# Physical intuition behind negative values for wave function?

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?

(I tried asking this question on physics.SE but it would seem that in physics you just don't talk about or use the wave function until after you square it)

• And imagine, the wavefunction can easily be even imaginary! (Or, better, complex-valued function). – ssavec Aug 19 '15 at 8:29
• Don't think about "negative" as being different than a "positive" wave function. What's really important is that the signs are either the same or opposite. It's the difference between signs in wavefunctions that give rise to the interesting stuff. – user19026 Sep 21 '15 at 5:05
• @ssavec, and it is not even that it can, rather that it is a complex-valued function. – Wildcat Oct 4 '15 at 20:23

The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum operator for the $n^\text{th}$ stationary state, you would get two solutions: $+\hbar k$ and $-\hbar k$, where $k = n\pi/L$. The plus and minus signs mean that there is a $50$% chance the particle could be moving from the left to the right and a $50$% chance of the particle moving from right to left.

In the case of summing of two wavefunctions to get the wavefunction of a molecular orbital, a negative wavefunction doesn't mean anything at all. The reason why when a positive and negative wavefunction is added, an antibonding orbital is formed is simply because when you add a positive and negative number, you get $0$ or a really small number. This is known as destructive interference. Therefore the new wavefunction will contain a region where the y values equal $0$. So when you square this new wavefunction of the molecular orbital you are going to get that there is very 0 probability of finding the electron in that position (known as a nodal plane), and hence it is called an antibonding orbital.However note that just because there is negative region in the wavefunction, doesn't mean that it will always add to form an anti-bonding MO. For example, if you add 2 negative wavefunction values (known as constructive interference), you get a large negative value. When you square becomes a very large positive value. Hence this means that there is a large probability of the electron being in that region and hence it will be a bonding MO. • but how can it make sense to say "a negative wavefunction doesn't mean anything at all" when it is the distinguishing feature between the combination of two wavefunctions being attractive or repulsive? – norlesh Aug 19 '15 at 11:23
• @norlesh The phase of the wavefunction (i.e. the sign) doesn't mean anything. Let me ask you this about the $2p_z$ orbital (can be any p orbital, but I choose this one to avoid potential nitpicking). In this orbital, there is a positive lobe and a negative lobe. What is the difference between them? Do you think there is a "correct" way to label which lobe is positive and which lobe is negative? – orthocresol Aug 19 '15 at 12:38
• @norlesh I'm going off soon so I'm just going to write, maybe a somewhat suitable analogy would be the two mirror images of a molecule, (R) and (S) isomers. They're different, but they react exactly the same way (let's assume their environment is achiral; our universe does not distinguish between positive and negative wavefunctions at all, just like an achiral environment does not distinguish between (R) and (S) isomers). – orthocresol Aug 19 '15 at 12:44
• @norlesh Let's say you react these two isomers with a different, enantiomerically pure, molecule to form (R,S) and (S,S) diastereomers. These are "different" molecules, they have different physical and chemical properties. These are analogous to the antibonding and bonding molecular orbitals. Only the relative phase matters; the absolute phase doesn't. As Nanoputian already said, the wavefunction $\psi$ itself actually has no physical interpretation; Max Born actually won a Nobel Prize for finally interpreting $|\psi|^2$ as the probability density (amongst other things, of course). – orthocresol Aug 19 '15 at 12:49

In addition to @Nanoputian's excellent description of constructive and destructive interference in the formation of MOs, I want to provide a more mathematical explanation for why the phase of the wavefunction does not matter.

# Finding the wavefunction

The time-independent Schrödinger equation, in one dimension, reads:

$$\hat{H}\psi(x) = E\psi(x)$$

It can be shown that, if a wavefunction $\psi = \psi(x)$ satisfies the above equation, the wavefunction $k\psi$ (with $k \in \mathbb{C}$) also satisfies the above equation with the same energy eigenvalue $E$. This is because of the linearity of the Hamiltonian:

\begin{align} \hat{H}(k\psi) &= k(\hat{H}\psi) \\ &= k(E\psi) \\ &= E(k\psi) \end{align}

There are several conditions that a wavefunction must satisfy for it to be physically realisable, i.e. for it to represent a "real" physical particle. In this discussion, the relevant condition is that the wavefunction must be square-integrable (or normalisable). In mathematical terms:

$$\langle\psi\lvert\psi\rangle = \int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x < \infty$$

This means that there has to exist a constant $N \in \mathbb{C}$ such that $N\psi$ is normalised:

$$\int_{-\infty}^{\infty}\!\lvert N\psi\rvert^2\,\mathrm{d}x = \lvert N \rvert^2 \!\!\int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x = 1$$

From this point onwards, we will assume that we will already have found the suitable normalisation constant such that the wavefunction $\psi$ is already normalised. In other words, let's assume $\langle\psi\lvert\psi\rangle = 1$, because we can. Now let's consider the wavefunction $-\psi$, which is equivalent to $N\psi$ with $N = -1$. Is this new wavefunction normalised?

\begin{align} \int_{-\infty}^{\infty}\!\lvert -\psi\rvert^2\,\mathrm{d}x &= \lvert -1 \rvert^2 \!\!\int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x \\ &= \int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x \\ &= 1 \end{align}

Of course it is. So, what I've written so far basically says: if $\psi$ is a normalised solution to the Schrödinger equation, so is $-\psi$.

In fact, you could go one step further. Using exactly the same working as above, you could show that if $\psi$ is a normalised solution to the Schrödinger equation, the wavefunction $(a + ib)\psi$ would also be one, as long as $a^2 + b^2 = 1$. (If you like exponentials, that's equivalent to saying $a + ib = e^{i\theta}$.) I've illustrated this idea on this diagram:

$\qquad\qquad\qquad\qquad\qquad\qquad$ If $\psi$ is a real-valued, one-dimensional wavefunction, you could plot it on a graph against $x$. The wavefunction $i\psi$ would then be exactly the same shape, just coming out of the plane of the paper ($\theta = 90^\circ$). You could have the wavefunction $(1+i)\psi/\sqrt{2}$. It would be pointing outwards of the plane of the paper by $\theta = 45^\circ$, exactly halfway in between $\psi$ and $i\psi$, but exactly the same shape. However, physics doesn't know where the plane of your paper is, so all these wavefunctions are equally admissible. From the point of view of the system, they are all the same thing.

# Using the wavefunction

"But wait! If the wavefunction is negative, what about the values of momentum, position, and energy that you calculate? Will they become negative?"

"Good question, myself!"

Well, for starters, one thing that you use the wavefunction for is to find the probability density, $P(x)$. According to Max Born's interpretation of the wavefunction, this is given by $P(x) = \lvert \psi \rvert ^2$. Let's say that the probability density described by the negative wavefunction $-\psi$ is a different function of $x$, called $Q(x)$:

\begin{align} Q(x) = \lvert -\psi \rvert ^2 &= \lvert -1 \rvert^2 \lvert \psi \rvert ^2 \\ &= \lvert \psi \rvert ^2 \\ &= P(x) \end{align}

So, the probability density described by the negative wavefunction is exactly the same. In fact, the probability density described by $i\psi$ is exactly the same as well.

Now let's talk about observables, such as position $x$, momentum $p$, and energy $E$. Every observable has a corresponding operator: $\hat{x}$, $\hat{p}$, and $\hat{H}$ respectively (the Hamiltonian has a special letter because it's named after William Hamilton). You use these operators to calculate the mean value of the observable. I'll give an example regarding the momentum. If you want to find the mean momentum, denoted $\langle p \rangle$, you would do the following:

\begin{align} \langle p \rangle &= \langle\psi\lvert\hat{p}\rvert\psi\rangle \\ &= \int_{-\infty}^\infty\!\psi^*\hat{p}\psi\,\mathrm{d}x \end{align}

I'm going to call the value of that integral $p_1$. Now, let's do the same thing. Let's assume that the mean momentum for the negative wavefunction is not necessarily the same value. Let's call the new mean momentum something else, like $p_2$.

