# Particle in the box eigenfunctions

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• fog37
In summary: Nice chat, but in summary, the conversation mainly revolved around finding eigenfunctions for the particle in a 1D box and the implications of the uncertainty principle. We discussed that the energy eigenfunctions are not eigenfunctions of the linear momentum operator, and that finding the momentum eigenfunctions involves making a guess and using properties of functions. We also briefly touched on the position eigenfunctions and the uncertainty in momentum when the energy is known exactly.
fog37
Hello Everyone,

For the particle in the 1D box of width ##L##, the time invariant Schrodinger equation is cast in the form of the Hamiltonian operator and automatically leads to the energy eigenfunctions

$$\Psi(x) = \sqrt{\frac 2L} sin(n\pi x/L)$$
I know that these energy eigenfunctions ##\Psi(x)## have each a well defined energy and are NOT simultaneous eigenfunctions of the linear momentum operator ##\hat p##. The energy eigenfunctions ##\Psi(x)## are simultaneous eigenfunctions of the magnitude squared of the momentum operator ##\hat p##.

To find the eigenfunctions ##\Phi## of linear momentum operator, should we simply apply the operator ##-i \hbar \frac \partial {\partial x}## to the energy eigenfunctions? The result would be a new function that represents the linear momentum eigenfunction ##\Phi(x)##...

What other eigenfunctions can we find for the particle in the box? What other operators can we apply?
I know that the particle in the box does not have any "rotational attributes" so the angular momentum operator is not applicable...But what if we applied the angular momentum operator (in position space) to the functions ##\Psi(x)##? What would we obtain?

Thanks!
Fog37

fog37 said:
To find the eigenfunctions ##\Phi## of linear momentum operator, should we simply apply the operator ##-i \hbar \frac \partial {\partial x}## to the energy eigenfunctions?

No. An eigenfunction of momentum would be a function ##\Phi## that satisfies ##\hat{p} \Phi = p \Phi## (notice no hat on the RHS, ##p## there is the eigenvalue, i.e., the value of the linear momentum). It should be easy to show that the energy eigenfunctions ##\Psi(x)## do not satisfy this equation.

fog37 said:
what if we applied the angular momentum operator (in position space) to the functions ##\Psi(x)##?

Have you tried it?

Given
$$\Psi(x) = \sqrt{\frac 2L} sin(n\pi x/L)$$

$$-i \hbar \frac \partial {\partial x} \Psi(x) = -i \hbar \sqrt{\frac 2L} (n\pi/L) cos(n\pi x/L)$$

So the linear momentum eigenfunction will be ##-i \hbar \sqrt{\frac 2L} (n\pi/L) cos(n\pi x/L)## which is clearly different from the energy eigenfunction ##\Psi(x)##

Fog37

fog37 said:
So the linear momentum eigenfunction will be ##-i \hbar \sqrt{\frac 2L} (n\pi/L) cos(n\pi x/L)##

No, it won't. Did you read my previous post?

PeterDonis said:
It should be easy to show that the energy eigenfunctions ##\Psi(x)## do not satisfy this equation.

To clarify further, it is also easy to show that the functions you get by applying ##\hat{p}## to the energy eigenfunctions don't satisfy the momentum eigenfunction equation either.

Sorry. I have read it completely just now and I am now clear on the fact that the energy eigenfunctions are NOT also eigenfunctions of the linear momentum operator.

Now, given that I know the expression for the linear momentum operator, which is ##-i \hbar \frac \partial {\partial x}##, and that its eigenfunctions must satisfy the eigenvalue equation ##-i \hbar \frac \partial {\partial x} \Phi(x) = p \Phi(x)##, which is how do we go about finding these functions ##\Phi(x)##?

Ok, I have made some progress:

We need to hypothesize that the linear momentum eigenfunction have a certain form, like ##e^{ikx}##. This function has a specific momentum eigenvalue ##k \hbar##.

The energy eigenfunction ##sin(n\pi x/L)## will be a linear combination of two momentum eigenfunctions: ##e^{ikx}## and ##e^{-ikx}##. When we measure the momentum we would obtain either ##- k \hbar## or ##k \hbar##

Is that correct?

fog37 said:
how do we go about finding these functions

You guess. There isn't any cookie cutter procedure to do it.

Knowledge of general properties of functions can help. For example, the eigenvalue equation is basically telling you that if you differentiate the function ##\Phi(x)## with respect to ##x##, you get the same function (multiplied by some constant). What kind of function has that property, that it stays the same when differentiated?

Knowledge of basic QM can also help. For example, if we express a momentum eigenfunction as a function ##\Phi(p)## of momentum ##p## instead of position ##x##, what should it be? That should be an easier question to answer than the question of what ##\Phi(x)## should be. Then once you have ##\Phi(p)##, you can just Fourier transform to get ##\Phi(x)##.

fog37 said:
We need to hypothesize that the linear momentum eigenfunction have a certain form, like ##e^{ikx}##.

Yes, that's a good guess. For extra credit, you could try the other method I suggested, of guessing a function ##\Phi(p)## that expresses a momentum eigenfunction as a function of momentum (instead of position), and then Fourier transforming it.

fog37 said:
The energy eigenfunction ##sin(n\pi x/L)## will be a linear combination of two momentum eigenfunctions

Yes, which is why it isn't itself an eigenfunction of momentum.

fog37 said:
When we measure the momentum we would obtain either ##- k \hbar## or ##k \hbar##

Yes. Can you see what this is telling you physically?

Ok, I can see that if the momentum eigenfunction ##\Phi(p)= \delta (p-p_0)##, we would obtain, by Fourier transform, the momentum eigenfunction ##e ^{ikx}## in position space.

As far as your question "Can I see what this is telling me physically?" I guess it means that...we cannot know the energy and the momentum exactly simultaneously. There is an uncertainty in momentum when the system is in a state of exact energy.

I guess the whole discussion would also apply to position eigenfunctions of the the particle in the box. By the uncertainty principle, momentum and position are uncertain. But now, what are the position eigenfunctions ##\Theta(x)## of the position operator ##\hat x##, i.e. ##\hat x \Theta(x)= x \Theta(x)## ?

fog37 said:
I can see that if the momentum eigenfunction ##\Phi(p)= \delta (p-p_0)##, we would obtain, by Fourier transform, the momentum eigenfunction ##e ^{ikx}## in position space.

Exactly.

fog37 said:
I guess it means that...we cannot know the energy and the momentum exactly simultaneously. There is an uncertainty in momentum when the system is in a state of exact energy.

That is true, but notice that the two possible momentum values, ##k \hbar## and ##- k \hbar##, have the same magnitude but opposite directions. (In 3 dimensions, the possible values would be 3-vectors with the same magnitude pointing in all possible directions.) So the magnitude of the momentum (more precisely, its square) is determined by the energy; what is not determined is its direction.

fog37 said:
what are the position eigenfunctions ##\Theta(x)## of the position operator ##\hat x##

Good question. Can you make a guess? (Hint: try a guess similar to the one you made for momentum eigenfunctions as functions of momentum.)

well, I know they are ##\delta(x-x_0)## but I am not sure these are the position eigenfunctions specific to the particle in the box.
In general, aren't the eigenfunctions of a specific operator specific to the particular physical situation and it all depends on the type of ##V(x)##? For instance, the position eigenfunctions for a free particle and the position eigenfunctions for the particle in the box may be different.

Assuming the ##\delta(x-x_0)## are indeed position eigenfunctions for the particle in the box, an energy eigenfunction like ##sin(kx)## would be a superposition of many position eigenfuctions...

fog37 said:
I am not sure these are the position eigenfunctions specific to the particle in the box

The position operator is the same as for any other case, so the eigenfunctions will be the same as well.

fog37 said:
In general, aren't the eigenfunctions of a specific operator specific to the particular physical situation

No. The specific operator itself, the one that is relevant to whatever physics you are trying to capture, is what depends on the particular physical situation. But given a particular operator, its eigenfunctions are what they are; that's a matter of math, not physics.

fog37 said:
it all depends on the type of ##V(x)##?

