I Orbital electrons in stationary states?

fox26

My textbook in elementary Q.M. stated that orbital electrons in an atom must have stationary state
wavefunctions. Was this just a simplification, the truth being maybe that their wavefunctions can be
nonstationary for a little while, but soon decay into stationary ones? I’ve seen an answer, somewhere,
that if an orbital electron were not in a stationary state, its wavefunction would be a superposition of
wavefunctions one of which would be a wavefunction which would decay into a lower energy
wavefunction. However, if such a nonstationary wavefunction ψ1 were equal to such a decay-prone ψ2 + some other ψ3, then any stationary ψ4 would also be a superposition of wavefunctions one of which would be such a decay-prone wavefunction, e.g., ψ4 = ψ2 + ψ5, where ψ5 = ψ4 - ψ2, so the same objection would apply to ψ4 being a stable wavefunction of an orbital electron. What is the correct answer?

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DrClaude

Mentor
My textbook in elementary Q.M. stated that orbital electrons in an atom must have stationary state
wavefunctions.
Can you quote the book, because this doesn't make much sense. Orbitals are stationary states. i.e., solutions of the time-independent Schrödinger equation (neglecting the coupling to the electromagnetic vacuum, which is responsible for spontaneous emission).

fox26

Can you quote the book, because this doesn't make much sense. Orbitals are stationary states. i.e., solutions of the time-independent Schrödinger equation (neglecting the coupling to the electromagnetic vacuum, which is responsible for spontaneous emission).
By "orbital electrons" I mean the electrons bound to an atom and not in the nucleus (maybe there are not any in the nucleus); I don't mean electrons whose wavefunction ψ by definition of "orbital electron" satisfies the stationary state condition (that the modulus of ψ at each point in space doesn't change with time). Its been a long time since I took Q.M., but one book that I've looked at stating the stationary state requirement for orbital electrons was by Messiah, and there were others. That the electron is in a stationary state is used to show, for example, that its angular momentum satisfies certain quantization rules, the Bohr rules, which satisfaction was one of the triumphs of the new wavefunction form of Q.M. that led physicists to consider it as a replacement for the old Bohr quantum theory, with which it agreed and which had been experimentally confirmed as far as what were the possible angular momentum states of electrons in an atom. "Orbital", I thought, referred to a group of possible orbits, with the lowest energy orbital consisting of the two lowest energy orbits, the next orbital consisting of the next eight higher energy orbits, etc., with the orbital structure of the highest energy orbital containing any electrons being important in determining the chemical valence of the atom. However, I never studied in detail the orbital structure of atoms. My concern is, why can't electrons be bound to an atom (outside the nucleus), at least temporarily, in an orbit with their ψ's modulus, as well as maybe its phase, at some points in space changing with time?

HAYAO

Gold Member
By "orbital electrons" I mean the electrons bound to an atom and not in the nucleus (maybe there are not any in the nucleus)
I honestly do not understand what this means. Like DrClaude said, you might want to quote the exact part of the text that actually says this.

fox26

I honestly do not understand what this means. Like DrClaude said, you might want to quote the exact part of the text that actually says this.
I'm uncertain what I can do to clarify my simple sentence quoted above your comment. The only word I imagine you likely would not understand is "bound". For an electron to be bound to an atom means that the electron cannot escape to infinity, which is usually, maybe always, due to the electron's not having enough kinetic energy to escape the net electrostatic attraction of the atom's positive nucleus and the other negative electrons between it and that nucleus. The quote from my post you request a quote from a textbook saying is about the meaning of "orbital electrons", whereas DrClaude requested a quote from a textbook stating that orbital electrons must be in stationary states. I don't have a Q.M. textbook handy, and can't give a quote satisfying either your or DrClaude's request, and I am not sure that I would supply one even it I did have such a text. The text's quotes would not be any clearer than my statements, and both my statement about the stationary state requirement for orbital electrons given in standard texts and the meaning of "orbital electrons" are well-known elementary facts about standard Q.M. texts and Q.M. terminology, respectively, and I don't feel any need to prove them by actual cited quotes.

