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Suppose we are given a wavefunction $$\psi(x) = Ae^{ikx} + Be^{-ikx}.$$

After some internet research I found that

When coupled to the usual time-dependent energy (phase) factors in the full TDSE solution, then the $e^{ikx}$ terms give right-going waves and the $e ^{−ikx}$ terms give left-going waves. Physically, the term $Ae^{ikx}$ is the incoming wave and $Be^{−ikx}$ is a reflected wave.

I know that $e^{ikx}=\sin(kx)+i\cos(kx)$, but this doesn't tell me anything about which way the wave is travelling in (positive or negative $x$?).

Why do the $e^{ikx}$ terms give right-going waves?


Edit:

I have been given a good answer by @SteveB which informed me about having to consider the real part of a wave with time dependence also and considering the consequences for values of $k$ greater than zero and less than zero for a positive $\omega$ $$f(x,t)=Re[e^{ikx-i\omega t}]=\cos(kx-\omega t)\tag{1}$$ In his answer he mentions that for $k\gt 0$ the wave will go right and for $k\lt 0$ the wave will go left.

I see by plotting graphs that the sign of $k$ flips the wave $(1)$ about the $y$-axis (or $x$-axis reflection as Cosine is even).


How can you tell the direction a wave is travelling in just by looking at which way it is oscillating about the $y$-axis?

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    $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Oct 9 '16 at 11:21
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    $\begingroup$ As far as I know, it's just convention to use the cartesian co-ordinate system to assign directions to waves. We can choose right or left, as long as we are consistent. So it maps to the normal x axis. $\endgroup$ – user108787 Oct 9 '16 at 11:26
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    $\begingroup$ It depends on whether the time dependence is e^+iwt out e^-iwt. Physicists traditionally use the latter; electrical engineers traditionally use the former! $\endgroup$ – Steve Byrnes Oct 9 '16 at 12:37
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    $\begingroup$ Waves always have time dependence ... but it's often implicit! (omitted to shorten formulas) $\endgroup$ – Steve Byrnes Oct 9 '16 at 13:24
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    $\begingroup$ @SteveB Are you hinting that if simply given $e^{ikx}$ there is no way to tell whether or not the wave moves left or rightward? $\endgroup$ – BLAZE Oct 9 '16 at 13:26
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If a wave is $f(x,t)=Re[e^{ikx-i\omega t}]$ with $\omega>0$, then it goes right if $k>0$ or left if $k<0$. Do you see why? Try picking two or three values of $t$ and making plots...

But,you can alternately write a wave as $f(x,t)=Re[e^{ikx+i\omega t}]$, and now it's reverse! $k>0$ is left-traveling. (In practice, it would be more common to write $f(x,t)=Re[e^{-ikx+i\omega t}]$ so that $k$ has the normal sign convention.)

So anyway, the key issue is whether the wave's time dependence is $e^{-i\omega t}$ or $e^{+i\omega t}$. It can be a tricky issue because sometimes this time-dependent factor is omitted to make the formulas look simpler. In general, one needs to look through the book or paper to see if the time-dependent factor is written down somewhere.

If they don't say it explicitly, the rule-of-thumb is: Electrical engineers almost always use $e^{+i\omega t}$ and physicists almost always use $e^{-i\omega t}$.

Luckily for you, there's one area that is completely unambiguous: The Schrodinger equation universally uses $e^{-i\omega t}$ (where $\omega=E/\hbar$). So you can count on that everywhere in quantum mechanics. That case is applicable to you.

(One could imagine an evil twin of the Schrodinger equation with the opposite sign on $i$, i.e. $-i\hbar \partial \Psi/\partial t = H\Psi$. It's objectively no less correct than the traditional Schrodinger equation, but uses $e^{+i\omega t}$ time dependence instead. Lucky for us, nobody has ever used this version, as far as I know!)

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Using the usual separation of variables method, the solution of the time-dependent Schrödinger equation is $$\Psi(x,t)=e^{-iEt}\psi(x)=e^{i(\pm kx-Et)},$$ where $\psi(x)$ is the solution to the time-independent equation and we've set $\hbar=1$.

Wave equations are characterized by having solutions of the form $f(x,t)=f(x-vt)$ where $v$ is the wave propagation speed. Imagine that the wave has a peak at $x=vt$. As time passes, the $x$ position of the peak increases, and therefore the wave goes to the right.

Comparing the general solution with your plane waves,$$\Psi(x,t)=e^{i(\pm kx-Et)}=e^{\pm ik(x \mp \frac{E}{k}t)}=f(x-vt)$$ it is clear that $e^{ikx}$ is the right-going wave with speed $v=E/k$, while $e^{-ikx}$ is the left-going one with velocity $v=-E/k$.

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$\vec{k}$ is a vector, in particular, the wave vector. It is related to the momentum, (e.g., in quantum mechanics, $\ \vec{p}=\hbar\vec{k}$). A positive momentum propagates in the positive direction, hence to the right if the axis is horizontal and the positive direction is to the right.

The reason that the time dependence is negative comes from the fact that $\, \vec{k}\cdot\vec{x}-\omega t$ is invariant. If $\vec{k}$ is a positive wave vector and you change the value of $x$ from, say, $0$ to $2$, then you also moved some distance forward in time and the overall value $\vec{k}\cdot\vec{x}-\omega t$ must not change.

The engineering convention is actually not different. You will notice that engineers use $j\omega t$ instead of $-i\omega t$ and $\ j\,=\,-i$. Remember there are 2 square roots of $-1$.

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  • $\begingroup$ Thanks for your answer (+1). Two questions for you: Why is $k$ a vector? I'm new to quantum mechanics so I ask that because I have never come across a vector notion for $k$. Secondly, Could you please elaborate on what you mean by "the time dependence is negative comes from the fact that $\, \vec{k}\cdot\vec{x}-\omega t$ is invariant"? Many thanks. $\endgroup$ – BLAZE Oct 10 '16 at 8:50
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    $\begingroup$ $\vec{k}$ is a vector because it is related to the momentum and momentum is a vector. If you are just beginning quantum mechanics, then you are probably doing everything in 1-d, so that fact is not apparent. What if you had momenta in two directions so that the wave was propagating at some angle between the $x$ and $y$ axis? Since $\vec{p}=\hbar\vec{k}$ you get $p_x = \hbar k_x.\, p_y=\hbar k_y$. $\endgroup$ – drs Oct 10 '16 at 8:59
  • $\begingroup$ Well said, and yes you're right we are doing everything in 1D at the moment! So lets just work in 1D for now; Why is $k\cdot x - \omega t$ invariant? I might be easier to edit your post to explain. Thanks again. $\endgroup$ – BLAZE Oct 10 '16 at 9:04
  • $\begingroup$ The scalar product of two vectors does not change just because you perform a coordinate transformation. Picking different coordinates still describes the same wave. Consider two vectors, $\vec{v}$ and $\vec{w}$, with some x and y components. Now, rotate your coordinates: $x' = x\cos(\theta) - y\sin(\theta),\, y'=y\cos(\theta) + x\sin(\theta)$. Multiply it out, and you will get the same answer in both coordinate systems. How you label coordinates doesn't change physical things. To include the time component, you can perform a Galilean transformation, $\vec{x'}=\vec{x}-\vec{v} t,\, t=t'$ $\endgroup$ – drs Oct 10 '16 at 9:30

protected by Qmechanic Oct 10 '16 at 13:30

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