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this is wave function: $$\Psi{(\vec r, t)}=\Psi_0 e^{i(\vec k \cdot \vec r-\omega t)}$$ $$\Psi{(\vec r, t)}=A e^{i(\phi + \vec k \cdot \vec r-\omega t)}$$.

  1. what is phase angle $\phi$ of wave function?

  2. is there any graph that shows what is phase angle of waves?

  3. how measure phase angle of light (double slit experiment)?

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if some one is interested can improve this question – user8784 May 31 '12 at 13:12
I'm not sure if this is homework but it certainly sounds like it might be so I added the homework tag. If I'm wrong, feel free to remove it. – Colin K May 31 '12 at 13:30
I'd guess this question is an extension of so it isn't homework. – John Rennie May 31 '12 at 14:22
@Colin K that is not homework but if you thought that is it, don't worry because homework tag is just a kind of general tag and doesn't make any sense – user8784 May 31 '12 at 14:23
Woah, somebody is a little cranky today. – Colin K May 31 '12 at 15:12

1 Answer 1

up vote 1 down vote accepted

To address your questions 1 and 2: this graph shows the real part of $\Psi{(\vec r, t)}=A e^{i(\vec k \cdot \vec r-\omega t)} $ in blue and the real part of $\Psi{(\vec r, t)}=A e^{i(\phi + \vec k \cdot \vec r-\omega t)} $ in purple. Obviously $\Psi$ is a function of two variables, so you can regard the graph either as keeping $\vec r$ constant and varying $t$ or keeping $t$ constant and varying $\vec r$.


The quantity $\phi$ is just the phase difference between the two waves e.g. the distance between the peaks shown by the arrow on the diagram.

The absolute value of $\phi$ has no physical significance because you can measure $\phi$ from any reference point you want. However the difference in $\phi$ between two wavefunctions has a very important physical meaning because it determines how the waves will interfere.

To address your question 3: actually the mention of the double slit experiment is spot on. The slits split the incoming light (or electrons or whatever) into two sources, call these $\Psi_a$ and $\Psi_b$, and if you take some point on the screen, this point will receive light from $\Psi_a$ and from $\Psi_b$, but the phase of the two waves, $\phi_a$ and $\phi_b$ won't be the same.

There isn't any physical meaning to the absolute phase of $\Psi_a$ and $\Psi_b$, $\phi_a$ and $\phi_b$, but if $\phi_a - \phi_b$ is an even multiple of $\pi$ ($2\pi$, $4\pi$, etc) the waves will be in sync and you'll get constructive interference and a bright area. If the phase difference is an odd multiple of $\pi$ ($\pi$, $3\pi$ etc) the waves will interfere destructively and you get a dark spot. This is exactly why you get the pattern of alternating bright and dark bands in the two slit experiment - it's because the phase difference, $\phi_a - \phi_b$, varies as you move along the screen.

So no experiment can measure the absolute value of $\phi_a$ or $\phi_b$, because the absolute value has no physical significance. However the double slit experiment can measure the phase difference $\phi_a - \phi_b$.

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phase angle $\phi$ has no physical significance wave Amplitude $A$ has no physical significance too! is this really physics!? i means it sucks me do you have another graph for wave wave Amplitude (that i really don't know why its constant!)? – user8784 May 31 '12 at 14:39
Remember it's the modulus of the wave function, $\Psi^*\Psi$, that has a physical significance. So it's true to say that the value of $A$ has no physical significance, but $A^2$ does! Also it's only the absolute value of the phase $\phi$ that has no significance. The phase difference, i.e. the differencve in $\phi$ for two or more waves, is exceedingly important and is the original of all interference phenomena. – John Rennie May 31 '12 at 14:45
ok what the hell, i just need to make sure that wave Amplitude is constant. do you know why unit of the modulus of the wave function $|\Psi|^2=A^2$ is $m^{-3}$? – user8784 May 31 '12 at 14:57
The modulus squared of the wave function gives the probability density i.e. the probability per unit volume. "Probability" is just a number and has no units, so the units of the modulus squared of the wave function are inverse volume. – John Rennie May 31 '12 at 15:06
The wavefuntion you are looking at, $A e^{i(\phi + \vec k \cdot \vec r-\omega t)}$ is an infinite plane wave. If you evaluate $\Psi^*\Psi$ for this wavefunction you'll find it is independant of $\vec r$ and $t$ i.e. it is constant everywhere. That's why the probability is constant everywhere. Different wavefunctions, e.g. the harmonic oscillator, give probabilities that are not constant everywhere. – John Rennie May 31 '12 at 15:23

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