# Is plane wave equation $\Psi(\mathbf r,t)=\Psi_0e^{i(\mathbf k \cdot \mathbf r-\omega t)}$ for quantum-mechanical wavefunctions really complex? [duplicate]

The equation given for plane sine wave (for instance used to "derive" the Schrödinger equation) is $$\Psi(\mathbf r,t)=\Psi_0e^{i(\mathbf k \cdot \mathbf r-\omega t)}$$

I would have assumed that it would be equal to the REAL part of this complex number, i.e. a cosine (or a phased sine). But all the explanations I looked at on the net seem to say that the complex representation is equivalent. As a matter of fact the derivation of Schrödinger's equation starts with the assumption that the wave equation is equal to the above, which yields to complex results in Schrödinger's equation.

I am totally confused. It's just as if we said that the Real part (or the Imaginary part) of this complex expression was equivalent to the full complex expression...

Please note that I am NOT asking why we use complex numbers in quantum mechanics, but why we say that a wave function can be expressed as a complex value, combining cos and sin in the complex plane, while it actually only is equal to a sin (or cos) function.

EDIT:
@probably_someone I think your answer/question points me in the right direction. Let me know if I am correct.

My confusion is that I have seen in a lot of instances on the web people saying that a plane progressive sine wave amplitude could be expressed as in the complex equation above. This is wrong (hence my troubles). The equation of a plane sine wave is purely real (i.e. with values in $$\Bbb R$$): $$w(x,r)=A.cos(\omega t-kx-\phi )$$.

However when we use this as the Real part of a complex number (using Euler formulas), it becomes easier to derive results such as, for instance, the probability of finding a 'particle' (using old QM terminology) in some place at some given time, by using the squared module of that complex number. Please confirm. Thanks.

• @Carla "why we say that a wave function can be expressed as a complex value, combining cos and sin in the complex plane, while it actually only is equal to a sin (or cos) function." Why do you think this is true in the context of wavefunctions, which are typically complex-valued in general? – probably_someone Oct 13 '18 at 10:58
• "It's just as if we said that the Real part (or the Imaginary part) of this complex expression was equivalent to the full complex expression" - I've read your question several times now and I just don't see how you arrived at this. I've voted to close this question for the reason that it is unclear. You may edit your question to add additional details to explain your claim that the wave function "actually only is equal to a sin or cos function" (which is simply not true AFAIK). – Alfred Centauri Oct 13 '18 at 12:23
• @probably_someone please see my EDIT above. Thanks. – Carla Oct 13 '18 at 12:55

but why we say that a wave function can be expressed as a complex value, combining cos and sin in the complex plane, while it actually only is equal to a sin (or cos) function.

It isn't. Wavefunctions in quantum mechanics really are complex-valued.

There are multiple places in physics where we use complex-valued waveforms of the form $$e^{i(kx-\omega t)}$$ in the understanding that the physical objects that they represent are the real parts of those waveforms (as explained in more depth here).

However, quantum mechanics is not one of those domains. When we say things like $$\Psi(\mathbf r,t)=\Psi_0e^{i(\mathbf k \cdot \mathbf r-\omega t)},$$ we really do mean that the wavefunction is the complex exponential, not just its real part.

• Ah, I see now, the OP was thinking about complex exponential representation of real plane waves. But, indeed, at the linked Wikipedia article, there is this: In quantum mechanics the solutions of the Schrödinger wave equation are by their very nature complex-valued and in the simplest instance take a form identical to the complex plane wave representation above. The imaginary component in that instance however has not been introduced for the purpose of mathematical expediency but is in fact an inherent part of the “wave”. – Alfred Centauri Oct 13 '18 at 14:47
• @AlfredCentauri Yes. Thank you for taking the time and trying to understand my point. I visibly did not make myself clear. I do understand that in Quantum Physics we use a complex wave function. My question was: "why do certain persons 'explain' how Schrodinger derived his equation by saying a plane harmonic (sine) wave can be expressed as $e^{ix}$ ? The answer is that he did not say that... – Carla Oct 14 '18 at 8:31

In Quantum mechanics, wavefunctions are complex: $$\psi:\Omega\to\mathbb{C}$$ Where $$\Omega$$ is a subset of the set in which you are working in.

This may sound strange, since it can be thought: "What does it mean for a wavefunction to be complex?" Well, wavefunctions are not Observables, so you can't detect them directly. What you can detect is the probability that a particle confined in a region $$\Omega$$ is found in the region $$\Sigma\subset\Omega$$ as: $$P(\Sigma)=\int_\Sigma |\psi|^2d^3x$$

In general, you can measure Observables (which are Self-Adjoint operators) by means of one of the axioms of quantum theory:

The probability that an execution of a measurement of an Observable $$Q$$, once gives a result contained in a certain numerical range $$X$$, is given by: $$p_Q(X)=\sum_r\int_{x_r(\rho)\in X}\frac{d\rho}{K_r(\rho)}|\langle r,\rho|\psi\rangle|^2$$ where $$K_r(\rho)$$ is a normalization and the kets $$|r,\rho\rangle$$ are a generalized orthonormal basis constituted by eigenkets of $$\hat{Q}$$ and the numbers $$x_r(\rho)$$ are the eigenvalues of $$\hat{Q}$$ corresponding to $$|r,\rho\rangle$$: $$\hat{Q}|r,\rho\rangle=x_r(\rho)|r,\rho\rangle$$

$$\def\br{{\bf r}} \def\ket#1{|#1\rangle} \let\up=\uparrow \let\down=\downarrow$$ I want to complement what has already been said.

Wavefunctions in quantum mechanics really are complex-valued.

Let's see some examples. The first is just the plane wave (eigenfuntion of momentum). Its probability density is constant all over space (not-normalizable function). It ought to be, since, as an eigenfunction of momentum, it may only change by a phase factor under translations.

If the real part were taken: $$\Re\,\Psi = \cos({\bf k}\cdot\br-\omega t)$$ then $$|\Psi|^2$$ would depend on $$\br$$ and also on $$t$$. Dependence on $$t$$ is unacceptable for a stationary state. As to $$\br$$ instead, a dependence would be possible for different problems.

Second example: unidimensional box. In this case the eigenfunctions of $$H$$ have to vanish at the boundaries ($$x=0$$, $$x=a$$), and solutions are $$\psi_n(x)=A \sin {n \pi x \over a}$$ which are real. I neglected time dependence, but if it should be written a complex exponential would appear again. Furthermore, if we want to study non-stationary states we have to take linear combinations of $$\psi_n$$'s, and to get all possible states the coefficients of the combination must be complex.

Third example: a spin 1/2 particle. If we are only interested in spin states, a basis can be taken $$\ket{\up{}} \qquad \ket{\down{}}$$

meaning eigenvectors of $$s_z$$ to eigenvalues $$+1/2$$, $$-1/2$$ respectively. It is well known that in order to get all spin states linear combination of $$\ket{\up{}}$$, $$\ket{\down{}}$$ with complex coefficients are requested. E.g. if the eigenvector for $$s_x=1/2$$ is $${1 \over \sqrt2}\,(\ket{\up{}} + \ket{\down{}})$$ then the eigenvector for $$s_y=1/2$$ is $${1 \over \sqrt2}\,(\ket{\up{}} + i\,\ket{\down{}}).$$

A general remark. I'm afraid it is too often said by physicists that "only real numbers have physical meaning". IMHO this is pure nonsense. Physics is free to use whichever mathematical entity and structure physicists find useful to build up theoretical explanations. Physical meaning of a mathematical object is defined by how it is used in the theory. Thus, if QM uses complex numbers as an essential ingredient of its structure, then complex numbers do have physical meaning.