# Wavefunctions, radial nodes and probability density oh my!

Discussion in 'Clinicians [ RN / NP / PA ]' started by Paseo Del Norte, 10.16.10.

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1. ### Paseo Del Norte 5+ Year Member

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I understand this question may be a bit unusual for the clinician forum; however, I am not premed nor am I studying for any examination such as DAT, MCAT and so on, therefore I am not exactly sure where to put this question. I simply think about concepts for the sake of pure mental masturbation. Moderators, please feel free to move and my apologies if this creates a problem.

I have been experiencing a dilemma and simply cannot wrap my head around two concepts that are creating a dichotomy in my brain and frankly, driving me nuts.

So, when looking at the solutions of the Schrodinger equation (assuming a hydrogen atom), I am having difficulty interpreting the physical meaning of nodes such as radial nodes.

For example, say I have solve the Schrodinger equation for a 2s wavefunction, square the value, obtain a probability density, and so on. So, when we do all this nifty stuff and graph the square of the wavefunction, we have a node where we have zero probability density of finding an electron. I think this node lives somewhere around r~ 1 Angstrom ( two times the Bohr radius).

My dilemma is this:

I can interpret this to say that I will "never" find an electron at this distance (r). However, this appears highly deterministic and I have always though in spite of having an area of very low probability, the electron could be found, well anywhere. It would just be highly improbable to find it certain areas. Yet, here I have something that says, you can "never" find it at a node.

My problem is making sense of what appears to be somewhat of a contradiction. Clearly, I am missing something here and would appreciate any guidance anybody has to offer.

3. ### fab4fanTiredRetiredRNGold Donor 10+ Year Member

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Holy cow. I'm challenged by balancing my checkbook, Paseo.

4. ### FutureDoc4 2+ Year Member

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Forgive me because Chem Gem is >7 years ago.

I don't think I see a "contradiction" in your logic. The probability density generated by squaring the wave functions determines an area that is most likely to have an electron occupy it for the atom to have its "lowest energy"

The key is that solving those wave functions make it statistically unlikely (how that is defined I am not sure, could be <0.000001, I don't know the math well enough) to find an electron there because the atom would no longer be in its most stable state if it were found elsewhere-->as we all know nature constantly heads toward equilibrium/lowest energy states

The nodes appear as a solution to Schrodinger's equation in polar coordinates & spherical harmonics (which I am not familiar with)--see here: http://en.wikipedia.org/wiki/Spherical_harmonics

Summary: this node is generated as probabilistic area that is more statistically unlikely to have an electron (due to the laws discussed above---lowest energy) occupy by its described wave functions which are solved in polar coordinates (that's a pretty good description)

....more than that you'd have to go to an advanced textbook and a better understanding of quantum mechanics/applied mathematics or find a physics major to help you

5. ### Paseo Del Norte 5+ Year Member

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Actually, I am not all that great at math and I can not actually "solve" the Schrodinger equation. My dilemma, I think is more about interpretation of the results. I am having a difficult time wrapping my head around this zero probability concept. Assuming the electron is bound to the nucleus, and it most certainly is in the scenario provided, does a radial Node absolutely and deterministically mean there is no chance of finding an electron, or does there exist a very small possibility of finding an electron at a node?

I guess I am not sure where the uncertainty principle fits in here and it obviously does because an orbital is nothing but probability density in the first place. Or, perhaps the node is the result of a boundary that we impose that may or may not actually exist?

6. ### Paseo Del Norte 5+ Year Member

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Okay, thanks a bunch. It would make perfect sense that a node is also probabilistic. I've always been told no chance of finding an electron at a node, but it does make sense that the entire function, node and all will be probabilistic.

7. ### zenmanSenior Member 7+ Year Member

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Way back yonder in school, I had a difficult weather related chem problem. I called up the local TV weatherman and he said, "Hell if I know, I only report the weather!"

8. ### Paseo Del Norte 5+ Year Member

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Sort of my situation, I am only making sense of values that have already been calculated or derived. However, I cannot let this one go.

9. ### group_theoryEX-TER-MIN-ATE!' SDN Administrator 10+ Year Member

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Remember that particles can also exhibit dual properties, not just light. Electrons are small enough and fast enough that you can think of it as a wave in addition to a solid particle (technically a baseball have the same properties). You are trying to picture an electron as a solid object rotating around a nucleus, while the math equation you are doing pictures electrons as a wave with the boundry condition being that it must fit into an orbit that makes it around the nucleus. These waves can only exist in these defined orbits because anywhere else, the waves cancel each other out. Think of guitar strings as they cross the plane when they vibrate up and down. Electrons (in wave form) are similar, except they exist in a 3D plane.

Electrons are small, so its momentum is small, and based on de Broglie's relationship, its wavelength is relatively large.

A baseball also have dual properties (as a particle and a wave). But the momemtum (p) of a baseball is large so its wavelength is small and very difficult to observe. For a 140 gm baseball being thrown at roughly 100 mph (roughly 50 m/s), its wavelength is 9.4 x 10^-35 meters (and hence why its hard to phantom a baseball as a wave). Hence it acts more like a particle than a wave.

10. ### Paseo Del Norte 5+ Year Member

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Absolutely I agree, the electron basically has the properties of a standing wave and it also has particle like properties and even a non zero rest mass (albeit very small). However, the Schrodinger equation does "treat" the electron as a particle in that you are using the de Broglie relationship when setting up the calculations (Hamiltonian Operator)? So, the electron is quantised, and the 'n" quantum number even falls out of the "E" when calculating the Schrodinger equation.

Edit:

A rather scary thought just entered my mind based on your post. Assuming the above is true, the electron is quantised and a consequence of that is that there will be energy levels (non integer quantities i.e.. n= 1.2, 1.4, etc.) that are simply not allowed. After thinking about this, the node would make perfect sense in that the nodes represent energies that are simply not allowed based on this quantisation?

Let me know if I am on the right track. If this is true, it seems I have made a very basic and rather amateur error in logic, as a condition of quantisation would be energies that are not allowed (non integer amounts of Planck's constant).

I appreciate all the help and it seems I am making some beginner mistakes, but thanks all for helping me out.

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12. ### Charles_CarmichaelModerator Emeritus 7+ Year Member

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Yes, only certain energies are allowed. The nodes represented by the probability density are points where you can never find an electron. I feel like the question you're asking is regarding how electrons can move back and forth if nodes exist. Is that what you're thinking? If so, you have to realize that you cannot describe things at the quantum level using classical terms. So that question is essentially meaningless. The electron is not flitting back and forth through the nodal area.

Edit: As an interesting aside, even at 0 K, particles at the quantum level will have some finite kinetic energy and momentum. Otherwise, the Heisenberg uncertainty principle breaks down.

13. ### Paseo Del Norte 5+ Year Member

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I have no problems with the classical breakdown of how does an electron "move" from one area to another when you have an area of zero probability between said areas. I have no problems with this and I understand that the electron is not fitting through a node. It's simply here, there or over there.

I want to understand why we have nodes in the first place. Assuming a hydrogen atom or any single electron atom I suppose, why should we have nodes within an orbital? After all, an orbital will essentially exist within a principle energy level "n." This being the case, does this whole quantised concept even come into play with the electron being that it already "lives" within a discreet level of energy.

I think the quantum jitters concept you mentioned is also known as zero point motion?

I appreciate the response.

14. ### Paseo Del Norte 5+ Year Member

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Okay, after thinking about this all day and looking through my notes, I think I have come to the following:

The formula for radial nodes is n-1 = number of nodes. So, n=1 has 0 nodes, n=2 has one node and so on. Then, I happened to go back through his thread and noticed Group Theorie's picture. Actually, I looked at it differently. In the n= 1 state, we have 1 half wave, n= 2, we have two and so on. The waves have a peak and a trough represent highest and lowest amplitude. So, a radial node is simply a representation of the waves trough, or it's lowest amplitude?
So, it seems I may have been neglecting the wave like characteristics to overlook this rather obvious relationship.

However, since this is still a relationship of probability, does a node literally mean no electron at all, ever?

15. ### group_theoryEX-TER-MIN-ATE!' SDN Administrator 10+ Year Member

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No, a radial node is the halfway point between the peak and trough. Stop thinking of an electron as a discrete particle that orbits around a nucleus like a planet orbiting a sun. Think of it as a wave. It's not just a probability density but the wave function of the electron. It exist simultaneously everywhere in that wave function. The radial node is just the transition point from peak and trough of the electron function (and at that infinite point in space, there is no wave function). The electron still exist, just that it's wave function is basically nill at that exact point in space (which is questionable whether it still exist). If you take the guitar strings, are there waves at the nodes? Just like if you give the strings more energy (and have more nodes and shorter wavelengths), when electrons have more energy, they occupy higher energy levels (and hence why S shells have no nodes, p shells have node in each plane, d shells have more, f shells have more than d shells etc)

16. ### Paseo Del Norte 5+ Year Member

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I think of it as a smeared out cloud of probability where you have areas that are more dense and less dense, not a planet orbiting. However, I guess I do keep running back to a particle analogy when thinking about this node concept? I guess I am having difficulty trying to visualise a node in a physical sense. I know it's probably not possible, but I would like to somehow consolidate both analogies. When I think about the guitar string analogy, the string still exists at a "node" in a physical sense and I guess that's sort of how I think of the electron. However, the electron does not even exist in our world or space so to speak...