By the way, DrClaude, "stationary state" doesn't mean a "solution of the time-independent Schrodinger equation". The Schrodinger equation is said to be time-independent if the potential in it does not depend explicitly on time, being dependent on at most position. A free particle's Schrodinger equation is time-independent, since the potential is constant in both time and space, but some solutions of it, such as the time evolution of a state initially localized to a finite region of space, or having a Gaussian position probability distribution, have a modulus at a fixed point in space which varies in time, so the solution is not a stationary state solution.

DrClaude

Mentor
By "orbital electrons" I mean the electrons bound to an atom and not in the nucleus (maybe there are not any in the nucleus); I don't mean electrons whose wavefunction ψ by definition of "orbital electron" satisfies the stationary state condition (that the modulus of ψ at each point in space doesn't change with time). Its been a long time since I took Q.M., but one book that I've looked at stating the stationary state requirement for orbital electrons was by Messiah, and there were others. That the electron is in a stationary state is used to show, for example, that its angular momentum satisfies certain quantization rules, the Bohr rules, which satisfaction was one of the triumphs of the new wavefunction form of Q.M. that led physicists to consider it as a replacement for the old Bohr quantum theory, with which it agreed and which had been experimentally confirmed as far as what were the possible angular momentum states of electrons in an atom.
That sounds like the old quantum theory, pre-Schrödinger. Overall, the requirement of stationary states comes from the observation that atoms, when left alone, are stationary.

"Orbital", I thought, referred to a group of possible orbits, with the lowest energy orbital consisting of the two lowest energy orbits, the next orbital consisting of the next eight higher energy orbits, etc., with the orbital structure of the highest energy orbital containing any electrons being important in determining the chemical valence of the atom.
An orbital is an eigenstate of the non-relativistic Hamiltonian describing a nucleus and an electron and the interaction between them. It is characterized by three quantum numbers (qn), the principle qn $n$, the angular momentum qn $l$, and the magnetic qn $m_l$. These orbitals are degenerate due to spin.

My concern is, why can't electrons be bound to an atom (outside the nucleus), at least temporarily, in an orbit with their ψ's modulus, as well as maybe its phase, at some points in space changing with time?
You can of course put an electron in a superposition of stationary states, in which case it is no longer stationary. Coming back to the first point above, the only requirement for stationary states is to explain the stability of atoms.

By the way, DrClaude, "stationary state" doesn't mean a "solution of the time-independent Schrodinger equation". The Schrodinger equation is said to be time-independent if the potential in it does not depend explicitly on time, being dependent on at most position. A free particle's Schrodinger equation is time-independent, since the potential is constant in both time and space, but some solutions of it, such as the time evolution of a state initially localized to a finite region of space, or having a Gaussian position probability distribution, have a modulus at a fixed point in space which varies in time, so the solution is not a stationary state solution.
I stand by my statement. A free particle is poses a problem as the eigenstates of the Hamiltonian are plane waves, which are not valid wave functions since they are not square integrable. The only valid states are superpositions of states of different momentum (wave packets), which are definitely non-stationary.

HAYAO

Gold Member
I'm uncertain what I can do to clarify my simple sentence quoted above your comment. The only word I imagine you likely would not understand is "bound". For an electron to be bound to an atom means that the electron cannot escape to infinity, which is usually, maybe always, due to the electron's not having enough kinetic energy to escape the net electrostatic attraction of the atom's positive nucleus and the other negative electrons between it and that nucleus. The quote from my post you request a quote from a textbook saying is about the meaning of "orbital electrons", whereas DrClaude requested a quote from a textbook stating that orbital electrons must be in stationary states. I don't have a Q.M. textbook handy, and can't give a quote satisfying either your or DrClaude's request, and I am not sure that I would supply one even it I did have such a text. The text's quotes would not be any clearer than my statements, and both my statement about the stationary state requirement for orbital electrons given in standard texts and the meaning of "orbital electrons" are well-known elementary facts about standard Q.M. texts and Q.M. terminology, respectively, and I don't feel any need to prove them by actual cited quotes.
What I was confused was indeed the way you use the word "bound", especially when you say "it's not bound to a nucleus but an atom". If I were to interpret this literally, then I would have to imagine that you think that an electrons are not attracted to the nucleus because of electrostatic interaction with the proton, but instead attracted to the atom (which includes other electrons) in some other unknown way. If you just said either "electrons are bound to a nucleus" or "electrons are bound to an atom", instead of mentioning both and then denying the former, then my misunderstanding wouldn't have happened. So no, your sentences are definitely not "simple". In fact, it's hard to understand what you are trying to convey, which is probably why DrClaude and I asked for you to quote what the text says.

