How can two atomic orbitals with the same n value be orthogonal?

Click For Summary
Two atomic orbitals with the same principal quantum number (n) can be orthogonal due to the presence of radial nodes, which change the sign of the wave functions. For example, the 2s orbital has one radial node while the 3s has two, allowing their integrals over all space to yield zero when considering positive and negative regions. This orthogonality is significant because it ensures that the wave functions do not overlap, maintaining distinct quantum states for electrons. The concept is similar to orthogonal vectors in 3D space, where different wave functions act as basis vectors in an infinite-dimensional space. Understanding this property is crucial for grasping the behavior of electrons in atoms.
fsci
Messages
6
Reaction score
0
I am wondering how two orbitals of same n values can be orthogonal, for example how are a 2s and 3s orbital orthogonal?
What I understand is a property of orthogonality is the product of the two wave functions integrate to zero over all space. I tried to look at this graphically and categorize overlapping regions as either positive or negative products and then cancel out positive and negative regions to yield zero, but what I am having trouble is that the 3s orbital is larger than the 2s orbital, so how can they possible integrate to zero?
If someone could also explain the significance/implications of all orbitals being orthogonal that would be helpful too! I do not understand the importance of orthogonality in orbitals!
Thank you!
 
Physics news on Phys.org
fsci said:
I am wondering how two orbitals of same n values can be orthogonal, for example how are a 2s and 3s orbital orthogonal?
What I understand is a property of orthogonality is the product of the two wave functions integrate to zero over all space. I tried to look at this graphically and categorize overlapping regions as either positive or negative products and then cancel out positive and negative regions to yield zero, but what I am having trouble is that the 3s orbital is larger than the 2s orbital, so how can they possible integrate to zero?
If someone could also explain the significance/implications of all orbitals being orthogonal that would be helpful too! I do not understand the importance of orthogonality in orbitals!
Thank you!

There are radial nodes that you usually cannot see in the higher n (n>=2) representations, as typically drawn. 2s has one radial node, 3s has two radial nodes. On either side of the radial node there is a change in sign for the wave function.

When you do the integral for any of the products you will get positive regions cancelling the negative regions.
 
Quantum Defect said:
There are radial nodes that you usually cannot see in the higher n (n>=2) representations, as typically drawn. 2s has one radial node, 3s has two radial nodes. On either side of the radial node there is a change in sign for the wave function.

When you do the integral for any of the products you will get positive regions cancelling the negative regions.

Ok thank you. I may be heading in the wrong direction... But how do the product of a 1s x 3s and then product of a 2s x 3s both integrate to zero when 1s and 2s orbitals are not equal to each other?
 
fsci said:
Ok thank you. I may be heading in the wrong direction... But how do the product of a 1s x 3s and then product of a 2s x 3s both integrate to zero when 1s and 2s orbitals are not equal to each other?

You might look into some Linear Algebra, Differential Equations textbooks. For the H-atom wave functions, you will have an infinite set of wavefunctions, that are solutions to the Schroedinger Eq. Each l=0 wavefunction will be "orthogonal" to the other, as long as n_1<> n_2.

This is kind of like the way that the unit vectors in 3D space are all orthogonal to one another. x_hat dot y_hat = 0, x_hat dot z_hat = 0, y_hat dot z_hat =0.

The solutions to the Schoredinger Eq for H are like these basis vectors in 3D space, except the space is infinite.
 
Thread 'Unexpected irregular reflection signal from a high-finesse cavity'
I am observing an irregular, aperiodic noise pattern in the reflection signal of a high-finesse optical cavity (finesse ≈ 20,000). The cavity is normally operated using a standard Pound–Drever–Hall (PDH) locking configuration, where an EOM provides phase modulation. The signals shown in the attached figures were recorded with the modulation turned off. Under these conditions, when scanning the laser frequency across a cavity resonance, I expected to observe a simple reflection dip. Instead...

Similar threads

  • · Replies 2 ·
Replies
2
Views
661
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
8
Views
4K