Quantum Mech. Si Wavefunction probability

[mkkrnfoo85]

Question:

If Si1 represents the wavefunction of a Px orbital in the hydrogen atom, and Si2 represents the wavefunction of a Py orbital in the same hydrogen atom. Si1 and Si2 are both normalized wavefunctions.

a) what is the value of integral:Si1*Si2*dt. ?
b) what is the value of integral:Si1*Si1*dt. ?

From what my teacher was saying during lecture, integral:(Si^2)dt = 1, which represents the probability that a particle has to be somewhere there. (I don't quite understand what this equation means). Would someone this to me? I'm pretty sure answer (b) is 1, because of what I've learned, but for (a), the answer is supposed to be 0. Please help. THanks

 

[mkkrnfoo85]

bump. please help.

 

[Janitor]

I think I know the answers, but when you use the variable 't' in this context, what physical quantity do you understand that to represent?

 

[codyg1985]

I think I know the answers, but when you use the variable 't' in this context, what physical quantity do you understand that to represent?

My guess would be that it stands for time. The wave function would spit out the probable position of the electron in that orbital.

 

[mkkrnfoo85]

'dt' stands for a VOLUME QUANTITY (sorry). I just don't know why 'integral:Si1Si2 = 0'

'Si' is a normalized wavefunction, and the integration is carried out over all space.


Can someone explain the physical meaning of that? I know it has something to do with probability of a particle being in that vicinity. Thanks.

 

[Janitor]

Okay, I was hoping dt stood for a 3D volume element. I've got a book on electronic structure by W.A. Harrison which says

It is known also that the existence of boundary conditions (such as the condition that the wave functions vanish outside a given region of space) will restrict the solutions [to the time-independent Schroedinger equation for a central potential] to a discrete set of eignevalues E, and that these different eigenstates can be taken to be orthogonal to each other [which makes your first integral vanish]. It is important to recognize that eigenstates are wave functions which an electron may or may not have. If an electron has a certain eigenstate, it is said that the corresponding state is occupied by the electron. However, the various states exist whether or not they are occupied.

Legendre polynomials are used in solving the stationary-state 3D central-potential TISE because Legendre's differential equation arises in the process of solving the equation. The Legendre polynomials are orthogonal functions.

I'm not sure that anything I have written really helps too much in explaining the "physical meaning of that," though.

 

[Janitor]

You might be able to get more of a physical picture of what is going on from this website: http://www.chemguide.co.uk/atoms/properties/atomorbs.html

 

[dextercioby]

The spherical harmonics are involved in the wavefunctions:

$$\psi_{nlm}(r,\vartheta,\varphi) =C_{nlm}R_{nl}(r)Y_{lm}(\vartheta,\varphi)$$

U have to discuss the case when

$$\{n,l,m\}=\{2,1,-1\} \ \mbox{and} \ \{n,l,m\}= \{2,1,0\}$$

I think u want to compute these numbers

$$I=:\iiint_{R^{3}} \psi^{*}_{2,1,-1}(r,\vartheta,\varphi) \psi_{2,1,0}(r,\vartheta,\varphi) \ r^{2} \ dr \ d\Omega$$

$$J=:\iiint_{R^{3}} \psi^{*}_{2,1,-1}(r,\vartheta,\varphi) \psi_{2,1,-1}(r,\vartheta,\varphi) \ r^{2} \ dr \ d\Omega$$

Keep in mind that

$$\iint Y^{*}_{l',m'}(\vartheta,\varphi) Y_{l,m}(\vartheta,\varphi) \ d\Omega = \delta_{l'l}\delta_{m'm}$$

and the hydrogenoid wavefunctions are orthonormalized.

Daniel.

 

[mkkrnfoo85]

Thanks, that sort of cleared up some of my concepts, esp. that site on atomic orbitals. I don't really understand dextercioby's explanation, I'm only in a freshman 2nd semester chem course, and haven't really covered double and trip. integrals in calc. yet. I may never get to your explanation any time soon either. But thanks.

 

[dextercioby]

Then it surely is something devious with your course.Speaking about wavefunctions for hydrogen without the calculus part fully covered...

Daniel.