Undergrad Wave Functions of Definite Momentum

[Jimmy87]

No idea what happened there - sorry!

 

[PeterDonis]

I'm talking about the wave function value - the one you said I need to know the answer to - "the complex value of the wavefunction" which I still don't know yet.

Once again: the answer is right in front of you.

I totally get that the wavefunction is useful because you can apply energy, momentum, position operators etc on it to get really useful information.

And in order to do that, you need to know the (complex) value of the wave function for all possible values of $x$ and $p$. That is the answer to your question.

you said "the complex value of the wavefunction is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else." I still can't find an answer to this.

Because knowing the complex value of the wave function, as a function of its input variables, is necessary in order to do anything else--because without it, you can't apply any operators to get any predictions for measurements.

 

[Jimmy87]

Once again: the answer is right in front of you.
And in order to do that, you need to know the (complex) value of the wave function for all possible values of $x$ and $p$. That is the answer to your question.
Because knowing the complex value of the wave function, as a function of its input variables, is necessary in order to do anything else--because without it, you can't apply any operators to get any predictions for measurements.

Oh I see now (I think). So when you plot a graph of Ψ against x and Ψ against p, are the values plotted on the graph the complex value of the wave function by doing separate calculations for different values of x and p? Or at least the real part anyway.

 

[PeterDonis]

when you plot a graph of Ψ against x and Ψ against p, are the values plotted on the graph the complex value of the wave function by doing separate calculations for different values of x and p?

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.

 

[vanhees71]

I don't know, whether you have ever told the OP that a pure quantum state which, up to a phase, is given by a normalizable vector in Hilbert space, $|\psi \rangle$ can be given in position and momentum representation. The eigenstates of position and momentum are in fact not true Hilbert-space vectors but they live in a larger space, the dual of the dense subspace where the position and momentum operators are defined, which are thus only "normalizable to a $\delta$ distribution",
$$\langle p'|p \rangle=\delta(p-p'), \quad \langle x'|x \rangle=\delta(x-x').$$
Further you can show that the generalized momentum eigenstate is given in position representation by
$$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$
I've used "natural units" such that $\hbar=1$, which makes things a bit easier.

Now the most important key to change from one basis, or equivalently from one representation, to another is the completeness relation for generalized position and momentum eigenstates,
$$\int_{\mathbb{R}} \mathrm{d} p |p \rangle \langle p|=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=\mathbb{1}.$$
This implies for the position wave function
$$\psi(x) = \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p),$$
where $\tilde{\psi}(p)$ is the wave function in momentum representation (representing up to a phase factor still the same state as the position wave function!). As you see, you get the position wave function from the momentum wave function by Fourier integration. Of course, you also immediately get the inverse transformation by the same trick of "inserting a unit operator":
$$\tilde{\psi}(p)=\langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x).$$
Here I have used that the generalized position eigenfunction in momentum representation is given by
$$u_x(p) = \langle p|x \rangle = \langle x | p \rangle^*=u_p^*(x)=\frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x).$$

 

[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!