Trouble understanding matter as waves

[khkwang]

For simplicity sake, I'm assuming an electron is traveling in just one direction (x).

From what I understand the square of the electron's wave function is proportional to the probability of finding it at x.

But there's a point in every period that the amplitude is zero, which means the probability of finding it there is zero as well.

So doesn't that mean that the probability of an electron existing over a range of x grows and shrinks back to zero periodically?

example for probability p(x):

p(<0)=0, p(0)=0, p(1)=1, p(2)=0, p(3)=4, p(4)=0, p(5)=9, p(6)=0, p(>6)=0

Am I interpreting this properly?

 

[Plaster]

To me you seem to be describing waves mathematically "in one direction". Waves don't travel in one direction. They 'wave'.

 

[khkwang]

That's the way my textbook is teaching me. I know it's simplifying things so that it'll be easier to understand, but I'm having trouble understanding the consequences displayed in the simplification.

 

[Plaster]

My advice would be to look at youngs slit experiment, and then look at it again, often.

Don't think that people understand it, and you are alone in finding it strange. No one understands it.

 

[epenguin]

In the introductory part of QM they seem to play formula games with the wave idea, just accept whatever results they get, especially those which can relate to experimental results.

Yes a wave does have amplitude 0 at some points. For instance in the diffraction of electrons you will see there are places where the amplitude is 0.

There are these uncomfortable feelings until you come to a situation required for most concrete situations where the electron is confined by a force. As e.g. in the infinitely high-walled square potential well. At the boundaries of the well the density is 0 because there is an infinite repulsive force on the electron there. But between the walls the waves of an electron of more than the minimum energy do have nodes where its density is 0 as you will see in textbooks. But even in a more realistic case without infinities too, e.g. in the (excited) hydrogen atom in the most symmetrical state the proton is surrounded by some spherical nodes where density is 0, and there are lots of other states with nodes of which you can see pictures in textbooks. (OK these are stationary waves.) At least when confined these waves begin to look like matter even if slightly bizarre.

 

[jtbell]

From what I understand the square of the electron's wave function is proportional to the probability of finding it at x.

It's not the simple square $\Psi^2$, but rather the "complex square" $\Psi^*\Psi$, where $\Psi^*$ is the complex conjugate of $\Psi$ (replace i with -i everywhere).

But there's a point in every period that the amplitude is zero, which means the probability of finding it there is zero as well.

The idealized plane-wave wave function is $\Psi = Ae^{i(kx-\omega t)}$. Therefore

[itex]\Psi^* \Psi = A^* e^{-i(kx-\omega t)} Ae^{i(kx-\omega t)} = A^* A[/tex]<br /> <br /> which is constant.[/itex]

 

[khkwang]

There are these uncomfortable feelings until [snip]

Thanks for the reply, but I get lost half way through your post... I'll re-read it when I get further into my textbook.

Edit: actually, I'm going to try to interpret what you said. So are you basically telling me, "yeah some points will have zero probability. This has been shown experimentally"? As in my original interpretation was correct? Or are you saying something else?

It's not the simple square $\Psi^2$, but rather the "complex square" $\Psi^*\Psi$, where $\Psi^*$ is the complex conjugate of $\Psi$ (replace i with -i everywhere).
The idealized plane-wave wave function is $\Psi = Ae^{i(kx-\omega t)}$. Therefore

[itex]\Psi^* \Psi = A^* e^{-i(kx-\omega t)} Ae^{i(kx-\omega t)} = A^* A[/tex]<br /> <br /> which is constant.[/itex]

$<br /> But wait, if the probability is constant A, then doesn't the value of x not matter? Then wouldn't the probability be the same everywhere?$

 

[jtbell]

In this example, yes. But this is an idealized example, a wave with a single definite wavelength, corresponding to a single definite momentum, which is impossible according to the Heisenberg Uncertainty Principle.

To get a more realistic wave function, you have to add together a lot of waves with different wavelengths (and frequencies) to form a wave packet. The probablility is maximum at some position (the "center" of the packet), and falls off smoothly towards zero (without "oscillating") as you go further and further away on either side.