It became clear after your above post that you were only trying to clarify that you are not meaning to say electron are literally glued with the nucleus. That will not happen. (If you collide electron and nucleus with sufficient kinetic energy, deep inelastic scattering occurs).

About the latter point you made, I will have to apologize that I didn't mean to say if what you said about "bound electrons" was in the textbook, but what DrClaude asked for in his post about stationary states. I initially quoted your entire post with replies for each parts, until I thought I would delete most of it to make my post shorter. Then the post got mixed up.

Otherwise, I agree with DrClaude. By looking at your OP once again I say the textbook is right under the circumstance that the atom is left alone (as DrClaude said). If however the electron is in the superposition of stationary states, then indeed it is not going to be a stationary state. You are talking about two different situation.

fox26

What I was confused was indeed the way you use the word "bound", especially when you say "it's not bound to a nucleus but an atom". If I were to interpret this literally, then I would have to imagine that you think that an electrons are not attracted to the nucleus because of electrostatic interaction with the proton, but instead attracted to the atom (which includes other electrons) in some other unknown way. If you just said either "electrons are bound to a nucleus" or "electrons are bound to an atom", instead of mentioning both and then denying the former, then my misunderstanding wouldn't have happened. So no, your sentences are definitely not "simple". In fact, it's hard to understand what you are trying to convey, which is probably why DrClaude and I asked for you to quote what the text says.

It became clear after your above post that you were only trying to clarify that you are not meaning to say electron are literally glued with the nucleus. That will not happen. (If you collide electron and nucleus with sufficient kinetic energy, deep inelastic scattering occurs).

About the latter point you made, I will have to apologize that I didn't mean to say if what you said about "bound electrons" was in the textbook, but what DrClaude asked for in his post about stationary states. I initially quoted your entire post with replies for each parts, until I thought I would delete most of it to make my post shorter. Then the post got mixed up.

Otherwise, I agree with DrClaude. By looking at your OP once again I say the textbook is right under the circumstance that the atom is left alone (as DrClaude said). If however the electron is in the superposition of stationary states, then indeed it is not going to be a stationary state. You are talking about two different situation.
HAYAO, I suggest you look at my thread again carefully to see what I really said, which isn't what you said that I said.

DrClaude and HAYAO, plane waves are indeed non-normalizable, non-square integrable solutions of the free particle Schrodinger equation, but the two types of waves I mentioned are also solutions, and in addition are normalizable and square integrable.

DrClaude, I am not certain of the point of the last paragraph in your last comment. You seem to be defending your statement that solutions of the time-independent Schrodinger equation are stationary states (and vice-versa). I agree that some are, but the two that I mentioned are not. You seem to disqualify them as counterexamples to your statement on the grounds that they are not eigenstates of the Hamiltonian but are superpositions of plane waves, which are such eigenstates. My two examples are just as much solutions as are plane waves, and maybe more physically realizable. Since the Schrodinger equation is linear, superpositions of solutions to the homogeneous equation are also solutions, so my two examples are solutions of the (really a) time-independent Schrodinger equation for a free particle which are not stationary, ergo, your statement is untrue. Also, what are you referring to when you say "This sounds like the old quantum theory, pre-Schrodinger."? In the quote just above that sentence, I was clearly talking about the new Schrodinger form of Q.M. compared with the old Bohr theory. I did say about the electron that its being in a stationary state was used to show that it satisfied the Bohr rules on angular momentum. "Stationary state" names a wavefunction concept; it doesn't mean anything in the old Bohr model. You have in one statement "...atoms, when left alone, are stationary." Surely this doesn't mean that atoms, when left alone, don't move. What does it mean? I am concerned with the textbook justification for assuming that orbital (in the sense of "in an orbit", an old pre-wavefunction phrase, but still commonly used for non-nuclear bound electrons in an atom) electrons are in a stationary state, which is assumed in sections of textbooks which are developing theory, without mention of any observations that atoms, when left alone, are "stationary". If, as you later say, the only requirement for stationary states is to explain the stability of atoms (I imagine you are referring to the fact that electrons don't spiral into the nucleus due to energy loss from their emitting electromagnetic radiation; is that what you mean by atoms being "stationary"?), can you show theoretically that this requirement is necessary (not just sufficient)? Such a demonstration would indeed show that electrons must be in stationary states, however based partially on the observed stability of atoms, not just on quantum theory (which of course itself was partially based on, and is confirmed by, observations).

