Wave Functions of Definite Momentum

[Jimmy87]

Basically, yes, but with some clarifications. If you plot $\Psi$ as a function of $x$, you are plotting the values you get when you evaluate the function $\Psi(x)$ for each different value of $x$. Which function $\Psi(x)$ that is will depend on what value you pick for $p$--or, more generally, what kind of wave function you want to work with (functions of the form $Ae^{ipx}$--leaving out the factor $\hbar$ for simplicity--are not the only possible kinds of wave functions).

Similarly, if you plot $\Psi$ as a function of $p$, you are plotting the values you get when you evaluate the function $\Psi(p)$ for each different value of $p$. Which function $\Psi(p)$ that is will depend on what kind of wave function you want to work with.

The relationship between the functions $\Psi(x)$ and $\Psi(p)$ is that they are Fourier transforms of each other. In the case we are working with, where the function $\Psi(x)$ is $A e^{ipx}$, the corresponding function $\Psi(p)$, obtained by Fourier transforming $\Psi(x)$, is $\delta(p)$, where $\delta$ stands for the Dirac delta function. (Which is not actually a "function", strictly speaking--but for our purposes here we can think of it as one, which is zero for any value of momentum except the single value $p$.)

So if you draw a graph of your $\Psi(x)$, for some particular value of $p$, you will find that it looks like a "corkscrew" winding around the $x$ axis (think of stacking an infinite series of complex planes, one for each value of $x$, and plotting the complex value of $\Psi(x)$ for each value of $x$ in the plane corresponding to that value of $x$). The particular value of $p$ that you pick determines how tightly the corkscrew winds--the higher the value of $p$, the tighter the windings (i.e., the more closely spaced they are).

And if you draw a graph of the $\Psi(p)$ that corresponds to the above $\Psi(x)$, you will find that it is a single "spike" at a particular value of $p$ on the $p$ axis--everywhere else it is zero.

Perfect - thank you! I understand that now. Thanks for all your help (and patience). One final question if you could - when you find the complex value of the wavefunction what position 'x' do you actually put it? Is it a number measured in metres for example? If so, how would you define its position? If you put in 1mm for example - you would have to put in some framework for what 1mm means wouldn't you? I have seen wavefunctions in terms of confined in a box but how do you relate the 'x' to where exactly your looking because 1mm could mean anything.

 

[jtbell]

I have seen wavefunctions in terms of confined in a box but how do you relate the 'x' to where exactly your looking because 1mm could mean anything.

In that case the wave function contains the length of the box L as a parameter. For example, the lowest-energy state of a particle in a box that extends from x = 0 to x = L is $$\psi(x) = \sqrt{\frac 2 L} \sin \left( \frac {\pi x} L \right)$$ You can use whatever units you like for x and L, so long as you use the same units for both. This means that $\psi$ itself has units corresponding to the length unit that you use for x, because its value depends on the number you use for L in the factor in front. This is OK, because when you use $\psi$, you have to use it in such a way that involves x again, and you have to use the same units there.

For example, if you want to calculate the probability that the particle lies between x = a and x = b, you evaluate the integral $$P = \int_a^b |\psi(x)|^2 \, dx$$ The units you use for a and b have to match the units that you use for x and L. The value of the probability P doesn't depend on the units, so long as a and b mark off the same "fraction" of the box in whatever units you're using.

In fact, in a calculation like this one, it should be possible to phrase it in such a way as to avoid using specific units for x, L, a and b. In calculating P above, for example, we can specify the limits as fractions of L, e.g. a = 0.25L and b = 0.9L.

 

[PeterDonis]

when you find the complex value of the wavefunction what position 'x' do you actually put it? Is it a number measured in metres for example?

You can choose any units you want for $x$, as long as the rest of the wave function's units are chosen to correspond--for example, if you measure $x$ in SI units of meters, you would need to use SI units for other things like momentum $p$ and $\hbar$ as well.

 

[Tio Barnabe]

@vanhees71 , why have you integrated the quantities over all $\mathbb{R}$ in post #35?

 

[vanhees71]

Because I thought you are talking about particles in free space.

 

[jtbell]

Some posts do talk about the "particle in a box", where the particle is confined to a certain region of space. However, this doesn't prevent us from writing the integrals over all space as vanhees71 did. The regions where $\psi = 0$ simply don't contribute any value to the integral.

 

[vanhees71]

Well, a "particle in a box" in the usual sense (rigid boundary conditions) forbid the definition of a momentum observable. You can invoke periodic boundary conditions, where this is possible, and this is often needed in some calculations, particularly in many-body QFT. This is a perfect example for a pretty useless conversation due to the fact that the problem to be answered is not specifically defined!