What is a wave function and what are the requirements for it to be well-behaved, i.e. for it to properly represent physical reality?

In quantum mechanics, we talk about the state of a system. One of the simplest examples is a particle that can be in an up or down spin, for instance an electron. When we measure the spin of a system, we either measure it to be up or down. A state by which we are certain of the outcome of the measurement, we call an eigenstate (one up state #uarr# and one down state #darr#).

There are also states where we are uncertain of the outcome of the measurement before we measure it. These states we call a superposition and we can write them down as #a*uarr+b*darr#. Here we have #|a|^2# the probability of measuring #uarr#, and #|b|^2# the probability of measuring #darr#. This means of course that #|a|^2+|b|^2=1#. We allow #a,b# to be complex numbers, the reason for this is not immediately clear from this example, but in the context of the wavefunction it will be more clear. The bottom line is that there are more states than one giving the same probabilities for measuring the spins.

Now we could try to assign a function to this spin state. Since there are only two outcomes of the measurement of the spin, we have a function that has only two possible inputs. If we call the function #psi# (this is a very conventional symbol used for a wavefuntion), we set #psi(uarr)=a# and #psi(darr)=b#.

Now we turn to the wavefunction. One aspect of a particle is of course its location. Just like in the case of spin, we can measure distinct values for the location, and we can have states in which the outcome of the measurement is not fixed beforehand. Since we have an uncountable infinity amount of locations where a particle can be, writing down this state as #a*"here"+b*"there"# won't do. However, the idea of the function that we have used above does. So for any location #x#, we have a complex value #psi(x)#. The probability density function of the particle is now given by #|psi(x)|^2#.

In all fairness, historically the idea of the wavefunction is older than that of the spin, but I think understanding the idea of spin to a certain degree helps in the understanding of the wavefunction.

Now first of all, why is the wavefunction complex valued? The first reason can be found in the idea of interference. The wavefunction of a particle can interfere with itself. This interference has to do with adding up wavefunctions, if the wavefunctions give the same absolute value at a certain point, then the probability of measuring a particle around that point is similar. However the function values can be different, if they are the same, adding them up will make the amplitude, or probability density 4 (#|2|^2#) times bigger (constructive interference), and if they differ by a sign they negate each other (destructive interference). However the can also differ by for instance a factor #i#, meaning that the probability density becomes #2# times bigger at that point. We know that all of these interferences can occur. So this points towards a complex valued wavefunction as described earlier.

The second reason can be found in the Schrödinger equation. Initially it was thought that these wavefunctions behaved just like classical waves. However, when Schrödinger tried to describe the behaviour of these waves, or at least their evolution through time, he found that the equation governing classical waves was not adequate. In order for it to work, he had to introduce a complex number into the equation, leading to the conclusion that the function itself has to be complex as well, and the order of the derivatives appearing in the equation differs from the classical wave equation.

This difference in the equations also answers your second question. Since the evolution of the wavefunction differs so much from that of classical waves, we cannot use the same methods we use in classical wave physics. There are of course geometrical arguments you can use, but it will not be enough to describe all of the phenomena in quantum physics. Besides, even though the wavefunction gives a lot of information about the state of a particle, it tells you nothing about its spin, since the observables spin and location have little to do with eachother.

Perhaps I'm interpreting what you mean by a geometric nature wrongly. Could you perhaps give an example of what you mean. Perhaps then I could help you further.

The wave function represents the state of a quantum mechanical system such as an atom or a molecule.

It can be represented as either #psi#, the time-independent wave function, or #Psi#, the time-dependent wave function.

Because the wave function evidently represents a system that behaves like a wave (it's no coincidence that it's called the wave function!), we would normally expect an unrestricted wave function to have no boundaries. Consider the fact that #sinx# and #cosx#, two functions that are clearly waves, have domains of #(-oo,oo)#.

EXAMPLE: THE WAVE FUNCTION FOR ORBITALS

However, let's take orbitals for example. There must be a set of boundary conditions for an orbital, because obviously orbitals aren't infinitely large.

A wave function can depict the linear combination of atomic orbitals to form molecular orbitals:

#color(blue)(psi_("MO")) = sum_(i) c_iphi_i^"AO"#

#= color(blue)(c_1phi_(1s) + c_2phi_(2s) + c_3phi_(2px) + c_4phi_(2py) + c_5phi_(2pz) + . . . )#

where #c_i# is the expansion coefficient indicating the contribution of each atomic orbital to the particular molecular orbital in question, and #phi_i^"AO"# is the experimental/trial wave function for each atomic orbital.

