Particles, waves, and mass

Classical Mechanics Level 3

We now toss a little quantum mechanics into the mix. The key aspect we will need from quantum mechanics is that particles, such as electrons, protons, or photons, can also be described by waves. In quantum mechanics, the probablity P of finding a particle at a certain location is given by P ( t , x ) = ∣ f ( t , x ) ∣ 2 P(t,x)=|f(t,x)|^2 where f ( t , x ) f(t,x) is a function in space called the probability amplitude . For free particles, i.e. those that are just traveling through empty space, f ( t , x ) f(t,x) takes the form of a wave. Therefore, there is such a thing as a free "electron wave" analogous to a free electromagnetic wave.

Such waves must satisfy the wave equation:

( 1 c 2 ∂ 2 ∂ t 2 − ∂ 2 ∂ x 2 + m 2 c 2 ℏ 2 ) f ( t , x ) = 0 (\frac{1}{c^2} \frac {\partial^2}{\partial t^2} - \frac {\partial^2}{\partial x^2} + \frac {m^2 c^2}{\hbar^2}) f(t,x)=0

where c c is the speed of light, ℏ \hbar is a constant (called Planck's constant), and m m is the mass. The solutions are just like the wave solutions you are familiar with from Newtonian mechanics,

f ( t , x ) = A cos ⁡ ( ω t − k x ) + B sin ⁡ ( ω t − k x ) f(t,x)=A \cos (\omega t - kx) + B \sin (\omega t - kx) .

If we plug this form for f ( t , x ) f(t,x) back into the wave equation, what must ω \omega be as a function of k k if the wave equation is to be satisfied?

ω 2 = c 2 k 2 + m 2 c 4 / ℏ 2 \omega^2=c^2 k^2 + m^2 c^4/\hbar^2 ω 2 = c 2 k 2 − m 2 c 4 ℏ 2 \omega^2=c^2 k^2 - m^2 c^4 \hbar^2 ω 2 = c 2 k 2 − m 2 c 4 / ℏ 2 \omega^2=c^2 k^2 - m^2 c^4/\hbar^2 ω 2 = c 2 k 2 + m 2 c 4 ℏ 2 \omega^2=c^2 k^2 + m^2 c^4 \hbar^2

David Mattingly by David Mattingly

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2 solutions

Chew-Seong Cheong
Jul 31, 2014

It is given that:

f ( t , x ) = A cos ⁡ ( ω t − k x ) + B sin ⁡ ( ω t − k x ) f(t,x) = A\cos{(\omega t-kx)}+B\sin{(\omega t-kx)}

Therefore,

∂ ∂ t f ( t , x ) = − ω A sin ⁡ ( ω t − k x ) − ω B cos ⁡ ( ω t − k x ) \frac{\partial}{\partial t}f(t,x) = -\omega A\sin{(\omega t-kx)}-\omega B\cos{(\omega t-kx)}

∂ 2 ∂ t 2 f ( t , x ) = − ω 2 A cos ⁡ ( ω t − k x ) + ω 2 B sin ⁡ ( ω t − k x ) \frac{\partial^2}{\partial t^2}f(t,x) = -\omega^2 A\cos{(\omega t-kx)}+\omega^2 B\sin{(\omega t-kx)}

Similarly,

∂ ∂ x f ( t , x ) = k A sin ⁡ ( ω t − k x ) + k B cos ⁡ ( ω t − k x ) \frac{\partial}{\partial x}f(t,x) = kA\sin{(\omega t-kx)}+kB\cos{(\omega t-kx)}

∂ 2 ∂ x 2 f ( t , x ) = − k 2 A cos ⁡ ( ω t − k x ) + k 2 B sin ⁡ ( ω t − k x ) \frac{\partial^2}{\partial x^2}f(t,x) = -k^2 A\cos{(\omega t-kx)}+k^2 B\sin{(\omega t-kx)}

Substitute into:

( 1 c 2 ∂ 2 ∂ t 2 − ∂ 2 ∂ x 2 + m 2 c 2 ℏ 2 ) f ( t , x ) = 0 \left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+\frac{m^2c^2}{\hbar^2}\right) f(t,x)=0

− ω 2 c 2 A cos ⁡ ( ω t − k x ) + ω 2 c 2 B sin ⁡ ( ω t − k x ) -\frac{\omega^2}{c^2}A\cos{(\omega t-kx)}+\frac{\omega^2}{c^2} B\sin{(\omega t-kx)} + k 2 A cos ⁡ ( ω t − k x ) − k 2 B sin ⁡ ( ω t − k x ) + m 2 c 2 ℏ 2 f ( t , x ) = 0 \quad +k^2 A\cos{(\omega t-kx)}-k^2 B\sin{(\omega t-kx)}+\frac{m^2c^2}{\hbar^2} f(t,x)=0

( k 2 − ω 2 c 2 ) A cos ⁡ ( ω t − k x ) − ( k 2 − ω 2 c 2 ) B sin ⁡ ( ω t − k x ) + m 2 c 2 ℏ 2 f ( t , x ) = 0 \left(k^2-\frac{\omega^2}{c^2}\right)A\cos{(\omega t-kx)}-\left(k^2-\frac{\omega^2}{c^2}\right)B\sin{(\omega t-kx)}+\frac{m^2c^2}{\hbar^2} f(t,x)=0

( k 2 − ω 2 c 2 + m 2 c 2 ℏ 2 ) f ( t , x ) = 0 \left(k^2-\frac{\omega^2}{c^2}+\frac{m^2c^2}{\hbar^2}\right)f(t,x)=0

⇒ k 2 − ω 2 c 2 + m 2 c 2 ℏ 2 = 0 \Rightarrow k^2-\frac{\omega^2}{c^2}+\frac{m^2c^2}{\hbar^2}=0

⇒ ω 2 = c 2 k 2 + m 2 c 4 ℏ 2 \Rightarrow \boxed {\omega^2=c^2k^2+\frac{m^2c^4}{\hbar^2}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kevin Patel
May 23, 2014

It can be solved instantaneously with given data..

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