Quantum Chemistry 3.11 - 3-D Particle in a Box

[@Creepjacker]

"whenever we have partial differential equations in this course, we have two options: We'll do separation of variables, or we won't solve it" I had a good laugh at that part.

[@TMPChem]

When in doubt, hand waving works magic.

[@gillesbaumann675]

Your videos are the best man! The overview is simply unique. Keep doing what you're doing cause your doing a great job!

[@TMPChem]

Thanks Giles. Much appreciated. Keep up the good work.

[@Harjeetkaur-bu7zz]

I searched alot for 3 D box Equation And i finnally found you Sir Thank you for providing this amazing content

[@MeLSL1]

You came just in time for my quantum chemistry class this semester. Thank you so much for these helpful videos!!

[@TMPChem]

Glad to help. You'll probably find a video for almost every topic you cover in that course. And probably quite a few you won't.

[@PunmasterSTP]

I know it's been half a decade, but I just came across your comment and was curious. How'd quantum chemistry end up going?

[@MeLSL1]

@@PunmasterSTP I got a B, but I remember feeling like the professor made the final exam easier and gave us more points out of pity

[@PunmasterSTP]

@@MeLSL1 I know the feeling, but I'm glad you passed, and I hope the rest of your academic career was successful.

[@walacross]

For some stupid reason I couldn’t figure out what I would do if it wasn’t a cubical box due to the equations in my textbook simply writing L^3, but you writing them separately as lx,ly,and lz made it make way more sense, thank you so much :D. Sometimes these simple concepts can be overlooked.

[@PunmasterSTP]

3D particle? More like "Terrific videos, all of em!" Thanks so much for making and sharing them.

[@frogfrog1993]

this is sooooo gooood

[@seitdrs2467]

Hi, what happened to the wave function depending on time? In the separation of variables why isn't there also a time component. Thank your for the comprehensive video guide, it is the best I have found so far.

[@TMPChem]

Great question. The effect of time is discussed in chapter 4, specifically in video 4.14 on the time dependence of the wavefunction. Spoiler alert: for pure eigenstates (such as this one) time doesn't matter, and the expectation values are constant in time.

[@dynamicdonut2442]

does this take into account the boundary conditions psi = 0 and |psi|^2 = 1? Also thank you for this video!

[@BlackFiredDragon]

So... what quantum numbers do n_x, n_y, and n_z represent? Like if I'm given the four quantum numbers of a specific particle (n, l, m_l, m_s) in a 3D box, which 3 numbers do I use in finding that particle's energy?

[@TMPChem]

I don't think the question makes sense. The set (n, l, m_l, m_s) describes the quantum numbers of an electron / orbital within a hydrogen atom. The set (n_x, n_y, n_z) describes the quantum numbers of a particle within a 3-dimensional box. These two sets are completely unrelated. If you want to find the energy of a particle in a 3-d box, you find / specify (n_x, n_y, n_z) and calculate the energy using those values.

[@chymoney1]

To actually answer your question n_x, n_y, and n_z are just arbitrary integers that we found after applying the boundary conditions onto our separated solutions and solving the strum liouville eigenvalue problem. Quantum numbers have to do with properties of particles like spin( this is where my knowledge dies off)

[@harishkoranga5257]

How energy get added afte the product of wave?

[@dps3]

6 years later still useful i have an exam tommorow lol

[@PunmasterSTP]

How'd your exam go?

[@dps3]

@@PunmasterSTP failed it,wrote it again recently, hope i pass. And this question didn't appear in none of them

[@PunmasterSTP]

@@dps3 Ah I'm sorry to hear that, and I hope you pass this time!

[@dps3]

@@PunmasterSTP thanks man

[@marybelarenas64]

Can you explain what psi* is one more time? I do not understand how/why it is added to the integral function

[@TMPChem]

Sure. Psi by itself has little physical meaning other than the mysterious claim that it is a function which represents how a particle is behaving like a wave. What does have physical meaning is psi_star * psi, (or | psi^2 | ), which represents the probability density of finding the particle at any given point in space. For example, if I have a wavefunction with a value of 0.5 at x=2, and it is approximately constant, then the probablility of finding of particle between x=1.9 and x=2.1 is |psi^2| * (x2 - x1), or 0.5 * 0.5 * (2.1 - 1.9) = 0.05. Thus, there is a 5% chance of finding this particle between x=1.9 and x=2.1.

[@aliaabdulaziz5589]

if the total energy is the sum of energy E_x, E_y and E_z, shouldn't the denominator then be 2m instead of 8m?

