Confused over complex representation of the wave

My quantum mechanics textbook says that the following is a representation of a wave traveling in the +$x$ direction:$$\Psi(x,t)=Ae^{i\left(kx-\omega t\right)}\tag1$$

I'm having trouble visualizing this because of the imaginary part. I can see that (1) can be written as:$$\Psi(x,t)=A \left[\cos(kx-\omega t)+i\sin(kx-\omega t)\right]\tag2$$

Therefore, it looks like the real part is indeed a wave traveling in the +$x$ direction. But what about the imaginary part? The way I think of it, a wave is a physical "thing" but equation (2) doesn't map neatly into my conception of the wave, due to the imaginary part. If anyone could shed some light on this kind of representation, I would appreciate it.

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If you have to visualize it, it would be as a cylindrical spiral going through space, where you have a 2D plane being the complex plane, and the third axis being x. –  NeuroFuzzy Feb 11 '13 at 10:05
The wavefunction itself isn't a "thing" that has a real only value everywhere in space. The physical thing is the probability, which is obtained by multiplying the wavefunction with its complex conjugate and integrating over the space under consideration. –  Mew Feb 11 '13 at 10:20
In quantum mechanics you can always multiply every wavefunction by a phase $\exp(i\phi)$ and all physical quantities are unchanged, so the real part is no more or less physical than the imaginary part. In fact, splitting the complex plane into real and imaginary parts is rather unphysical, and not very useful most of the time. You can see immediately that equation (1) is a wave travelling in the $+x$ direction because it has the form $\Psi=f(kx-\omega t)$ for the function $f(\cdot)=A\exp(\cdot)$. Any such expression is a travelling wave. A wave does not have to be a $\sin$ or $\cos$. –  Michael Brown Feb 11 '13 at 10:25
More generally in quantum mechanical scattering theory, concerning the fact that a plane wave $Ae^{i\left(kx-\omega t\right)}$, with a stationary time-independent probability density $|A|^2$, is interpreted as a right mover, see also this Phys.SE post. –  Qmechanic Feb 11 '13 at 13:31

What if I told you the wave equation was given by:

$$\Psi(x,t)=A \cos(kx-\omega t)\tilde{i}+A\sin(kx-\omega t)\tag2\tilde{j}$$

where $i$ and $j$ represent the unit vectors in the x and y directions?

If so, you could think about the wave oscillating in two separate spatial dimensions.

Now the wave equation is actually instead:

$$\Psi(x,t)=A\left[ \cos(kx-\omega t)+i\sin(kx-\omega t)\right]\tag2$$

But what's the difference? In vectors you must keep $i$ and $j$ components separately when doing equations; similarly, in complex numbers you solve equations keeping real parts equal, and complex parts equal. You can thus think of the wave equation as having two dimensions, a real dimension and a complex dimension.

In vectors, you obtain the square of the magnitude by adding the squares of the x-component and the y-component.

$\text{Magnitude}^2 = a^2 + b^2$ if a is the x-component and b is the y-component of a vector.

Similarly, to obtain the physically meaningful result of probability in quantum mechanics you multiply the wave function and its complex conjugate:

$$\text{Probability density} = \Psi(x,t)\times\Psi^{\dagger}(x,t) = (a+bi)\times(a-bi) = a^2 + b^2$$ where $b$ is the complex part and $a$ is the real part of the wavefunction. So probability is effectively the square of the magnitude of the "wave-vector", which has components in the real dimension and the complex dimension.

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The wave function itself is not a "real" thing. I.e. it is not an observable quantity. What's "real" is the probability distribution which is associated with the wave function. The probability of finding the particle between points $x=a$ and $x=b$ (restricting to one dimension for simplicity) is given by:

$$P(a\leq x\leq b)=\int_a^b |\Psi|^2 \mathrm{d}x$$

where $|\Psi|^2=\Psi^* \Psi$ and $\Psi^*$ is the wave-function's complex conjugate. $|\Psi|^2$ is a real-valued function (i.e. its imaginary part is zero). It isn't particularly useful to think of the wave function itself as being a physical wave. What matters is the magnitude of the wave function.

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Also the mathematical wave nature gives rise to intereference patterns. The wavefunctions (both real and complex components) interfere, and this interference is seen physically after performing the probability calculation above. –  Mew Feb 11 '13 at 11:17

I think the question is more about the physical intepretation of the complex expression

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

than the mathematical meaning of it. For the physical meaning of it, we think of the probability amplitude like a rotating arrow, which rotates as the particle travels in space. The rotation frequency of the arrow is determined by the energy (frequency) of the particle (photon.) This arrow has been given the name 'phasor' because the argument $\phi =kx-\omega t$ is an angle (in wave mechanics it is called 'phase' of the wave). This phase tells us how many degrees the arrow has rotated from the moment the particle has been created until it reaches the point $x$ at time $t$ of its journey.

This complex number representation is very convenient, not only because it shows the phase of the wave but it also shows the direction (if the wave travels in 3-D.) However its importance in QM comes from the need to combine (add) waves comming from different sources at some point in space. This is not a simple algebraic addition because the angles involved make the problem geometrical, and the complex number representation does this very neatly. In a way the fasors add like vecors do (the real with the real, and the imaginary with the imaginary and its done!)

The calculation of the probabilities follows rules that are also geometrical. For example, let us think of two waves comming from the two slits in the DS experiment as:

from slit 1 $S_1: \psi_1(x_1,t)$ and from slit 2 $S_2: \psi_2(x_2,t)$.

The $x_1$ and $x_2$ show the distances the two phasors (waves) traveled by the time they reach some point P on the screen. When these two waves arrive at the screen, they will be added to get the total amplitude first

$A=\psi_1(x_1,t)+\psi_2(x_2,t)$

and then the probability will be the 'square of the modulus' of the total amplitude as

$P=|A|^2= |\psi_1 (x_1,t)|^2+ |\psi_2 (x_2,t)|^2 + 2|\psi_1 (x_1,t)|\times|\psi_2 (x_2,t)|\cos(\theta)$

The thrird term in the equation above, shows the real need for the complex representation of the wave functions in QM, as well as the need for finding first the total probability amplitude, and then finding the probability as the square of the total modulus. This term is the root of all beautiful interference phenomana we observe in the quantum mechanical world. I hope this helps a little.

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In short, “a wave traveling in the $+x$ direction” has nothing to do with actual motion of a wave packet. In spite of some mathematical similarities, wave function isn’t anything physically like a gravity wave on water surface. In quantum “waves”, there is no water (or gas, other 3D continuum, string, or anything else that can convey a physical meaning to values of $Ψ(x,t)$).

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Substantiation for downvoting? Retaliation for criticism of an own posting (or a buddy’s posting, perhaps), Ī guess. –  Incnis Mrsi Nov 21 '14 at 12:18