# Having problems interpreting a Reduce answer can any one help

Posted 9 years ago
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 I have been working on finding intersections with wave equations of hydrogen orbitals and I got the intersections with Root; However I'm having problems singling the answers out due to the complexity of the answers. I was hoping that someone could help me find the needle in the hay stack. Attachments:
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Posted 9 years ago
 It's a volume integral in spherical coordinates, so it has r^2 dr Sin[theta] dTheta dPhi. I didn't integrate over Theta and Phi because the r integral gave 0.
Posted 9 years ago
 In the code you typed why is there an r^2?
Posted 9 years ago
 You get a number, not a function.
Posted 9 years ago
 I just have one more question with taking the integral of the wave equations. When you take the integral of the product of two different wave equations you get the overlap correct? However is it just going to give zero or is it going to give me other functions describing the spaces the wave equations overlap
Posted 9 years ago
 OK that make sense with the Graphs I made. Thank you very much you both were a great help.Sincerely,Zachary Nichols
Posted 9 years ago
 I'll expand a bit on what Frank has said. These hydrogen orbitals have been found as solutions to Schrodinger's equation for a single electron bound by the potential well of the Coulomb force exerted by the proton on the electron. When this LINEAR differential equation is solved, we find that there are solutions only for discrete energy values, which are eigenvalues of the linear system. (This is true as long as the electron is bound, meaning it has insufficient energy to escape.) These are denoted by the energy quantum number.Next, these eigenvalues are used to solve for the eigenfunctions associated with them. These are the associated wave functions. Here we arrive at the first orthogonality condition. Eigenfunctions associated with different eigenvalues are orthogonal.For the ground state ( n = 1 ) we found just one eigenfunction associated with the energy level. That is the 1S orbital. (Which can be occupied by 2 electrons of opposite spin, but that goes a bit farther.) But for n=2, we find 4 eigenfunctions, all having the same energy level: 2S, 2Px, 2Py, 2Pz. (There are other considerations in multi-electron atoms, which lower the S orbital energies compared to the P orbitals.) So here we have a second case of orthogonal functions. We have CHOSEN to construct orthogonal wavefunctions for mathematical convenience, For this case, if we consider a hydrogen atom in the n=2 energy state, the atom would have that energy in any of these states. But, more than that, these functions are the solutions to a linear differential equation. This means that any linear combination is also a solution. This set of orbitals plays the same role in forming the wavefunction that is played by the traditional i, j, k unit vectors to describe a vector in 3-space. We don't require every vector to be either i or j or k. Rather we write any vector as xi+yj+zk. In the same way, any linear combination of 2S, 2Px, 2Py, 2Pz orbital is a legitimate wavefunction for the n=2 quantum state, so long it is normalized so that the Integral of Y* Y over all space is equal to 1. (Since there is exactly probability=1 of finding the electron somewhere.)In fact, when considering these orbitals being involved in bonding between atoms to form a molecule, it is often the case that some linear combination offers a better bond -- meaning lower energy. Then we see things like SP hybrid orbitals involved in bonding.
Posted 9 years ago
 Back to Mathematica. Ignoring normalization constants, this shows that that the 1s and 2s orbitals have zero overlap: In[1]:= psi1s[r_] = Exp[-r/a0]; psi2s[r_] = (2 - r/a0) Exp[-r/(2 a0)]; In[4]:= Integrate[psi1s[r]*psi2s[r] r^2, {r, 0, \[Infinity]}] Out[4]= ConditionalExpression[0, Re[a0] > 0] since a0 is a positive constant.
Posted 9 years ago
 One of the weird things about quantum mechanics is that an electron can partially be in one state and partially in another state at the same time, like Schrodinger's cat. The overlap between two states of different energy is zero is because the mathematical operator which determines the states and the energies has certain symmetry properties.
Posted 9 years ago
 Ok now I think i'm starting to understand. The overlap has to be zero because an electron can't be in two different states at the same time.
Posted 9 years ago
 I don't understand when you said "Wave functions having the same energy are chosen so their overlap is zero". Does that mean wave functions that have same energy levels have 0 probability to exist is one of two states (thinking of the Schrodinger's cat thought experiment). In that case how does phase play in all this. Isn't phase just changing the angle at which the system is orientated.
Posted 9 years ago
 The hydrogen orbitals you're working with are like the basis vectors of a coordinate systems, which are chosen to be at right-angles. This is similar to the orbitals having zero overlap. Schrodinger's cat is in a mixed state, a combination of the alive state and the dead state. The alive state and the dead state have zero overlap.
Posted 9 years ago
 The absolute value squared of wave functions are probability densities. The overlap of the different wave functions of a system should be zero. This is necessary if they represent different energy states. Wave functions having the same energy are chosen so that their overlap is zero.
Posted 9 years ago
 OK so what I'm getting is that the over lap of the two probability densities of the two systems is found by the integral of Y1*×Y2, and that since the functions are just a way to describe a set of values the functions can't be treated as a normal function in the sense of f(x,y,z)=0. Meaning the overlap of the functions would describe the probability of electron existing as either one of two systems.
Posted 9 years ago
 The wave functions are not curves or surfaces. (technically, I mean that they are not 2D manifolds embedded in 3-space.) They are complex-valued scalar functions, which take on a value everywhere in space. They contain everything that can be known about the system they describe. In the full sense, they are time-dependent as well as varying in space.But it is often meaningful to work with time-independent functions. These take on complex values everywhere. (They are continuous, so they can't just stop.) One of the things they determine is the probability of finding a particle in some region of space. This density function is given by Y* Y. (Where I use Y for psi, and * denotes the complex conjugate..) The probability of finding a particle in a region of space is given by the integral of Y* Y over that region.So these are density functions. You can picture their value as a cloud in space of varying density. When these orbitals are plotted, it is this density function than is being represented. So, just for visualization, we plot some contour of constant density, but it is not a surface so far as the orbital is concerned. There is not much in the way of a precise meaning in the intersection of these contours. There is, however, a meaning in how much the orbitals "overlap" in all of space. Just like Y1* Y1 represents a density for Y1, Y1* Y2 represents the the extent to which the orbitals overlap, and the integral Y1* Y2 over all space represents the total overlap of the Y1 and Y2 orbitals.So, in short, the wave functions are not an equation of the form f(x,y,z)=0, which could define a surface which could have an intersection with another surface. Rather, they are of the form f(x,y,z) -- not an equation -- where the values taken on vary from one wave function to another, and it is the way this variation in space occurs that contains information about the system.
Posted 9 years ago
 The wavefunctions are defined for all space so all coordinates are common. Perhaps you want the overlap, the integral of their product.
Posted 9 years ago
 The intersections are points where the different wave functions share a common coordinates. What I'm trying to do is find those common coordinates. and I'm having trouble reading the answers I got back.
Posted 9 years ago
 I'm not sure what you mean by the "intersection" of orbitals.