 

[khkwang]

In this example, yes. But this is an idealized example, a wave with a single definite wavelength, corresponding to a single definite momentum, which is impossible according to the Heisenberg Uncertainty Principle.

To get a more realistic wave function, you have to add together a lot of waves with different wavelengths (and frequencies) to form a wave packet. The probablility is maximum at some position (the "center" of the packet), and falls off smoothly towards zero (without "oscillating") as you go further and further away on either side.

Oh yeah, but I'm talking about a wave packet here.

But even wave packets have points of value zero. At least how my textbook displays it.

[PLAIN]http://vedantic-physics.com/web_images/wave_packet.jpg

How I'm interpreting this picture is that this wave squared is the probability of finding an electron at position x on the axis. Looking at the second wave packet in the picture, there are 15 points on the x-axis in which there is ZERO probability of finding it, not including the end points of the packet. Once again I'm assuming a 1 dimensional system here.

By labeling the points with max/min amplitude, 1 through 16, it would seem that points 8 and 9 have the highest probability, while all points between the max/mins have zero probability (i.e. 1.5, 2.5, 3.5, etc.).

 

[jtbell]

You're still thinking of real wave functions, not complex ones. When you use the complex versions and find $\Psi^*\Psi$, the "wiggles" disappear. The probability distribution ends up looking like this:

 

[khkwang]

ahh ok, I see.

I don'e know the reasoning behind it yet, but at least I know there's an explanation up ahead.

Thanks.

 

[jtbell]

The wave packets that you illustrated have real wave functions which might look like this:

$$\Psi = A(x,t) cos (kx - \omega t)$$

A(x,t) gives the shape of the "envelope" that encloses the oscillatory wiggles. The "wiggles" come from the cos() part.

The actual complex wave function might look like this:

$$\Psi = A(x,t) e^{i(kx - \omega t)} = A(x,t) [cos (kx - \omega t) + i sin (kx - \omega t)]$$

The imaginary part makes the difference. For the first one, the probability distribution is

$$\Psi^*\Psi = [A(x,t)^2] cos^2 (kx - \omega t)$$

which is zero at a number of points inside the packet, as you noticed. But for the second one, the probability distribution is $\Psi^*\Psi = [A(x,t)]^*A(x,t)$ which doesn't have the zeroes in the middle, although it does tail off to zero at large distances on either side.

 

[zonde]

The wave packets that you illustrated have real wave functions which might look like this:

$$\Psi = A(x,t) cos (kx - \omega t)$$

A(x,t) gives the shape of the "envelope" that encloses the oscillatory wiggles. The "wiggles" come from the cos() part.

The actual complex wave function might look like this:

$$\Psi = A(x,t) e^{i(kx - \omega t)} = A(x,t) [cos (kx - \omega t) + i sin (kx - \omega t)]$$

The imaginary part makes the difference. For the first one, the probability distribution is

$$\Psi^*\Psi = [A(x,t)^2] cos^2 (kx - \omega t)$$

which is zero at a number of points inside the packet, as you noticed. But for the second one, the probability distribution is $\Psi^*\Psi = [A(x,t)]^*A(x,t)$ which doesn't have the zeroes in the middle, although it does tail off to zero at large distances on either side.

This is a bit puzzling.
Its clear that $$\Psi$$ in the first case differs from $$\Psi$$ in the second case.
But in first case for oscillation to exist there should be some conserved quantity. So let's say that for this wavefunction:

$$\Psi = A(x,t) cos (kx - \omega t)$$

we define complimentary wavefunction such that:

$$\Psi' = A(x,t) sin (kx - \omega t)$$

and now we find conserved quantity by adding squares of both wavefunctions:

$$C=\Psi^{2}+\Psi'^{2}=A(x,t)^{2}$$

And we get the same result (if A(x,t) is not complex) as in the second case. This C that is constant seems very much like energy of oscillator.
So does it mean that we can draw parallels between energy of classical wavefunction and measured particle of QM wavefunction?