DrClaude

Mentor
You seem to be defending your statement that solutions of the time-independent Schrodinger equation are stationary states (and vice-versa).
Indeed, I maintain that stationary states are eigenstates of the Hamiltonian, and vice versa.

I agree that some are, but the two that I mentioned are not. You seem to disqualify them as counterexamples to your statement on the grounds that they are not eigenstates of the Hamiltonian but are superpositions of plane waves, which are such eigenstates. My two examples are just as much solutions as are plane waves, and maybe more physically realizable. Since the Schrodinger equation is linear, superpositions of solutions to the homogeneous equation are also solutions, so my two examples are solutions of the (really a) time-independent Schrodinger equation for a free particle which are not stationary, ergo, your statement is untrue.
I completely agree that superposition of eigenstates are valid solutions to the Schrödinger equation. But superposition of states are not eigenstates, and only eigenstates are stationary states. (Caveat: I'm assuming that we are talking about superpositions of non-degenerate energy eigenstates.)

A state is stationary if and only if $| \langle \psi(t) | \psi (t') \rangle |^2 = 1 \forall t, t'$. This is only possible if $H \psi = E \psi$.

(By the way, a single plane wave is also a stationary state, since it is an eigenstate, but not a wave packet.)

Also, what are you referring to when you say "This sounds like the old quantum theory, pre-Schrodinger."? In the quote just above that sentence, I was clearly talking about the new Schrodinger form of Q.M. compared with the old Bohr theory. I did say about the electron that its being in a stationary state was used to show that it satisfied the Bohr rules on angular momentum.
The use of the expression "Bohr rules on angular momentum" is old quantum theory. I've never hear this in the context of modern QM.

You have in one statement "...atoms, when left alone, are stationary." Surely this doesn't mean that atoms, when left alone, don't move. What does it mean?
I should have clarified that I'm talking about internal states, i.e., as seen from the rest frame of the atom (or the relative electron-nucleus motion).

I am concerned with the textbook justification for assuming that orbital (in the sense of "in an orbit", an old pre-wavefunction phrase, but still commonly used for non-nuclear bound electrons in an atom)
"Orbital" is the modern nomenclature, to replace orbit in the Bohr model or Bohr-Sommerfeld theory. And like @HAYAO, I don't understand what you mean by "non-nuclear bound electrons".

(I imagine you are referring to the fact that electrons don't spiral into the nucleus due to energy loss from their emitting electromagnetic radiation; is that what you mean by atoms being "stationary"?),
Correct.

can you show theoretically that this requirement is necessary (not just sufficient)? Such a demonstration would indeed show that electrons must be in stationary states, however based partially on the observed stability of atoms, not just on quantum theory (which of course itself was partially based on, and is confirmed by, observations).
I'm not sure what kind of proof you want here. Considering the simplest case of a single electron, you get that the lowest energy state is a stationary state with the electron a certain distance (on average) from the nucleus. Hence, the electron will not collapse to the nucleus.

PeterDonis

Mentor
You seem to be defending your statement that solutions of the time-independent Schrodinger equation are stationary states (and vice-versa). I agree that some are, but the two that I mentioned are not.
The two that you mentioned are not solutions of the time-independent Schrodinger Equation; that equation is the eigenvalue equation $H \psi = E \psi$. Only eigenstates of the Hamiltonian, i.e., stationary states, are solutions of that equation. Superpositions of (non-degenerate) stationary states are not eigenstates of the Hamiltonian, and are therefore not solutions of the eigenvalue equation. They are solutions of the time-dependent Schrodinger Equation (since all valid states are), but that's not the same thing.

I completely agree that superposition of eigenstates are valid solutions to the Schrödinger equation.
They are valid solutions to the time-dependent Schrodinger Equation. They are not valid solutions to the time-independent Schrodinger Equation. See above.

