Wavefunctions and the de Broglie wavelength

[Owen Loh]

The position wavefunction makes a spatial probability amplitude wave right?
And it is the combination of different frequencies
My question is that if these frequencies are the spatial frequencies in the debrogile relations

 

[jfizzix]

The short answer, is yes.

The position wavefunction is expressible as a sum (an integral, really) over spatial frequencies, where each spatial frequency times \hbar gives a value of momentum, as in the deBroglie relation.

So, just as the wavefunction gives a spectrum of possible position values, it also gives a spectrum of possible momentum values as well.

 

[Owen Loh]

Thank you very very much I am trying to explain HUP liddat

 

[Owen Loh]

The short answer, is yes.

The position wavefunction is expressible as a sum (an integral, really) over spatial frequencies, where each spatial frequency times \hbar gives a value of momentum, as in the deBroglie relation.

So, just as the wavefunction gives a spectrum of possible position values, it also gives a spectrum of possible momentum values as well.

Btw is there any way to explain that spatial freq iz related to momentum without math

 

[jfizzix]

Btw is there any way to explain that spatial freq iz related to momentum without math

Not that I can think of, but for a more conceptual understanding, you can go to DeBroglie's original argument for why the momentum is equal to \hbar times the spatial frequency (or just h divided by the wavelength).

So, in the early days of quantum physics, physicists had good experimental reasons to believe that the energy of a particle such as a photon was equal to h times its frequency (in time) f. The math gets a little technical, but using this fact, together with special relativity, you can prove that if E=hf then p=h/\lambda.

The idea comes from the relativity of simultaneity, which says that two clocks in sync when at rest will appear to be out of sync when in motion. The farther away from each other they are, the more out of sync they will be to us. When this idea is applied to the wavefunction of a moving particle, we find that over a large enough distance, two different points of the wave will be so out of sync that they will be one period apart. This separation distance will be the wavelength we see in our reference frame. The faster the particle moves, the stronger this relativistic effect becomes, and a smaller distance is enough to get one full cycle out of sync. The faster the particle moves, the shorter its wavelength.

 

[Owen Loh]

1. The idea comes from the relativity of simultaneity, which says that two clocks in sync when at rest will appear to be out of sync when in motion.

( but if they are in the same linear motion shouldn't they seem like they are at rest, or are they moving at diff directions)

2. The farther away from each other they are, the more out of sync they will be to us.

(Is this related to the motion thing,
Or just at relatively at rest and it appears out of sync because light takes time to travel)

3. When this idea is applied to the wavefunction of a moving particle, we find that over a large enough distance, two different points of the wave will be so out of sync that they will be one period apart.

(I thought the wavefunction isn't real, and we can't just allow one point of the wave to 'observe' another point of the wave.
They only appear out of sync for observers.I juz din kno you could observe a wavefunction)

4.This separation distance will be the wavelength we see in our reference frame.

(Umm...why? And separation distance of what?)

5.The faster the particle moves, the stronger this relativistic effect becomes, and a smaller distance is enough to get one full cycle out of sync. The faster the particle moves, the shorter its wavelength.

(Ill probably understand this after i understand the ones on top)

Sorry if i asked too many questions, i just want to get this right.

 

[jfizzix]

1. The idea comes from the relativity of simultaneity, which says that two clocks in sync when at rest will appear to be out of sync when in motion.

( but if they are in the same linear motion shouldn't they seem like they are at rest, or are they moving at diff directions)

2. The farther away from each other they are, the more out of sync they will be to us.

(Is this related to the motion thing,
Or just at relatively at rest and it appears out of sync because light takes time to travel)

They are in the same linear motion. They are at rest with respect to each other, but moving with the same velocity in the same direction with respect to us. With respect to each other, their clocks are in sync. With respect to us, they are out of sync by an amount that grows with the distance between them and how fast they are moving relative to us. This is not an optical illusion or due to how long it takes light to meet our eyes (unusual, but true).

This is a good video on the subject:

3. When this idea is applied to the wavefunction of a moving particle, we find that over a large enough distance, two different points of the wave will be so out of sync that they will be one period apart.

(I thought the wavefunction isn't real, and we can't just allow one point of the wave to 'observe' another point of the wave.
They only appear out of sync for observers.I juz din kno you could observe a wavefunction)

Whether or not the wavefunction is "real" is a hard philosophical question that I don't know the answer to. People can measure it (see, for example https://arxiv.org/abs/1112.3575), which is usually real enough for me,but others only think of it as a mathematical object. For the sake of education, we can imagine it as a real thing existing in space, and oscillating with frequency f.