I do appreciate all the help.

17. ### BuddhaFW

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Ok So first off a a probability of finding an electron at any particular point is zero. Remember that the probability of finding the electron is based on the area under the rdf, so you can not calculate the area for one particular point, such as a node, or the maximum or anywhere.

18. ### group_theoryEX-TER-MIN-ATE!' SDN Administrator 10+ Year Member

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A radial node occurs when the radial part of the wavefunction vanishes, and an angular node occurs when the angular part vanishes.

Nodes are intrinsic features of electron/atom wavefunctions, and are coupled to the orthogonality between wavefunctions for different solutions of the Schrodinger equation.

The example of the guitar strings (a good example of a 1-dimensional box) indicates how the phase of the different solutions of the underlying differential equation changes at various nodal points, and how this ensures the mutual orthogonality of the solutions.

EDIT: In trying to explain what a radial node is, I completely forgot to answer the original question. At the radial node, an electron cannot exist. It can be on either side of the node but not on the node itself. Doesn;t violate the Heisenberg principle because you still don't know where the electron is, just where it is not. Just like we know the electron cannot be at the nucleus (with r=0), we know that electrons cannot be at radial nodes.​

Last edited: 10.22.10
19. ### Paseo Del Norte 5+ Year Member

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20. ### Paseo Del Norte 5+ Year Member

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"EDIT: In trying to explain what a radial node is, I completely forgot to answer the original question. At the radial node, an electron cannot exist. It can be on either side of the node but not on the node itself. Doesn;t violate the Heisenberg principle because you still don't know where the electron is, just where it is not. Just like we know the electron cannot be at the nucleus (with r=0), we know that electrons cannot be at radial nodes."

This helps quite a bit because that was my first point. I was thinking a node may be a violation of the uncertainty principle because it seemed so deterministic.

21. ### Paseo Del Norte 5+ Year Member

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Sort of a thread resurrection here; however, I want to run this be you all to see if I am interpreting this properly:

I have been playing around with an app for the iPad/iPhone that was adapted from a teaching programme called atom in a box. (Clearly, a funny take on the "particle in a pox" problem.) It solves the wave equation for the hydrogen atom and displays the 140 or so Eigenvalues. It produces the orbitals we all know and love and even shades in different colours to differentiate the differences in phase. You can put in the quantum numbers and it will show a picture of the probability density for you. In addition, at the bottom of the screen, it solves the wave equation, will give a graph of the most probable radius and calculate the binding energy for each "n" number.

My question relates back to "p" orbitals. We have all been taught they are dumbell shaped and align along the various axes (x,y & z). At least, I have ben taught this in the chemistry courses that I have taken thus far.

So, when I plug in the quantum numbers for a 2pz orbital (aligned with the nuclear axis), I get a picture that looks like this:

Cool, no problem.

However, when I plug in numbers for the other two orbitals, I find the following:

Clearly, we have a problem.

I've done some of my own research and understand that to create the other two orbitals, linear combinations are used to create the "p" orbitals we all know and love. Apparently, as far as Eignenvalues go, this is kosher.

My question is why do we do this? From what I have read, I can only conclude we do this because it is allowed by quantum mechanics and it's more ascetically pleasing? I also suspect that it is easer for people to have intuition when they hybridise "s" and "p" orbitals for molecular bonding. So, is the answer really as simple as the commonly accepted visualisation is prettier?

I understand this whole orbital business is rather arbitrary as the orbitals don't really exist in reality and only represent arbitrary boundaries that hold most of an electron's probability density. However, I want to make sure I at least have a rough understanding of what is going on.

22. ### group_theoryEX-TER-MIN-ATE!' SDN Administrator 10+ Year Member

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Sorry - I don't usually read this particular forum on a regular basis.

Not sure what happened with your program.