• DrClaude and bhobba

PeterDonis

Mentor
Since the Schrodinger equation is linear, superpositions of solutions to the homogeneous equation are also solutions
This is true of the time-dependent Schrodinger equation, yes. It is not true of the time-independent Schrodinger Equation, because a superposition of two (non-degenerate) solutions of that equation is no longer time-independent.

• bhobba

HAYAO

Gold Member
HAYAO, I suggest you look at my thread again carefully to see what I really said, which isn't what you said that I said.
Well then I also have to suggest you write in a more organized and concise fashion. No offense intended, because I am also not quite a good writer (thus my writings are also subject to improvements), but you said you wrote it "simple" when it was certainly not.

Anyhow, I am starting to suspect (sorry if I am wrong) that fox26 is also confused about Schrodinger equation having the property that any linear combination of the solutions is also a solution, as is usually written in textbooks. To be precise, time-dependent Schrodinger equation has the property that any linear combination of the solutions is also a solution. Time-independent Schrodinger, on the other hand, does not. This is actually very easily testable.

Once you decide that you are going to talk about superposition, you are talking about it in the realm of time-dependent Schrodinger equation. They are certainly not stationary. However, orbital electrons in an atom (for example hydrogen atom) can be solved from time-independent Schrodinger equation, and the wavefuctions of these orbitals are stationary. So like I have said before, you are talking about two different situation.

• DrClaude

fox26

Well then I also have to suggest you write in a more organized and concise fashion. No offense intended, because I am also not quite a good writer (thus my writings are also subject to improvements), but you said you wrote it "simple" when it was certainly not.

Anyhow, I am starting to suspect (sorry if I am wrong) that fox26 is also confused about Schrodinger equation having the property that any linear combination of the solutions is also a solution, as is usually written in textbooks. To be precise, time-dependent Schrodinger equation has the property that any linear combination of the solutions is also a solution. Time-independent Schrodinger, on the other hand, does not. This is actually very easily testable.

Once you decide that you are going to talk about superposition, you are talking about it in the realm of time-dependent Schrodinger equation. They are certainly not stationary. However, orbital electrons in an atom (for example hydrogen atom) can be solved from time-independent Schrodinger equation, and the wavefuctions of these orbitals are stationary. So like I have said before, you are talking about two different situation.
DrClaude, PeterDonis, & HAYAO:

Our disagreements are partially (but only partially) due to confusions of terminology, as quite a few in PF seem to be. I will try to clarify what I mean by certain terms. First of all, by the "Schrodinger equation" (SE) I mean the equation that is the basic equation of quantum mechanics, in its non-relativistic, single particle form, which is:

iħ(∂/∂t)ψ(r,t) = [-ħ22/2m + V(r,t)]ψ(r,t). (Eq. 1)

This is sometimes called the "time-dependent Schrodinger equation". The time-independent Schrodinger equation (TISE) can be derived from the
SE under the assumption that V(r,t) = V(r), that is, the potential is time-independent (which is the defining assumption for the TISE to be applicable), which is sometimes expressed by "The system is stationary." (Which is one possible source of terminological confusion, together with the following.) A stationary state of a particle with wavefunction ψ, on the other hand, is one for which the modulus of ψ, and so its probability density, at each point in space, is independent of time. This will be the case, for a system whose potential is time-independent, if and only if Hψ = Eψ, where H is the operator on the right hand side of Eq. 1, and E is the (time-independent) energy of the system (as DrClaude pointed out, except he omitted the qualification that the potential be time-independent).

DrClaude, below your quote of my statement concerning you: "You seem to be defending your statement that solutions of the time-independent Schrodinger equation are stationary states (and vice-versa)" (Call that statement of yours DC1), you say "Indeed, I maintain that stationary states are eigenstates of the Hamiltonian, and vice versa" (Call that statement you say you maintain DC2). DC2 is true for particles in a system whose potential is time-independent, but neither implication in it is true in general; there are simple examples that show this. However, even for particles in a time-independent potential, DC1 is not equivalent to DC2; if it were, all solutions of the time-independent Schrodinger equation would be eigenstates of the Hamiltonian, which isn't true. Your and PeterDonis's belief that it is true probably is due to a misunderstanding of the solution of the TISE by separation of variables, which I will discuss below. You can see your error before looking at that discussion by considering the following: Wave packets, which you agreed are definitely non-stationary, and which are not eigenstates of the Hamiltonian, and which both of my examples of wavefunctions are, can obviously be solutions of the TISE, since they can exist in a potential which is everywhere and always zero (or is everywhere and always a constant), i.e., be wavefunctions of free particles, and the TISE certainly is true of such particles, since by definition it is the SE for particles in a time-independent potential, which a potential which is everywhere and always zero undoubtedly is. Thus these free wave packets are solutions of the TISE which are not stationary state solutions and are not eigenstates of the Hamiltonian.