[@TMPChem]

Hi Alia. The energy is indeed a sum of E_x, E_y, and E_z, but each of them individually has a common factor of h^2 / 8*m which distributes into all three terms. If lx = ly = lz, *and* nx = ny = nz, then we could factor out a factor of 3, giving us 3/8 instead of 1/8. There may be some confusion here between the 3-d wavefunction (which is a *product* of 3 1-d wavefunctions), and the 3-d energy (which is a *sum* of 3 1-d energies).

[@aliaabdulaziz5589]

TMP Chem aah ok thanks for your prompt reply! Just realised you substituted the h bar for h which explains why the pi disappears and why it's 8 instead of 2. Thanks anyway! I like your use of colours 😊 btw, why is it a product for the wavefunction but sum for energy when doing separation of variables?

[@TMPChem]

That's the math of how separation of variables ends up working out for the Schrodinger equation in multiple dimensions. Separation of variables starts by assuming that the wavefunction can be "separated" into a product of three 1-dimensional functions. When we do this, the partial derivatives in the Laplacian operator each act on only one part of the wavefunction, and we can separate the Schrodinger into three separate equations. Each individual second derivative contributes separately to the energy. The total energy is a sum of each of their contributions, just as the Laplacian is a sum of the individual derivatives.

[@mariohernandez1111]

Is there a way to know when a multidimensional function can be decomposed into a product of unidimensional components?

[@TMPChem]

Other than attempting to do the algebra and succeeding or failing, I'm not aware of a systematic method.

[@mariohernandez1111]

TMP Chem But for doing the Algebra you would need to know the function itself right?

[@TMPChem]

Yes, but if you can do it for the Hamiltonian then you can do it for the wavefunction too. So it's really a matter of being able to factor the Hamiltonian into a sum of unidimensional components.

[@mariohernandez1111]

It's so obvious now, unless I'm wrong. As the Hamiltonian is a linear operator (sum of Del, which is formed by linear operators; and V(x), which by nature, being a potential function it only depends on the components), the application of the operator can be decomposed into three linear applications which in term mean that the eigenvalue of the multidimensional operator is the sum of eigenvalues of the unidimensional components, in this case the sum of energies. Does this make sense?

[@TMPChem]

Yes. One small correction: For the case of the particle in a box, the Hamiltonian is only *kinetic* energy, and the kinetic energy operator is always separable, as del is a sum if partial derivative components, as you noted. Once we add a non-zero potential energy function, that function must also be separable as well. So if the potential P(x, y, z) = P(x) + P(y) + P(z), then we're good to go. Any system with no potential energy will always be separable.

[@karin585]

Can someone explain how to draw the specific shape of the particle in the box? For example for 111 the particle would be a sphere?

[@TMPChem]

Hi Ruby. The particle itself (being a point particle) is an individual, structureless point in space which is infinitely small and occupies no space. What we're discussing here is the probability distribution of where that particle is likely to be found during a measurement. The result is the 3-d generalization of the general 1-d particle in a box probability. The 111 state is a half-sine wave in all three dimensions, x, y, and z. This gives maximum probability of finding the particle in the center of box, and no probability at the edges, with intermediate (but not quite spherical) values in between.

[@karin585]

Thank you so much for replying. But my professor asked us to determine the shape of the orbital within the box. The example being 111 would be a sphere, 121 would be a sphere cut into two along the x axis. I don't understand how he can determine that?

[@TMPChem]

I wouldn't completely agree with the description of such a function as a sphere for a few reasons. 1) you have to assume that all three sides of the box are of equal length, which is a special case and not true in general. 2) it's not the wavefunction itself, but an isosurface (2-d surface where all points have the same function value) which would be spherical in such a case. 3) The wavefunction isosurface has such a property in that special case, but the probability density is the wavefunction squared, which is a sin squared function, which does not have spherical isosurfaces. Otherwise the general description is qualitatively correct. There is a single region of density with highest density towards the middle in the 111 state (half sine wave is a maximum in the middle for all three dimensions). For 211, one of the three is a full sine wave (with a node in the middle), cutting the blobs of density into two separate regions along the x-axis, where we have a full sine wave (i.e. 2 half sine waves).

[@karin585]

Thank you so much for replying. The explanation of the spherical shape makes a lot more sense now!

[@josephineamala5792]

What is lx,ly,lz is.it angular momentum

[@TMPChem]

Hi Josephine. In this video, Lx, Ly, and Lz are the length of the box in the x, y, and z dimensions. Each is a constant value.

[@williamng7887]

Are the Axis labeled wrong?

[@TMPChem]

Yes. ly and lz should be switched.

[@Aefryn]

your x and y axis are switched on the graph

[@TMPChem]

I would agree that my y and z axes are switched in the graph, but x appears to be correct.

[@Aefryn]

+TMP Chem Ops, that's what I meant, was tired sorry. Also great videos!

[@jacobvandijk6525]

A 3D particle! Hahaha.