The potential ##V(x)## will affect the energy operator (the Hamiltonian); a different ##V(x)## means a different Hamiltonian. So there will be different energy eigenfunctions, because the Hamiltonian operator is different.

Thanks PeterDonis.

Also, once we find the energy eigenfunctions of the particle in the box problem with infinite potential, we need to multiply each function by the time dependent part
$$e^{-iEt/ \hbar}$$

By linear superposition, i.e. by a weighted summation of stationary states each multiplied by their time-dependent function, we can generate a general solution to the time-dependent Schrodinger equation. But Isn't the particle in a box a time-invariant problem that is described only by the time independent Schrodinger equation? I guess not, since, our starting point was to assume solutions of the form ##\Psi(x) \Phi(t) ## so the particle in a box still looks for solutions to the time dependent Schrodinger equation which are just separable and are obtained through the time independent equation. The point is that all wavefunctions will be functions of the type ##\Psi(x,t)## but the probability density function may or may not depend time depending on the problem...

fog37 said:
Isn't the particle in a box a time-invariant problem that is described only by the time independent Schrodinger equation?

No. The particle's state does not have to be time-independent.

You are right. I thought about it deeply today.

We start with the full time dependent Schrodinger equation, we assume we can find separable solutions like ##\Psi(x,t)= \Phi(x) \Theta(t)## and insert that into the TDSE. From there we obtain two ODEs. One of them is the TISE since it only applies to the function ##\Phi(x)## and there is not time parameter in it. By solving the TISE we find the functions ##\Phi## which we later multiply by ##\Theta(t) = e^{i\omega t}##.
The full and final solutions ##\Psi(x,t)= \Phi(x) \Theta(t)## are the stationary, bound solutions because their associated probability function is time invariant.

In essence, the particle in a 1D box problem is eventually about finding solutions to the time-dependent Schrodinger equation like any other problem in QM. The TISE is just an intermediate step...

fog37 said:
we assume we can find separable solutions

This is not just an arbitrary assumption. If you want to find a solution whose probability function is time independent, it must be separable in this way (plus, the function ##\Theta(t)## must have the specific form you give later on). Can you see why?

Ok, the fact that the solution is separable leads to to an ODE for the time part that has ##\Theta(t)## solutions that are time-harmonic, i.e. of the form ##e^{i\omega t}##.
The product ##\Psi(x,t)^* \Psi(x,t)## is the probability density function which does not depend on time ##t## if the function ##\Theta(t)## is a complex exponential ##e^{i\omega t}##. Is that what you meant?

fog37 said:
The product ##\Psi(x,t)^* \Psi(x,t)## is the probability density function which does not depend on time ##t## if the function ##\Theta(t)## is a complex exponential ##e^{i\omega t}##.

That's part of it; but it assumes that ##\Psi(x, t) = \Phi(x) \Theta(t)##. But that, i.e., separability, also can be proved given the requirement that the probability density function does not depend on ##t##; you don't have to assume it.

Ok. Thanks.

Why all eigenfunctions of various operators do not depend of time and can have a position space representation like ##\Psi(x,y,z)## that does not involve time? Is it because the operators don't depend on time themselves?

fog37 said:
Why all eigenfunctions of various operators do not depend of time and can have a position space representation like ##\Psi(x,y,z)## that does not involve time?

What do you mean by "do not depend on time"? Consider: eigenfunctions of energy and momentum are left invariant by time evolution--i.e., they time evolve into themselves--but eigenfunctions of position, for example, are not--they do not time evolve into themselves. Yet we can express any given eigenfunction of position as a function of position, not time--it's just ##\delta(x - x_0)##, which is a function of ##x## only.

fog37 said:
Is it because the operators don't depend on time themselves?

