# I How is e^ikx a plane wave? (1 Viewer)

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#### Phys12

In these notes, https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/lecture-notes/MIT8_04S16_LecNotes11.pdf, in the middle of page 5, it is mentioned:

We will be interested in bound states namely, energy eigenstates that are normalizable. For this the energy E of the states must be negative. This is readily understood. If E > 0, any solutions in the region x > a where the potential vanishes would be a plane wave, extending all the way to infinity. Such a solution would not be normalizable.

I'm guessing that the requirement that bound states are energy eigenstates that are normalizable is by definition. I also get why E>0 leads to a solution of e^ikx, as given in earlier sections, which extends to infinity will not be normalizable and hence, won't be a bound state (by definition). But why is e^ikx a plane wave? On page 10, it looks as though a plane wave, represented at places with zero potential, is just an exponential decay, why do we have the complex i in the equation? Also, in wikipedia, it says that a plane wave is mathematically represented as a cosine or a sine, not an exponential. I don't understand what's happening here...

#### jtbell

Mentor
is just an exponential decay, why do we have the complex i in the equation? Also, in wikipedia, it says that a plane wave is mathematically represented as a cosine or a sine, not an exponential.
Look up Euler's formula.

#### Nugatory

Mentor
Also, in wikipedia, it says that a plane wave is mathematically represented as a cosine or a sine, not an exponential. I don't understand what's happening here...
We have the identity $e^{i\theta}=\cos\theta+i\sin\theta$; that's how Wikipedia can say that plane waves are represented by sines and cosines while your textbook says that they're represented by a complex exponential. (If you want to see where that identity comes from, just look at the Taylor series for $\sin x$, $\cos x$, and $e^x$).

#### Phys12

We have the identity $e^{i\theta}=\cos\theta+i\sin\theta$; that's how Wikipedia can say that plane waves are represented by sines and cosines while your textbook says that they're represented by a complex exponential. (If you want to see where that identity comes from, just look at the Taylor series for $\sin x$, $\cos x$, and $e^x$).
I knew about the identity, sorry, I forgot to include my main question, here it is:

So, to get the plane wave, do we just take the real part of the exponential? And in the diagram on page 10, at 0 potential, it doesn't look like a sinusoidal wave, it looks more like an exponential decay/growth... (here exponential meaning there's no imaginary part in the exponent, of the form e^kx or e^-kx)

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#### Nugatory

Mentor
So, to get the plane wave, do we just take the real part of the exponential?
You don't need to; just accept this as the the definition of a complex-valued wave. The real part alone, or the complex part alone, are not particularly interesting when we go to calculate physical observables from the wave function.
And in the diagram on page 10, at 0 potential, it doesn't look like a sinusoidal wave, it looks more like an exponential decay/growth... (here exponential meaning there's no imaginary part in the exponent, of the form e^kx or e^-kx)
That is because in the two regions outside the well, the solution to the Schrodinger equation for a bound state (energy is less than the depth of the well) is an exponential of a real number and the solution for an unbound state (energy greater than the depth of the well) is an exponential of a complex number. We're solving an equation of the form $\frac{d^2\psi}{dx^2}=\pm k^2\psi$; the solutions will be of the form $e^{\pm kx}$ or $e^{\pm ikx}$ according to whether we need the left-hand side to come out positive or negative.

#### Phys12

You don't need to; just accept this as the the definition of a complex-valued wave. The real part alone, or the complex part alone, are not particularly interesting when we go to calculate physical observables from the wave function.

That is because in the two regions outside the well, the solution to the Schrodinger equation for a bound state (energy is less than the depth of the well) is an exponential of a real number and the solution for an unbound state (energy greater than the depth of the well) is an exponential of a complex number. We're solving an equation of the form $\frac{d^2\psi}{dx^2}=\pm k^2\psi$; the solutions will be of the form $e^{\pm kx}$ or $e^{\pm ikx}$ according to whether we need the left-hand side to come out positive or negative.
Oh, I see. So you get a real valued exponential when the energy is less than the potential and a complex valued when the energy is greater than the potential, correct?

#### atyy

So, to get the plane wave, do we just take the real part of the exponential? And in the diagram on page 10, at 0 potential, it doesn't look like a sinusoidal wave, it looks more like an exponential decay/growth... (here exponential meaning there's no imaginary part in the exponent, of the form e^kx or e^-kx)
In plain English it is not a plane wave, because there is no plane in 1D. However, by similarity to the equation in 3D, this is called a plane wave by convention (I usually say sinusoidal wave in 1D). http://farside.ph.utexas.edu/teaching/qmech/Quantum/node16.html

For physical plane waves, one has to take only the real parts of the solution. Here the notes are being a little bit informal, assuming you know the background well enough to understand his analogy even though the quantum mechanical wave function is complex.

• Phys12

#### hilbert2

Gold Member
I think the name "plane wave" comes from the fact that a function $\psi (x,y,z)=exp(-i\mathbf{k}\cdot\mathbf{x})$ is constant in any plane normal to the wavevector $\mathbf{k}$.

• dextercioby and vanhees71

#### Phys12

I think the name "plane wave" comes from the fact that a function $\psi (x,y,z)=exp(-i\mathbf{k}\cdot\mathbf{x})$ is constant in any plane normal to the wavevector $\mathbf{k}$.
So it's not a plane wave in terms of actual planes, but just a sinusoidal wave?

#### Phys12

In plain English it is not a plane wave, because there is no plane in 1D. However, by similarity to the equation in 3D, this is called a plane wave by convention (I usually say sinusoidal wave in 1D). http://farside.ph.utexas.edu/teaching/qmech/Quantum/node16.html

For physical plane waves, one has to take only the real parts of the solution. Here the notes are being a little bit informal, assuming you know the background well enough to understand his analogy even though the quantum mechanical wave function is complex.
If you take the real part of the e^ikx solution, wouldn't you get a sinusoidal wave and not an exponential decay/growth as with equation e^kx?

#### Mentz114

Gold Member
So it's not a plane wave in terms of actual planes, but just a sinusoidal wave?
Have you seen this Wiki article ? It could not be simpler.
A homogeneous plane wave is one in which the planes of constant phase are perpendicular to the direction of propagation $\vec{n}$.

• vanhees71

#### atyy

If you take the real part of the e^ikx solution, wouldn't you get a sinusoidal wave and not an exponential decay/growth as with equation e^kx?
Yes.

#### Phys12

Ok, I think I have my issue solved. Plane waves cannot be normalized and that's why the energy of a bound state will be negative and the other problem has the answer at the end of page 5 and beginning of page 6. Thank you all!

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