Before we go on, I'm going to establish that the momentum operator $\hat{p} = -i\hbar\frac{\mathrm{d}}{\mathrm{d}x}$ is also linear. If you doubt it, you can test it out using the definition of linearity in the very first link I posted. In fact, all quantum mechanical operators corresponding to observables are linear. Therefore $\hat{p}(-\psi) = -\hat{p}\psi$ and so:

\begin{align} p_2 &= \langle -\psi\lvert\hat{p}\lvert-\psi\rangle \\ &= \int_{-\infty}^\infty\! (-\psi)^*\hat{p} (-\psi)\,\mathrm{d}x \\ &= (-1)^2\!\!\int_{-\infty}^\infty\! \psi^*\hat{p}\psi\,\mathrm{d}x \\ &= \int_{-\infty}^\infty\! \psi^*\hat{p}\psi\,\mathrm{d}x \\ &= p_1 \end{align}

So, if we talk about the ground state of the particle in a box of length $L$, no matter whether you use the positive wavefunction

$$\psi_1 = \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$

or the negative wavefunction

$$-\psi_1 = -\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$

or the complex wavefunction

$$i\psi_1 = i\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$

you'll get exactly the same values for average position $(= L/2)$, average momentum $(= 0)$, and average energy $(= h^2/2mL^2)$ (the word average is redundant here, since this is a stationary state, but whatever).

Everything that I have said so far can be easily generalised to three dimensions. It can also be generalised to linear combinations of stationary states, i.e. solutions of the time-dependent Schrödinger equation.

# A note about molecular orbitals

"Okay, but what happens when you combine atomic orbitals to make molecular orbitals? You have constructive interference from the positive + positive, and destructive interference from the positive + negative, but what about the negative + negative combination?"

"Good question, myself!"

Let's talk about the $\ce{H2}$ molecule. The proper way to find the molecular orbitals is to solve the Schrödinger equation for the entire system, which is really difficult to do. One way to find approximate forms of the MOs is to make linear combinations of atomic orbitals; this method is called the LCAO approximation. Let's call the 1s orbital of the hydrogen on the left $\phi_1$ and the 1s orbital of the hydrogen on the right $\phi_2$. From the previous sections, we have already established that as far as the hydrogen atom is concerned, the individual phases of $\phi_1$ and $\phi_2$ do not matter. So, let's assume for simplicity's sake that their phases are both positive.

Now, from what you already know, you can get two molecular orbitals $\psi_1$ and $\psi_2$:

\begin{align} \psi_1 &= \phi_1 + \phi_2 \\ \psi_2 &= \phi_1 - \phi_2 \end{align}

These are the bonding and antibonding orbitals respectively (at least, to within a normalisation constant, which I'm not going to care about here because the details are irrelevant). Now let's talk about those combinations that we missed out.

\begin{align} -\phi_1 - \phi_2 &= -\psi_1 \\ -\phi_1 + \phi_2 &= -\psi_2 \end{align}

We already said that $\psi_1$ and $\psi_2$ are (approximations of) solutions to the Schrödinger equation. That means that, from what we've talked about earlier, $-\psi_1$ and $-\psi_2$ must also be (approximations of) solutions to the Schrödinger equation. They must have the same energies as $\psi_1$ and $\psi_2$. In fact, as far as the molecule knows (and cares), they are the same thing as $\psi_1$ and $\psi_2$.

Now, since the individual phases of the atomic orbitals do not matter, if you really wished to, you could declare to the whole world that you define:

$$\phi_3 = \phi_1 \text{ and } \phi_4 = -\phi_2$$

i.e. left hydrogen 1s orbital, $\phi_3$, is positive and right hydrogen 1s orbital, $\phi_4$, is negative. In that case, you can construct the molecular orbitals:

\begin{align} \psi_1 &= \phi_3 - \phi_4 \\ \psi_2 &= \phi_3 + \phi_4 \end{align}

The coefficients of the atomic orbitals would have to be different, since you insisted on having them in different phases - however, the outcome is the same! You get one bonding MO and one antibonding MO.

