# How Light or Water Intensity is equal to square modulus of wave function of Light or Water Waves $I=|\psi|^2 \,$?

I've seen the Wave Function as a psi $\Psi$ $\psi$. And always heard that the wave function is the Complex Number as Imaginary and real number. But I've never seen it I've never seen components of wave function But I have seen it everywhere in the book of Quantum Mechanics.

i want to know how wave function could be intensity! in double slit experiment sometimes called Young's experiment,

that is relation: $$I=I_1+I_2$$ $$I=|\psi_1+\psi_2|^2=|\psi_1|^2+|\psi_2|^2+2Re(\psi_1 * \psi_2)$$ i want to know how Intensity could be square modulus of wave function $|\psi|^2$ $$I=|\psi|^2$$

$$I=I_1+I_2$$ $$I=I_1+I_2+2\sqrt {l_1 l_2} cos\delta$$

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The short answer is: the wave function is the amplitude, not the intensity. The square of the wave function is the intensity. Intensities don't add up, amplitudes do. So $I=I_1+I_2$ is in general wrong. –  Raskolnikov May 12 '12 at 8:16
I corrected your wiki link to young's experiment. There was a funny en.m.wikipedia link which seems nonsense in its first page. –  anna v May 12 '12 at 15:45
@anna v Thank You –  user8784 May 12 '12 at 16:21

I can't explain QM here. It takes a lot of reading and working things out for yourself. For this particular question, however, an analogy might help (this may be far below your level, in which case apologies). QM is very often about "simple harmonic oscillators" (SHOs), for which the oldest prototype is the pendulum (approximately, if the amplitude is small). For a pendulum, if we want to know how far it will go from the vertical, we can wait to see how far it goes on each cycle. An alternative way is to measure how fast the pendulum goes when it passes through the vertical. For any given speed, there is a corresponding farthest distance from the vertical. We can equate these two, in a notional sort of way, by choosing units just so, $s_0=d_1$, the speed at its maximum is the same as its farthest distance from vertical. [if you don't want to choose such helpful units, write $s_0=Kd_1$.]

Now, suppose that we measure the speed and the distance at some intermediate point, for which we obtain $s_t,d_t$. For a simple harmonic oscillator, and approximately for a pendulum if its oscillations are small, we obtain $\sqrt{s_t^2+d_t^2}=s_0=d_1$. The square root $\sqrt{s_t^2+d_t^2}$ is an invariant quantity of the coordinates $(s_0,0)$ and $(0,d_1)$, which in general are $(s_t,d_t)$. Anything that is a function of the square root $\sqrt{s_t^2+d_t^2}$ is also a function of $s_t^2+d_t^2$, so we can work with whichever is more convenient.

The effects of a given SHO on other systems ---or of a system that contains many SHOs on other systems--- are determined both the phases and by the amplitudes, but the amplitude often determines the more obvious properties, with differences of phase causing important but often more subtle effects, which we typically might call interference (but there are many other words, such as "caustics", or even, in a New Age sort of way, "sympathetic vibrations"!).

The effects of a given quantum mechanical system are, at an elementary mathematical level, sui generis with a classical SHO or system of SHOs, but quantum mechanics describes the ways that the probabilities of discrete events evolve over time, instead of describing the evolution of a trajectory. The introduction of probability as an essential property makes QM a discussion of a higher order mathematical object. Especially different is the fact that we can no longer talk about velocities, because individual events do not have velocities (if we are determined to talk in terms of particles we cannot in general be sure which individual events go with which particle), however it's useful to introduce a notional object that we call momentum, which allows us to model patterns that we observe in the evolution of the probabilities as interference effects (whether that is what they are not, we can model the patterns in the probabilities using patterns of varying phases and amplitudes). The mathematical quantity that we call momentum is, however, sufficiently different from the classical momentum that is associated with a particle trajectory that the analogy breaks down in various mathematically significant ways.

I can't see how to address the final aspect of this that occurs to me, for now, at least not well. The much touted linearity of quantum mechanics is a consequence of the fact that QM describes the evolution of probabilities of individual measurement events. The object we call momentum is closely related to the mathematics of Fourier transforms of probability distributions, which is essentially associated with a squared modulus like $s_t^2+d_t^2$. One consequence of that is noncommutativity of the algebra of observables.

This is a quick and very vague writing down of a lot of experience, without much editing, so take it with a pinch of salt and with a lot of other reading of what other people have to say about the hard questions that quantum mechanics poses for us. I hope you find it more useful than confusing, but hey, I can take a few downvotes, and it's been oddly useful to me to write this down in this somewhat wild way. In fact, if you can see the ways in which this answer is related to your question, you understand QM pretty well already.

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"an analogy might help (this may be far below your level, in which case apologies)." :D i think so. My problem actually is wave function, The origin of all surprises, Wave function for me is still undefined –  user8784 May 12 '12 at 13:48
OK, Bad Boy (and thanks for not downvoting my riff). If you want a definition, a wave function is a vector in a Hilbert space. To take a different from textbook approach to get a Hilbert space, introduce an algebra that models the properties of observables relative to each other and a map from that algebra to expected values (to model observed statistics), use the GNS construction, express expected values using the representation of the algebra on the resulting Hilbert space and (crucially, because it's bilinear), the inner product of the Hilbert space. –  Peter Morgan May 12 '12 at 14:46
$\psi (\vec r,t)=A\exp^{i\phi} + B\exp^{-i\phi}$ this is simple of wave function i heard that $A$ and $B$ are constants, When we say something is constant That does not change by the time, for me constant means speed of the light $c=300 m/s$ or planck's one. But I can not see the meaning of the constant for such wave function that even sometimes is time-independent . As long as i'm not familiar with the components of the wave function. i completely know that without knowing wave function i am not able to learn QM. –  user8784 May 12 '12 at 16:14