4.This separation distance will be the wavelength we see in our reference frame.

(Umm...why? And separation distance of what?)

So at rest, the whole wave oscillates together up and down, but when moving, the front of the wave will be out of sync with the back of the wave because of the relativity of simultaneity. If you could take a freeze-frame of this moving wave, you would see that because the time lag changes continuously from the front to the back of the wave, the wave would look like it is oscillating over its length instead of just in time. The distance over which we would see one full cycle of oscillation in this freeze frame is the wavelength of this wave.

 

[Owen Loh]

Ok i understand the relativity part and the wavelength geta shorter the faster you moveTo the point on the wave themselves, do they see the wave as a straight line ossicilating up and down through probability amplitude as time evolution occurs? Or do they see some other shape of wave.

But if we move with the same velocity as the particle and be at the middle of the wavefunction,wouldnt we see like only half a cycle.( a valley or mountain)

And does it mean that the wavefunction changes as we change our refrence frame.

 

[jfizzix]

To the point on the wave themselves, do they see the wave as a straight line ossicilating up and down through probability amplitude as time evolution occurs? Or do they see some other shape of wave.

That is a very interesting question!
Because the wavefunction gives a spectrum of position values, it also gives a spectrum of momentum values, and therefore velocity values. By going at the same average velocity as the particle, you would be in as close to the wave's own perspective as is physically possible, but my current understanding of quantum physics doesn't allow me to do better than that.

If we are moving at the same velocity as the average velocity of the particle, then its wavefunction would look like that of a particle at rest (on average). I don't have a particularly good description of this, other than the probability amplitude looking like a cloud centered around a point. More exact descriptions will vary depending on the circumstances.

Trying to visualize the oscillations of the particle on its own is a difficult task, particularly because any measurement we do only tells us about the probability rather than the probability amplitude. It is possible for the probability amplitude to oscillate while the probability itself stays constant.

That said, if we were to perform an experiment that tested something that depended on the wavelength of the particle (like how wide it diffracts after passing through a thin slit), you would be able to see that the faster the particle is moving relative to us, the shorter its wavelength becomes.

And does it mean that the wavefunction changes as we change our reference frame.

The wavefunction does change with our reference frame, yes.
The average values of the properties of a particle (position, momentum, energy, etc) mostly obey Newton's laws (see https://en.wikipedia.org/wiki/Ehrenfest_theorem for a better description), and they certainty change depending on our reference frame.

 

[Owen Loh]

That is a very interesting question!
Because the wavefunction gives a spectrum of position values, it also gives a spectrum of momentum values, and therefore velocity values. By going at the same average velocity as the particle, you would be in as close to the wave's own perspective as is physically possible, but my current understanding of quantum physics doesn't allow me to do better than that.

If we are moving at the same velocity as the average velocity of the particle, then its wavefunction would look like that of a particle at rest (on average). I don't have a particularly good description of this, other than the probability amplitude looking like a cloud centered around a point. More exact descriptions will vary depending on the circumstances.

Trying to visualize the oscillations of the particle on its own is a difficult task, particularly because any measurement we do only tells us about the probability rather than the probability amplitude. It is possible for the probability amplitude to oscillate while the probability itself stays constant.

That said, if we were to perform an experiment that tested something that depended on the wavelength of the particle (like how wide it diffracts after passing through a thin slit), you would be able to see that the faster the particle is moving relative to us, the shorter its wavelength becomes.
The wavefunction does change with our reference frame, yes.
The average values of the properties of a particle (position, momentum, energy, etc) mostly obey Newton's laws (see https://en.wikipedia.org/wiki/Ehrenfest_theorem for a better description), and they certainty change depending on our reference frame.

Hey dude thanks a lot ypuve been reaally really helpful

 

[Owen Loh]

by the way, i realized that the

YOU SAID:
The idea comes from the relativity of simultaneity, which says that two clocks in sync when at rest will appear to be out of sync when in motion. The farther away from each other they are, the more out of sync they will be to us. When this idea is applied to the wavefunction of a moving particle, we find that over a large enough distance, two different points of the wave will be so out of sync that they will be one period apart. This separation distance will be the wavelength we see in our reference frame. The faster the particle moves, the stronger this relativistic effect becomes, and a smaller distance is enough to get one full cycle out of sync. The faster the particle moves, the shorter its wavelength.

is just basically length contraction...right?
check out: