The range of (psi) squared is ?

[peeyush_ali]

the range of (psi) squared is ??

(psi) of the schrodinger's wave equation is displacement? ( or, what are it's SI units??)

(psi) squared gives the "probability" of finding a particle in space..does this mean that,

0<= ((psi) sqaured) <=1 ??

 

[henry_m]

|\psi|^2 is usually interpreted as a probability density. That is to say, not the probability of finding your particle at a point but the probability per volume of finding it in a small box at the point (small enough that psi is roughly constant for the whole box).

The probability of finding the particle in a finite (or infinite!) region R would be

\int_R \mathrm{d}^3 x |\psi(x)|^2

In particular, this must be 1 is R is all of space. This is the condition that psi is normalised.

This must mean that |\psi|^2 has units (length)^(-3), so the wavefunction itself has units (length)^(-3/2). Psi can take any complex value at all, and |\psi|^2 can take any nonnegative value, subject to the normalisation constraint.

 

[Matterwave]

The units of psi are funny in that they change depending on the dimension you are working in. Charge density, for example, has a surface density, or line density, and similarly for the wave-function...but they are all called psi.

 

[peeyush_ali]

|\psi|^2 is usually interpreted as a probability density. That is to say, not the probability of finding your particle at a point but the probability per volume of finding it in a small box at the point (small enough that psi is roughly constant for the whole box).

The probability of finding the particle in a finite (or infinite!) region R would be

\int_R \mathrm{d}^3 x |\psi(x)|^2

In particular, this must be 1 is R is all of space. This is the condition that psi is normalised.

This must mean that |\psi|^2 has units (length)^(-3), so the wavefunction itself has units (length)^(-3/2). Psi can take any complex value at all, and |\psi|^2 can take any nonnegative value, subject to the normalisation constraint.

(


hmm...(probabilty/volume) i wonder how this quantity can take a "non real" value..! how do u think we need to interpret and understand this?? anyone?? please! much appreciated!

 

[peeyush_ali]

The units of psi are funny in that they change depending on the dimension you are working in. Charge density, for example, has a surface density, or line density, and similarly for the wave-function...but they are all called psi.

umm..dimension?? the one that schrodinger's wave equation has to tell is about charge density?! what i feel is that, it is a quantity which we use to describe discrete yet continuous therefore "probabilistic" experience of nature to define THE POSITION (charge density! :?)
of a particular free particle..

 

[henry_m]

(
hmm...(probabilty/volume) i wonder how this quantity can take a "non real" value..! how do u think we need to interpret and understand this?? anyone?? please! much appreciated!

It can't; the probability density is the modulus squared of the wavefunction, a nonnegative real number, exactly as we need for any kind of probabalistic interpretation. The other bit of data in the wavefunction is the phase. The way the phase changes through space encodes data about the momentum of the particle.

If the ideas of modulus and phase aren't familiar, check out http://en.wikipedia.org/wiki/Complex_number#Polar_form" on wikipedia, particularly the bit on 'polar form'.

 

[jtbell]

hmm...(probabilty/volume) i wonder how this quantity can take a "non real" value..!

It doesn't. \psi is complex, in general, but |\psi^2| = \psi^*\psi[/itex] is always real. (A complex number times its conjugate is always real.)

 

[JDoolin]

(


hmm...(probabilty/volume) i wonder how this quantity can take a "non real" value..! how do u think we need to interpret and understand this?? anyone?? please! much appreciated!

Actually, the unit for |psi^2| is real, and represents a probability. The units for psi is complex, and you have to multiply psi by its complex conjugate to get the real value.

This might not be exactly right, but:

psi = e^(i omega t)
= cos(omega t) + i sin(omega t)

This has a wavelike real and an imaginary part, but the magnitude of the wave, you can get by taking it times its complex conjugate

psi* = cos(omega t) - i sin(omega t)

When you multiply (psi)(psi*) you get cos^2(omega t) + sin^2(omega t) which is 1.

As far as a density, though, you can have something like

kg/m for a linear density
kg/m^2 for an area density or
kg/m^3 for a volume density.

When you're just talking about one particle though, you can't exactly talk about "particles per meter^3)" You're not talking about a bunch of particles. You're just talking about one particle. So you discuss in terms of probability per meter, per square meter, or per cubic meter.