  • $\begingroup$ Think of the correspondences this way: the wavefunction corresponds to the wave height. But it's the wave's height squared that determines where its energy is, and likewise it's the wavefunction squared that determines where the particle is. So we never have a "negative" probability since we always square it. $\endgroup$
    – knzhou
    Aug 19, 2015 at 4:29
  • $\begingroup$ Or to say it another way: the magnitude of the wavefunction tells you where the particle is. But it can still interfere with other wavefunctions depending on its phase. For example, $+1$, $-1$, and $i$ all have the same magnitude but different phases. $\endgroup$
    – knzhou
    Aug 19, 2015 at 4:30
  • $\begingroup$ But we DO have a negative probability - were relying on it to distinguish between bonding and anti-bonding which is a very important distinction. Is it just accepted that this phase difference is something that magically works inside the mathematical black box? $\endgroup$
    – user263399
    Aug 19, 2015 at 4:47
  • 1
    $\begingroup$ There never is a negative probability. Anti bonding orbitals just have lower probability in the middle. Again, squaring always gives a positive result. $\endgroup$
    – knzhou
    Aug 19, 2015 at 4:56
0
$\begingroup$

As John Rennie has pointed out, the wavefunction $\psi$ itself isn't observable, and any expectation value is calculated squaring the wavefunction $$\langle A \rangle = \int \textrm{d}x \psi^*(x) A \psi(x)$$ Therefore, you are allowed to change the overall phase of the wavefunction without changing the expectation values: $$\psi(x)\to \psi'(x) = e^{i\theta} \psi(x) \\ \langle A \rangle \to \langle A'\rangle = \int \textrm{d}x \psi'^*(x) A \psi'(x) = \int \textrm{d}x e^{-i\theta}\psi^*(x) A e^{i\theta}\psi(x) = \int \textrm{d}x \psi^*(x) A \psi(x) = \langle A \rangle$$

The relevant case for your answer is $\theta=\pi$, so $\psi'(x) = -\psi(x)$. As you can see, changing the sign to the wavefunction has no physical relevance. So negative values of the wavefunction are by no means different to positive values.

John Duffield discusses weak measurements, and I think that it is (moderately) related to the question. First of all, should be noted that weak measurement is a clever technique to gather additional information about the wavefunction using minimally-invasive not-so-precise measurements. As such, lies within the rules of quantum mechanics, and therefore can never achieve the complete measurement of the wavefunction. In weak measurements two wavefunctions are used, $\psi_i$ is the pre-selected state and $\psi_f$ the post-selected state. $\psi_i$ is evolved along a (quantum) measurement device, and then it's projected into $\psi_f$. The weak measurement is defined as $$A_w = \frac{\int \textrm{d}x \psi_f^* (x) A \psi_i(x)}{\int \textrm{d}x \psi_f^* (x) \psi_i(x)}$$ Again, if you change the sign to $\psi_i$ and/or $\psi_f$, you will get the same weak measurement $A_w$ than before.

Strong measurements, weak measurements or whatever you use, the overall sign (in fact, the overall phase) of the wavefunction has no physical meaning attached.

$\endgroup$
-3
$\begingroup$

Physical intuition behind negative values for wave function?

Take a look at Jeff Lundeen's website. He and a team have done weak measurmeent work where they measure wavefunction directly. See his semi-technical explanation where you can read this:

"With weak measurements, it’s possible to learn something about the wavefunction without completely destroying it. As the measurement becomes very weak, you learn very little about the wavefunction, but leave it largely unchanged. This is the technique that we’ve used in our experiment. We have developed a methodology for measuring the wavefunction directly, by repeating many weak measurements on a group of systems that have been prepared with identical wavefunctions. By repeating the measurements, the knowledge of the wavefunction accumulates to the point where high precision can be restored. So what does this mean? We hope that the scientific community can now improve upon the Copenhagen Interpretation, and redefine the wavefunction so that it is no longer just a mathematical tool, but rather something that can be directly measured in the laboratory."

Also see In praise of weakness which you can access from Aephraim Steinberg's website. These guys are saying photon wavefunction is something real. And since we make electrons and positrons out of photons in pair production, we maybe have a clue to answering your question.

So a positive and a positive wave function create a bonding orbital... while a positive and a negative create an anti-bonding orbital... leading to a repulsion.

Note that electrons and positrons demonstrate attraction, electrons and electrons demonstrate repulsions, and positrons and positrons demonstrate repulsion. Also note that electrons and positrons have the opposite chirality. Now take a look at what Percy Hammond says here: "We conclude that the field describes the curvature that characterizes the electromagnetic interaction". Curvature.

My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?

I think it's to do with negative curvature.

$\endgroup$
1
  • $\begingroup$ Note that we typically detect a photon with an electron, and eac has its wavefunction, hence we have the product $\psi^*\psi$. $\endgroup$ Aug 21, 2015 at 11:49

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