Whether or not the operators depend on time depends on which picture you are working in. We have been working in the Schrodinger picture, where the operators stay the same and the states evolve in time. But there is also the Heisenberg picture, where the states stay the same and the operators evolve in time; and the interaction picture, which is sort of in between the two.

I think the op may be confused about time dependence.

The separability came out because the Hamiltonian for this problem is not time dependant.

If I am getting the discussion.

Thank you.

What I trying to say (poorly) is that to find the eigenfunctions of a certain operator we solve an eigenvalue ODE and obtain functions that do not depend on that. That led me to say that eigenfunctions don't depend on time. For example,
$$\hat H\Psi(x)= E \Psi(x)$$
For momentum, $$\hat p \Theta(x)= p \Theta(x)$$
For position, $$\hat x \Phi(x) = x \Phi(x)$$

Same goes for all other operators. All the functions ##\Psi(x)## (energy eigenfunctions), ##\Theta(x)## (momentum eigenfunctions), ##\Phi(x)## (position eigenfunctions), etc. do not depend on time.

the system is in its abstract state ##| \Psi >## before we do anything to it. The system then becomes one of the eigenfunctions of the operator (observable) we are considering.

PeterDonis, I am not really clear the meaning of "eigenfunctions of energy and momentum are left invariant by time evolution--i.e., they time evolve into themselves--but eigenfunctions of position, for example, are not--they do not time evolve into themselves."

What does it mean that the functions evolve into themselves and are not time evolved by the time evolution (operator)?

Thanks!

fog37 said:
to find the eigenfunctions of a certain operator we solve an eigenvalue ODE and obtain functions that do not depend on that

You left out a key qualifier: to find eigenfunctions that do not depend on time, we solve an eigenvalue ODE that does not depend on time. If you write out the operators ##\hat{H}##, ##\hat{p}##, and ##\hat{x}##, you will see that they are all time-independent. So the eigenvalue ODE is time-independent. But that is not always the case. It happens to be the case in this particular problem, the "particle in a box", but there are many other problems where it is not the case.

fog37 said:
the system is in its abstract state |Ψ>|Ψ>| \Psi > before we do anything to it. The system then becomes one of the eigenfunctions of the operator (observable) we are considering.

This is only true in a collapse interpretation. You should not be assuming a collapse interpretation if you are trying to understand what the QM math tells you. If you insist on using interpretations, at a minimum you should be able to describe things using both collapse and no collapse interpretations (e.g., Copenhagen and MWI).

fog37 said:
What does it mean that the functions evolve into themselves and are not time evolved by the time evolution (operator)?

For a system with a time-independent Hamiltonian, such as the "particle in a box", the time evolution operator is ##e^{- i \hat{H} t}##. If you apply this operator to any eigenstate of energy or momentum for this case, you will see that it leaves the state unchanged (except for a possible shift in phase, which does not change the probability density function). But this is not true for an eigenstate of position.

Thanks as always, PeterDonis.

In classical mechanics, a mechanical system has certain modes. To find the future time evolution of the system (say, an elastic drum membrane), we need to find the modes (aka eigenfunctions) and express the initial condition, which is the shape of the membrane at time ##t_0##, as a linear superposition of weighted and phased eigenfunctions (modes). The system then evolves on its own. Maybe in this case the total energy of the system (Hamiltonian) is constant and the membrane, although oscillating in time, keeps the same shape and does not evolve to something else, i.e. to some other shape. If damping was present, the total energy would not be constant and the shape of the oscillating membrane would change in time. That could be addressed giving the eigenfunctions in the superposition different amplitudes with different rates of attenuation...
In the case of the oscillating membrane, we solve a time-independent ODE to find the eigenfunctions of the system. The eigenfunctions (modes) are a signature of the system. The eigenfunctions of linear classical systems evolve in time but their shape remains the same. I follow what you are saying, that for a time-independent Hamiltonian, the wavefunction time evolution is trivial and only due to a phase change.