• Upvoting simply for "Good question, myself!" (well, and it's a good answer, too!) – hBy2Py Oct 4 '15 at 16:05

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?

Your confusion for its most part stems from two things:

1. You are using the wrong terms such as "positive" or "negative" wave function which shows that you don't quite understand what is going on here mathematically. This is especially important taking into account the second point.

2. You are looking for some physical intuition behind a purely mathematical model which has very little to do with physical reality.

Let us for the sake of argument define a positive function of a single real variable $x$ as a real-valued function that takes only positive values, i.e. such that $f(x) > 0$ for all $x$, and a negative function as a function that takes only negative values, i.e. such that $f(x) < 0$ for all $x$. I think this is something OP has in his mind talking about "positive" and "negative" wave functions. But then the way OP uses these adjectives to describe what he has trouble with is surely wrong, since given a wave function $\psi(x)$ how do you know is it a positive or a negative one? Obviously, you don't know that, and consequently, you can't tell either if $-\psi(x)$ is positive or negative.

These whole LCAO-MO business has simply nothing to do with wave functions being positive or negative, rather it is all about forming two orthogonal linear combinations of two atomic orbitals (AO) to form two essentially different molecular orbitals (MO). Nanoputian described this simple LCAO-MO formalism for the particular case of $\ce{H2}$ molecule in details in his answer, and here I just want to warn OP (and others) about the danger of taking this picture literally.

MOs are formed from AOs in a mathematical and not a physical sense. It is just a game of numbers: take AOs and write MOs as linear combinations of them. One should not think that this primitive LCAO-MO model with bonding and anti-bonding MOs formed as a result of constructive and destructive interference of the corresponding AOs describes real physical processes. As I said, this model has very little to do with reality.

In addition to other answers and very briefly, a bond is formed when there is a lot of electron density between the nuclei, this lowers the overall potential energy. Conversely when there is not much electron density between atoms the potential energy is higher and this is anti-bonding. We identify anti-bonding by looking for nodes between atoms. As a rule of thumb, the more nodes the higher the energy hence more anti-bonding.

When adding or subtracting or multiplying wavefunctions together the normal rules of maths apply because wavefunctions are represented by normal mathematical equations. We have to remember that often they are presented in polar coordinates instead of the usual x,y,z so may not seem so familiar, e.g. shapes of s, p, d orbitals.

The sign of a wavefunction $\psi$ is just part of its mathematical description; some wavefunctions have positive and negative parts, e.g. spatial part of p orbitals, some are represented by complex numbers e.g. some d orbitals. We can let $-\psi$ be the same as $\psi$ as the probability of finding the particle in a small region of space $\tau$ to $\tau+d\tau$ is $\psi^2d\tau$.
The measured value of some property, X, e.g. position, is given by the expectation (or average) value; $<X>=\int \psi^* X \psi d\tau$, so again the sign of $\psi$ does not mater. ($\psi^*$ is the complex conjugate, which is only important if $\psi$ is a complex number.) When adding wavefunctions so as to make a linear combination e.g. $\psi = s_1-s_2+s_3$ is the same as $-\psi=-s_1+s_2-s_3$.

Chemists often use diagrams with the sign of the wavefunction labelled as $\pm$ or coloured in a particular way to help understand bonding. Its a very useful shortcut compared to calculating things out. The figure below shows some examples. The p orbitals add 'in phase' as $p_z+p_z$ which is bonding and out of phase as $p_z-p_z$ which is anti-bonding. The shading on the orbitals shows the 'sign' of the wavefunction. In $p_z-p_z$ the difference is not zero everywhere because the orbitals have different origins since they are displaced along the bond axis. The difference is zero at the midpoint, however, and this is called a node.
The $s+p_z$ is non-bonding as the s orbital has overlap with both parts of the p orbital and the two overlaps cancel.
The three s orbitals can add, all three the same shade make bonding (grey or white its the same, $s_1+s_2+s_3$ is equivalent to $-s_1-s_2-s_3$) or subtract orbitals say as $s_1-s_2+s_3$ which is anti-bonding.