Since a wave function must be able to represent an orbital, it must have a positive radius (#r > 0#) and the wave function must be single-valued, closed, continuous, orthogonal to all related wave functions, and normalizable.

In other words, it must pass the vertical line test, have a finite area under the curve, have no jumps/discontinuities/asymptotes/breaks, and satisfy the following two equations:

#int_"allspace" psi_A^"*"psi_Bd tau = 0#
(the integral of a wave function and its complex conjugate is #0# if the wave functions are different)

#int_"allspace" psi_A^"*"psi_Ad tau = 1#
(the integral of a wave function and its complex conjugate is normalized such that it equals #1# if the wave functions are the same besides the sign of #pmi#)

One example equation for the wave function in spherical coordinates for the hydrogen atom is:

#color(blue)(psi_(2pz)(r,theta,phi)) = R_(21)(r)Y_(1)^(0)(theta,phi)#

#= color(blue)(1/(sqrt(32pi))(Z/(a_0))^("3/2")((Zr)/(a_0))e^(-Zr//2a_0)costheta)#

To think, I actually spent time to normalize this. I even took the time to check for orthogonality with the other two #2p# wave functions. :P

Just in case, here is an appendix of what I have linked above in Scratchpads.

#" "##" "##" "#APPENDIX

Normalization of the #2p_z# wave function

The #2p_z# atomic orbital wavefunction is:

#psi_(2pz)#

#= R_(nl)(r)Y_(l)^(m)(theta,phi) = R_(21)(r)Y_(1)^(0)(theta,phi)#

#= 1/sqrt(32pi) (Z/(a_0))^(3/2) (Zr)/(a_0)e^(-(Zr)/(2a_0))costheta#
(McQuarrie)

Is the #2p_z# wavefunction really normalized? LET'S FIND OUT!

#\mathbf(int_(0)^(oo) R_(nl)^"*"(r)R_(nl)(r)r^2dr int_(0)^(pi) Y_(l)^(m)(theta,phi) sintheta int_(0)^(2pi) dphi stackrel(?)(=) 1)#

#[1/sqrt(32pi) (Z/(a_0))^(5/2)]^2 int_(0)^(oo) e^(-(Zr)/(a_0)) r^4dr int_(0)^(pi) sinthetacos^2thetad theta int_(0)^(2pi) dphi stackrel(?)(=) 1#

#color(green)(1/(32pi) (Z/a_0)^5 int_(0)^(oo) e^(-(Zr)/(a_0)) r^4dr stackrel(= "2/3")(overbrace(int_(0)^(pi) sinthetacos^2thetad theta)) stackrel(= 2pi)(overbrace(int_(0)^(2pi) dphi)) stackrel(?)(=) 1)#

Now, examining only the radial part, which is the crazy part... let the quadruple Integration by Parts commence!

EVALUATION OF THE RADIAL COMPONENT OF THE WAVE FUNCTION

Part 1

#int_(0)^(oo) e^(-(Zr)/(a_0)) r^4dr#

Let:
#u = r^4#
#dv = e^(-(Zr)/(a_0))dr#
#v = -(a_0)/Ze^(-(Zr)/(a_0))#
#du = 4r^3dr#

#= -(a_0)/Ze^(-(Zr)/(a_0))r^4 - int -(a_0)/Ze^(-(Zr)/(a_0))4r^3dr#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 - 4int e^(-(Zr)/(a_0))r^3dr}#

Part 2

Let:
#u = r^3#
#dv = e^(-(Zr)/(a_0))dr#
#v = -(a_0)/Ze^(-(Zr)/(a_0))#
#du = 3r^2dr#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 - 4[-(a_0)/Ze^(-(Zr)/(a_0))r^3 - 3int -(a_0)/Ze^(-(Zr)/(a_0))r^2dr ]}#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 + (4a_0)/Z[e^(-(Zr)/(a_0))r^3 - 3int e^(-(Zr)/(a_0))r^2dr ]}#

Part 3

Let:
#u = r^2#
#dv = e^(-(Zr)/(a_0))dr#
#v = -(a_0)/Ze^(-(Zr)/(a_0))#
#du = 2rdr#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 + (4a_0)/Z[e^(-(Zr)/(a_0))r^3 - 3[-(a_0)/Ze^(-(Zr)/(a_0))r^2 - 2int -(a_0)/Ze^(-(Zr)/(a_0))rdr]]}#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 + (4a_0)/Z[e^(-(Zr)/(a_0))r^3 + (3a_0)/Z[e^(-(Zr)/(a_0))r^2 - 2int e^(-(Zr)/(a_0))rdr]]}#