PeterDonis, pay attention, since your claim that my two examples of wavefunctions of free particles are not solutions of the TISE is clearly false on its face. What could be more trivially a solution of the TISE than a free particle, which is in a potential which is obviously time-independent at every spatial point, since it is zero (or a fixed constant) at every point in space and every time (see also my similar comments to DrClaude above)? You are a Physics Mentor. How could you think that, as you claim, wavefunctions of free particles, which my two are by their specifications (definitions), are not solutions of the TISE? If they are not, what is? You are maybe a victim, with DrClaude, of drawing some unjustified conclusions from certain results in discussions of solution of the TISE by separation of variables. In particular, you seem to think that solutions of the TISE must themselves be time-independent. This (what you think, not my sentence) is not true, as shown below. I will now discuss solutions of a PDE, specifically the TISE, by separation of variables.

In the solution of the TISE by separation of variables, there is a constant E which the two separated sides of the equation, the side which depends on only the spatial coordinates, and the side which depends on only the time coordinate, must each equal, and each solution with such an E is a solution of the eigenvalue equation Hψ = Eψ. However, there is no one fixed E which this must be. E can take any non-negative value, and different values of E give different solutions, which will be solutions of the eigenvalue equation with that E. A general solution of the TISE is a sum of any solution with the given potential V(r) with any solution to the homogeneous TISE, i.e., with V(r) = 0. Since the free particle SE is a TISE with V(r) = 0 (or a constant), any square-summable superposition of plane-wave solutions of it is a solution of it. Thus free wave packets whose shapes change in time and whose position probability amplitudes at certain points in space change can be and are solutions of the TISE, but are not eigenstates of the Hamiltonian or stationary state wavefunctions.

I will maybe discuss further disagreements with you two and with HAYAO later. Right now I will post what I have written so far and go to breakfast.

blue_leaf77

Homework Helper
PeterDonis, pay attention, since your claim that my two examples of wavefunctions of free particles are not solutions of the TISE is clearly false on its face.
What is apparent to me, as is to you as you already suspected in the beginning of your above post, is that Peter must use a different definition of time-independent Schroedinger equation (TISE). Your TISE is
$$i\hbar\partial_t \psi = (-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r))\psi$$
A free electron wavepacket is a solution of the above equation when $V(\mathbf r)=0$ so yes you are correct within this definition of TISE.
On the other hand Peter's definition of TISE as mentioned in post #10 is
$$H\psi = E\psi$$
A free electron wavepacket is not a solution of the above equation and hence he is also correct within his definition of TISE.
PeterDonis, pay attention, since your claim that my two examples of wavefunctions of free particles are not solutions of the TISE is clearly false on its face.
Since you already suspected that the disagreement is mainly due to different definition of some terms, especially that on TISE, you could have realized that it would be too early for you to drop the discernment that he and some others are wrong.

• bhobba, HAYAO and PeterDonis

Mentor

PeterDonis

Mentor
PeterDonis, pay attention, since your claim that my two examples of wavefunctions of free particles are not solutions of the TISE is clearly false on its face.
No, it isn't, because, unlike you, I am using the term "time-independent Schrodinger Equation" the way the physics community uses it. @blue_leaf77 has already pointed out the difference between that usage and your usage. Sorry, but you don't get to just make up your own definitions of standard physics terms. You have to use the standard ones. Particularly when you start a thread with a question about something you read in a textbook, which of course is going to use the standard terminology. Basically, this whole thread amounts to you making up your own definition of a term and then wondering why your textbook says something that doesn't seem to satisfy that definition.

And with that, this thread is closed.

• bhobba, HAYAO and dlgoff

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