Each p subshells should contain 3 orbitals (m[sub]l[/sub] = -1, 0, +1). The p[sub]x[/sub] orbital has a symmetric double lobed shape directed along the x-axis, the p[sub]y[/sub] orbital has a symmetric double lobed shape directed along the y-axis, and the p[sub]z[/sub] orbital has a symmetric double lobed shape along the z-axis.

The two lobes are separated by a nodal plane that cuts through the nucleus and arises from the angular wavefunction Y(&#952;,&#981. 2p[sub]z[/sub] should be

(for the sake of my sanity, assume a = a[sub]0[/sub])

&#936; = {½(1/6a[sup]3[/sup])[sup]½[/sup] x (r/a)e[sup]-r/2a[/sup]} x {(3/4&#960[sup]½[/sup] cos &#952;} =

(1/32&#960;a[sup]5[/sup])[sup]½[/sup] r cos &#952; e[sup]-r/2a[/sup]

The math should be the same for each subshell, they just have to be orthogonal to each other (and hence why they should look the same but oriented along different axis)

Last edited: 02.18.11
23. ### Paseo Del Norte 5+ Year Member

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Thanks again for your replies, you have been very helpful. I agree, or at least this is what I've always thought. I've done additional studying and am a little confused as always. From what I understand, physicists will take linear combinations of wavefunctions to create the p orbitals we all know and love. I presume this is because we are dealing with complex numbers and the like. However, I have always thought that we take the square of the absolute value of &#936;? (if looking at position versus and time -->|&#936;|2 ) I've also seen the complex conjugate:&#936; times &#936;*. So, at my current level of understanding, I am not sure why linear combinations are needed as we can take the square of the absolute value and interpret as probability density because all of the complex stuff is real stuff so to speak, or can at least be interpreted in a physical sense.

Clearly, there are other methods that can be used such as matrix mechanics and this is actually where the Eigenvalues come from.

Anyway, I think I will contact the publisher of the app with a few questions, I just do not want to ask questions when an answer is clearly obvious and it's simply because I lack understanding or or the intellectual tools to understand.

24. ### Paseo Del Norte 5+ Year Member

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Just to verify, the Px and Py orbitals are in fact complex wavefunctions. For those who don't know, some wavefunctions have two components, a real and a complex part. The real part has real numbers we can use; however the complex or imaginary part has undefined numbers. In that, I don't mean they don't necessarily exist, they just cannot be defined. A simple thing you can do is to take -1 and put it into your calculator. Then hit the square root button. Your calculator will most likely give you a WTF? answer. That is because the square root of -1 is undefined (this is often called "i"). There is no known number when multiplied by itself will give -1. However, the actual operation of multiplying some undefined number by itself to get a defined number is actually very useful.

So, to get a wavefunction we can work with, we in fact have to take linear combinations of complex functions to get "real" functions. This is the case with the Px and Py orbitals.

Therefore, I am still assuming that perhaps the circular orbital I posted is how the application's programming interprets the original, complex p orbital wavefunctions? That is just my guess?

25. ### SimannBeats and Rhythm 2+ Year Member

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How on God's green earth does any of that apply to healthcare?

I know the MCAT guru's will argue, but to be frank, I have never ever dealt with a problem in any modality of medicine that is that perplexing.

26. ### Paseo Del Norte 5+ Year Member

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Roughly speaking, the physical world can be defined by this material. Medicine falls into the physical world, therefore, I believe medicine can be defined by this material.

However, I think I clearly stated that I am not going into "medicine" at the beginning of this thread and much of this is simply an exercise in mental masturbation for me. However, I always ask myself "why." After taking another semester of chemistry a couple of years back, I asked my self that same question. I wanted to know more of the "why" and "how" behind what I was taught. With a rough understanding of calculus and a bit of Chem and physics I've done some self study and hopefully presented half way intelligent questions?

All that aside, I think the ability to look out side the box so to speak is essential in health care and hopefully these exercises I do on my own help me develop critical thinking skills that will aid in overall problem solving, decision making, and the ability to appreciate a bigger and more complex picture.

Plus, this is one of a select few sites where I feel like I can present questions and have a discussion without people treating me like an idiot. The responses have been very helpful, in spite of everybody knowing I'm well out of my league regarding this material.

27. ### UNMorBUSTMystery Man 5+ Year Member

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OP your screen name sticks out to me.

28. ### Paseo Del Norte 5+ Year Member

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I hope that is a good thing.

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Not so much the name, but the avatar. It still scares me.

30. ### Paseo Del Norte 5+ Year Member

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What? Uncle Ruckus...scary?

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I didn't know that was a character from a cartoon. I had to google it.