Part 4

Let:
#u = r#
#dv = e^(-(Zr)/(a_0))dr#
#v = -(a_0)/Ze^(-(Zr)/(a_0))#
#du = dr#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 + (4a_0)/Z[e^(-(Zr)/(a_0))r^3 + (3a_0)/Z[e^(-(Zr)/(a_0))r^2 - 2{-(a_0)/Ze^(-(Zr)/(a_0))r - int -(a_0)/Ze^(-(Zr)/(a_0))dr}]]}#

#= -(a_0)/Z {e^(-(Zr)/(a_0))r^4 + (4a_0)/Z[e^(-(Zr)/(a_0))r^3 + (3a_0)/Z[e^(-(Zr)/(a_0))r^2 + (2a_0)/Z{e^(-(Zr)/(a_0))r - int e^(-(Zr)/(a_0))dr}]]}#

EXPANSION/SIMPLIFICATION

#= -(a_0)/Ze^(-(Zr)/(a_0))r^4 - 4((a_0)/Z)^2[e^(-(Zr)/(a_0))r^3 + (3a_0)/Z[e^(-(Zr)/(a_0))r^2 + (2a_0)/Z{e^(-(Zr)/(a_0))r + (a_0)/Ze^(-(Zr)/(a_0))}]]#

#= -(a_0)/Ze^(-(Zr)/(a_0))r^4 - ((a_0)/Z)^2 e^(-(Zr)/(a_0))4r^3 - 12((a_0)/Z)^3[e^(-(Zr)/(a_0))r^2 - (2a_0)/Z{e^(-(Zr)/(a_0))r + (a_0)/Ze^(-(Zr)/(a_0))}]#

#= -(a_0)/Ze^(-(Zr)/(a_0))r^4 - ((a_0)/Z)^2 e^(-(Zr)/(a_0))4r^3 - 12((a_0)/Z)^3e^(-(Zr)/(a_0))r^2 - 24((a_0)/Z)^4{e^(-(Zr)/(a_0))r + (a_0)/Ze^(-(Zr)/(a_0))}#

#= -(a_0)/Ze^(-(Zr)/(a_0))r^4 - ((a_0)/Z)^2 e^(-(Zr)/(a_0))4r^3 - ((a_0)/Z)^3e^(-(Zr)/(a_0))12r^2 - ((a_0)/Z)^4e^(-(Zr)/(a_0))24r - 24((a_0)/Z)^5 e^(-(Zr)/(a_0))#

EVALUATION-READY FORM

#= |[-e^(-(Zr)/(a_0))[(a_0)/Z r^4 + 4((a_0)/Z)^2 r^3 + 12((a_0)/Z)^3 r^2 + 24((a_0)/Z)^4 r + 24((a_0)/Z)^5]]|_(0)^(oo)#

First half cancels out to be #0#:

#= cancel({-e^(-(Zoo)/(a_0))[(a_0)/Z oo^4 + 4((a_0)/Z)^2 oo^3 + 12((a_0)/Z)^3 oo^2 + 24((a_0)/Z)^4 oo + 24((a_0)/Z)^5]})^(0) - {-e^(-(Z(0))/(a_0))[(a_0)/Z (0)^4 + 4((a_0)/Z)^2 (0)^3 + 12((a_0)/Z)^3 (0)^2 + 24((a_0)/Z)^4 (0) + 24((a_0)/Z)^5]}#

Second half simplifies down to be #1*(0+0+0+0+24((a_0)/(Z))^5)#:

#= cancel(e^(-(Z(0))/(a_0)))^(1)[cancel((a_0)/Z (0)^4)^(0) + cancel(4((a_0)/Z)^2 (0)^3)^(0) + cancel(12((a_0)/Z)^3 (0)^2)^(0) + cancel(24((a_0)/Z)^4 (0))^(0) + 24((a_0)/Z)^5]#

#= 24(a_0/Z)^5#

Now, let us re-examine the wave function as a whole...

#psi_(2pz)#

#= 1/(32pi) (Z/a_0)^5 (24(a_0/Z)^5) (2/3) (2pi) stackrel(?)(=) 1#

#= 1/(cancel(32)cancel(pi)) cancel((Z/a_0)^5) (cancel(16)cancel((a_0/Z)^5))(cancel(2)cancel(pi)) stackrel(?)(=) 1#

#color(blue)(1 = 1)#

YES! ONE DOES EQUAL ONE! I mean...