Why would an eigenstate of the position operator change in time even when the Hamiltonian is time-independent?The time evolution of a state is determined by the Schrodinger equation once the initial condition , i.e. ##\Psi(x,t_0)## is given. What is the difference between the Schrodinger equation, as the machine that evolves states in time, and the time evolution operator that applied to the initial state ##\Psi(x,t_0)##? Are they exactly equivalent approaches?

I am also still mixing the 2 versions of QM (I think there are more versions): Schrodinger wave mechanics, Heisenberg matrix mechanics. In my books, the two versions seem to be mixed the together since we are talking about PDEs and ODEs and functions (which is Schrodinger wave version) and operators and states as vectors (which seems to be Heisenber version). When we talk about operators are we actually using Heisenberg version of QM?

fog37 said:
In classical mechanics, a mechanical system has certain modes.

Correction: in classical mechanics, some mechanical systems have certain modes. (Basically, the ones where the Hamiltonian is time-independent, the same as in QM. Hamiltonians are not limited to QM; you can do classical mechanics in the Hamiltonian formulation, and if you're not familiar with that, I would highly recommend learning it. It makes it much clearer what the actual differences are between classical and quantum mechanics.)

You appear to be focusing on particular properties of a certain class of systems, without realizing that they do not generalize.

fog37 said:
Why would an eigenstate of the position operator change in time even when the Hamiltonian is time-independent?

Instead of asking, why don't you actually do the math? Take an eigenstate of position, apply the operator ##e^{- i \hat{H} t}## to it, and see what happens? (Hint: it won't stay the same, not even allowing for a shift in phase.) Then compare that to what happens when you apply the same operator to an eigenstate of energy or momentum. (Hint: they will stay the same, except for a shift in phase.)

Once you understand what the math actually says, it might help you in formulating better questions.

fog37 said:
Are they exactly equivalent approaches?

Have you looked at the math? What does it tell you? (Hint: I said that the time evolution operator is ##e^{- i \hat{H} t}##. Where did I get that from?)

fog37 said:
When we talk about operators are we actually using Heisenberg version of QM?

The two versions are mathematically equivalent, so it's not really a big deal to mix them in your ordinary language descriptions.

Mathematically, the term "operator" can apply to both versions: in the Heisenberg version, an operator is described by a matrix, whereas in the Schrodinger version, it is described by a function, or more properly an operation that we do on a function. For example, the momentum operator ##\hat{p}## is described in the Schrodinger version by ##- i \hbar \nabla##, which is an operation that you can do on a function. But it's still an operator.

Similarly, the term "state vector" can apply to both version: in the Heisenberg version, it's a row or column vector, whereas in the Schrodinger version, it's a function. For example, a momentum eigenstate is described in the Schrodinger version as a function ##e^{ikx}##, but it can still be considered a vector. That is because the space of all functions, or more properly all square integrable functions, on the real line, or over real three-dimensional space, is a vector space, so individual functions can be thought of as vectors. The difference is that this vector space has an infinite number of dimensions, and the "vector index" ##x## is continuous, not discrete, so it's not as intuitively obvious that the functions are vectors. But if you think of a function ##\Psi(x)## as a list of an infinite number of "vector components", one for each value of ##x##, you can see the correspondence.

Thanks PeterDonis,

I am working on it and reading about Hamiltonian mechanics. Not a light topic but I will get there :)

I read that in the Schrodinger picture the states evolve while in the Heisenberg picture the state never evolves and the operators evolve.

Also, the time evolution operator, which is unitary, is strictly related to the time-dependent Schrodinger equation. To run an analogy, the Schrodinger equation and the evolution operator are related to each other the same that ## \frac {\partial v}{ \partial t} = a## (a differential equation) and ##v_f = v_0 +a t## are related to each other. So the evolution operator derives from SE.

