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A wavefunction can be split by separation of variables and solved for the time independent Schrödinger equation:

$$-\frac{(h/2\pi)^2}{2m}\frac{d^2u}{dx^2} + Vu = Eu,$$ from $$\psi(x, t) = u(x)T(t).$$

On p. 101 in the book: Alastair I.M. Rae & Jim Napolitano, Quantum Mechanics, Sixth Edition, Boca Raton, United States, Crc Press Llc, 2015. There is a diagram describing $u(x)$ (the time independent component of $\psi(x, t)$) for a particle in a potential well that has sufficient energy to escape via quantum mechanical tunnelling.

Quantum Mechanics 6th edition

This shows negative values. I am aware that when calculating the corresponding probabilities at each point, the time dependent component should simply be a phaseshift and thus $u(x) \to \psi(x, t)$ for $|\psi(x, t)|^2$ and thus the probability at each point will always be real valued and positive.

But how would I interpret $u(x)$ on it's own without the copenhagen interpretation of the wavefunction? I am unaware of it having any physical meaning in of itself other than just a mathematical representation?

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  • $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/9109. $\endgroup$ Commented Aug 14 at 14:43
  • $\begingroup$ "I am aware that... the time dependent component should simply be a phase shift and thus... the probability at each point will always be real valued and positive." The probability will be real valued and non-negative at each point regardless of whether or not the time dependence is just a phase shift (e.g., when the time-independent Schrodinger equation is obeyed) or not just a phase shift (e.g., when time-dependent Schrodinger Equation is obeyed). $\endgroup$ Commented Aug 22 at 16:44
  • $\begingroup$ (The real and non-negative nature just follows directly from the fact that an absolute square $|\psi|^2$ has to be real and non-negative.) $\endgroup$ Commented Aug 22 at 16:47

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In QM the (in general) complex wave function in itself doesn't have any direct physical interpretation. As you mentioned, it has to be regarded as a mathematical tool to calculate the probability of measurements. According to the Born interpretation, the square of the modulus of the single particle wave function is interpreted as the probability density of finding the particle at a certain location. In chemistry this density is sometimes interpreted as the charge density of electrons. But the wave function as such, whether it is real, positive, negative or complex, has no direct physical significance.

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  • $\begingroup$ Is there any indirect significance? Etc. are there cases where the wavefunction could be guaranteed to be (for all x) positive, negative or complex? Or are there things known to affect say, the integrable ratio of area above the x axis to below (ratio of positive to negative values)? $\endgroup$ Commented Aug 17 at 21:01
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    $\begingroup$ @Kepler7894i - Examples: Bound stationary states, like a particle in a 1-D box, or an electron in an hydrogen atom, can be described by real wave functions. Their nodes are related to the quantum numbers. See QM textbooks. $\endgroup$ Commented Aug 20 at 9:36
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There is nothing special about a wave having a negative value. A point in an ocean wave has a positive height if it is above the average ocean height, and negative if it is below. A sound wave has a positive pressure if it is above the average pressure and negative if below.

In QM, waves are complex values, so positive and negative don't really apply in the same way. The real and imaginary components at a point can be positive or negative. The magnitude of the wave is non negative, and the phase is $0$ to $2\pi$. Or equivalently, $-2\pi$ to $0$.

It sounds like "What is the wave made of?" might be what you really want to know. In classical mechanics, you think of an electron as a point particle. The QM comes along and now the electron has mysterious properties like a spread out nature so that it can go through two slits and interfere with itself. And yet it has a particle like nature that hits one atom of the screen on the other side.

The wave is a useful abstract mathematical function. It can be used to describe the behavior of the electron. It isn't the electron itself.

The wave has a magnitude. The square of the magnitude at a point is the probability that a measurement would find the electron there. The wave must be complex valued so the phase can describe interference.

The spread out nature of the wave is different from the spread out nature of the electron. Part of the wave is present at one slit, and part at the other. The electron itself is not partially here and partially there. An electron has no internal parts. If it hits a screen, it hits one place. If you measure which slit it goes through, it only goes through one. The wave is a description of where it might be if measured.

The electron is like nothing classical. If you apply classical thinking and try to understand how a particle can go through both slits, it just makes a confusing situation worse.

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TL;DR when we take a real part of a complex exponent, it has negative values. Note that we cannot consider wave functions as real in this case, since the state is not really a bound one. (More precisely, we necessarily have two degenerate functions corresponding to the same energy, but only one is shown.)

IMHO there is some problem with the solution provided or at least with the picture/interpretation. Since the potential is piecewise with constant values in every region determined by adjacent pairs of points from $-\infty, -b, -a, a, b, +\infty$, the solutions can be trivially obtained for every region (though connecting them at borders is tedious and produces transcendental equations.)

Thus, for $x<-b$ and $x>b$ we have solutions $e^{\pm ikx}$, where $k=\frac{1}{\hbar}\sqrt{2m(E-V_1)}$:

  • for $E<V_1$ these are decaying/growing exponents (of which we choose one, so that the solution is finite in infinity). Manifestly, this is not the case in the picture
  • for $E>V_1$ we have complex exponents, with the probability density $|Ae^{ikx}+Be^{-ikx}|^2$. Obviously, if we take simply a real part of a complex exponent, it will be a sin/cosine-like function, which can take negative values... but taking the real part does not tell the whole story, since the state is not really bound.

Remark
As a trivial example I suggest pondering the case of no potential at all. The wave functions can be chosen as eigenfunctions of energy and momentum $$ e^{\pm ikx} $$ or as eigenfunctions of energy and "parity operator" (here - reflection in respect to $x=0$): $$ \cos(kx), \sin(kx) $$ The latter would result in negative values just like in the Q., though the probability density will be smooth and positive: $$ |\cos(kx)|^2= \frac{1+\cos(2kx)}{2} $$ So upon some consideration, what the book presents is not outright wrong, but presents a somewhat odd choice (though justified, e.g., if we are interested in decay of a quasibound state and outgoing only solutions, aka Gamow formula)

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  • The overall phase of the (possibly complex) wavefunction is unphysical/immaterial.

  • For a bound state its wavefunction can be chosen to be real, cf. e.g. this & this Phys.SE posts.

  • For a bound ground state in 1D its wavefunction can be chosen to be positive, cf. the node theorem in 1D QM.

  • More generally, for the $n$th bound state in 1D its real wave function has $n-1$ zeros/nodes, cf. the node theorem in 1D QM.

  • The wavefunction in Fig. 5.7 of Rae & Napolitano describes a scattering state with energy $E>V_1$, so in the asymptotic regions with potential $V_1$ the wavefunction is oscillatory. If the wavefunction is real$^1$, it therefore necessarily has infinitely many zeros, and hence both positive & negative values, cf. OP's question.

$^1$ Whether or not the wavefunction for a scattering state can be chosen to be real depends on the asymptotic boundary conditions of the scattering process.

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The sign of a wave function is a special case of its phase. Relative phases are detectable in interference experiments and so it can be relevant both in terms of explaining and predicting the behaviour of quantum systems.

In general the Born rule isn't sufficient for understanding the evolution of quantum systems. The square amplitudes of wave functions in general don't obey the rules of probability in interference experiments

https://arxiv.org/abs/math/9911150

The reason the Born rule works for many measurements is that copying information out of a quantum system suppresses interference, as explained by the literature on decoherence:

https://arxiv.org/abs/1911.06282

A tunnelling experiment involves interference so decoherence is weak or absent and interference is important, see

https://arxiv.org/abs/2507.19105

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