The wave function is indeed normalized! :D

Proving mutual orthogonality for the 2p wave functions

Let us choose the following wavefunctions:

#psi_(2px) = 1/(sqrt(32pi))(Z/(a_0))^"3/2" (Zr)/(a_0)e^(-"Zr/"2a_0)sinthetacosphi#

#psi_(2py) = 1/(sqrt(32pi))(Z/(a_0))^"3/2" (Zr)/(a_0)e^(-"Zr/"2a_0)sinthetasinphi#

#psi_(2pz) = 1/(sqrt(32pi))(Z/(a_0))^"3/2" (Zr)/(a_0)e^(-"Zr/"2a_0)costheta#

To show they are orthogonal, we need to show at least one of them:

#int_("all space") psi_(2px)^"*" psi_(2pz)d tau = 0#

And from induction we can imply the rest since the radial components are identical. In other words:

#\mathbf(int_(0)^(oo) R_(nl,2px)^"*"(r)R_(nl,2pz)(r)r^2dr int_(0)^(pi) Y_(l)^(m)(theta)sintheta int_(0)^(2pi) Y_(l)^(m)(phi)dphi stackrel(?)(=) 0)#

#color(green)(1/(32pi)(Z/(a_0))^5 int_(0)^(oo) e^(-"Zr/"a_0) r^4dr int_(0)^(pi) sin^2thetacosthetad theta int_(0)^(2pi) cosphidphi stackrel(?)(=) 0)#

The radial portion turns out to be #24 ((a_0)/Z)^5#. So, let us evaluate the angular portions.

The #theta# portion:

#color(green)(int_(0)^(pi) sin^2thetacosthetad theta)#

Let:
#u = sintheta#
#du = costhetad theta#

#= int_(0)^(pi) u^2du#

#= 1/3*|[sin^3theta]|_(0)^(pi)#

#= 1/3*[sin^3(pi) - sin^3(0)]#

#= 1/3*[0 - 0] = color(green)(0)#

And now the #phi# portion:

#color(green)(int_(0)^(2pi) cosphidphi)#

#= |[sinphi]|_(0)^(2pi)#

#= sin(2pi) - sin(0)#

Let:
#u = sintheta#
#du = costhetad theta#

#= int_(0)^(pi) u^2du#

#= 0 - 0 = color(green)(0)#

Therefore, we have overall:

#color(blue)(1/(32pi)(Z/(a_0))^5 int_(0)^(oo) e^(-"Zr/"a_0) r^4dr int_(0)^(pi) sin^2thetacosthetad theta int_(0)^(2pi) cosphidphi)#

#= cancel(1/(32pi)(Z/(a_0))^5 (24)((a_0)/Z)^5 (0)(0))^(0)#

#= color(blue)(0)#

Since

#int_("all space") psi_(2px)^"*" psi_(2pz)d tau = 0#

the #2p_z# and #2p_x# atomic orbitals are orthogonal.

Really, the main difference with using the #2p_y# equation is that you instead get:

#color(green)("Constants" int_(0)^(oo) "Same stuff"dr int_(0)^(pi) sin^3thetad theta int_(0)^(2pi) sinphicosphidphi stackrel(?)(=) 0)#

And so:

#color(blue)(int_(0)^(2pi) sinphicosphidphi)#

#= 1/2|[sin^2phi]|_(0)^(2pi)#

#= 1/2[sin^2(2pi) - sin^2(0)] = color(blue)(0)#

From multiplying #0# by the other integrals, thus the whole integral disappears and:

#int_("all space") psi_(2px)^"*" psi_(2py)d tau = 0#

thus, the #2p_x# and #2p_y# atomic orbitals are orthogonal.

Finally, for the #2p_y# vs. the #2p_z#:

#color(green)("Constants" int_(0)^(oo) "Same stuff"dr int_(0)^(pi) sin^2thetacosthetad theta int_(0)^(2pi) sinphidphi stackrel(?)(=) 0)#

We know the #theta# integral from before:

#color(blue)(int_(0)^(pi) sin^2thetacosthetad theta)#

#= 1/3*|[sin^3theta]|_(0)^(pi)#

#= 1/3*[sin^3(pi) - sin^3(0)]#

#= 1/3*[0 - 0] = color(blue)(0)#

And so the whole integral disappears again, and indeed the #2p_y# and #2p_z# orbitals are orthogonal as well!