As far as the time evolution of energy eigenstates, we know the evolve in a trivial manner (simple oscillatory factor). I need another hint on the position eigenstate. How do I apply the evolution operator, which depends on time, to ##\delta(x-x_0)## which does not depend on time at all?

fog37 said:
How do I apply the evolution operator, which depends on time

Not in the Schrodinger picture. So you can use that version of the operator.

fog37 said:
to ##\delta(x-x_0)## which does not depend on time at all?

"Does not depend on time" is tricky; that's what I've been trying to get across. The fact that this does not appear to be an explicit function of ##t## is deceiving. Try applying the Schrodinger picture time evolution operator to it.

Hello,

I am still thinking about this problem, i.e. the time evolution of the position eigenstates of the particle in the box. The position eigenstates for the 1D particle in the box are the function ##\delta(x-x_0)## where ##x_0## is any ##x## in the interval ##0\le x\le L##. The spectrum of the position operator for the 1D particle in the box is not discrete but continuous in a finite range.

So we know that the Schrodinger equation is:

$$\frac {\partial {\Psi(x,t)} } {\partial {t}} = \frac {\hat H(t)}{i \hbar} \Psi(x,t)$$
The solution to this equation is $$\Psi(x,t) =e^{-\frac {i}{\hbar} \hat H(t) (t-t_0)} \Psi(x, t_0)$$

So far so goo. If the Hamiltonian is time-independent, then ##\hat H \neq \hat H(t)##, the equation to find the future state of ##\Psi(x)## becomes

$$\Psi(x,t) =e^{-\frac {i}{\hbar} \hat H(t-t_0)} \Psi(x, t_0)$$

In the position representation, the energy eigenstates for the particle in the box are ##\Psi(x,t)= [\sqrt \frac {2}{L} sin(n\pi x)] e^{i\omega t}##. If we substitute this expression in the equation above we get that

$$\Psi(x,t) =e^{-\frac {i}{\hbar} \hat H (t-t_0)} [\sqrt \frac {2}{L} sin(n\pi x)] e^{i\omega t_0}$$

$$\Psi(x,t) = [\sqrt \frac {2}{L} sin(n\pi x)] e^{-\frac {i}{\hbar} \hat H (t-t_0)+i\omega t_0}$$

I am struggling with the term ##e^{-\frac {i}{\hbar} \hat H (t-t_0)+i\omega t_0} ## in the expression above with the operator ##\hat H## at the exponent. How does it become equal to ##e^{i\omega t}## leading to

$$\Psi(x,t)= [\sqrt \frac {2}{L} sin(n\pi x)] e^{i\omega t}$$

?

Once I get clear on this, then I may have a chance at understanding how the eigenstates of other operators evolve in time for the particle in the box problem...

Thanks for any hint.

fog37 said:
The solution to this equation is

No, a solution if the Hamiltonian is time-independent is... If ##\hat{H}(t)## actually does change with ##t##, then when you take the time derivative, you get an extra term in ##d\hat{H} / dt## that is not present in your solution. This doesn't affect the rest of your post since you are only considering the time-independent case; but as I've said before, you need to be careful not to generalize beyond the actual domain of validity of the expressions you're using.

fog37 said:
I am struggling with the term ##e^{-\frac {i}{\hbar} \hat H (t-t_0)+i\omega t_0}## in the expression above with the operator ##\hat H## at the exponent. How does it become equal to ##e^{i\omega t}##

If ##\Psi## is an eigenstate of ##\hat{H}##, then ##\hat{H} \Psi = E \Psi##. So ##(1 / \hbar) \hat{H} \Psi = (1 / \hbar) E \Psi = \omega \Psi##. You should be able to get from that to the result you are looking for.

(Btw, you also need a minus sign in the exponent in your final expression, i.e., ##e^{- i \omega t}##.)

fog37 said:
Now, given that I know the expression for the linear momentum operator, which is ##-i \hbar \frac \partial {\partial x}##, and that its eigenfunctions must satisfy the eigenvalue equation ##-i \hbar \frac \partial {\partial x} \Phi(x) = p \Phi(x)##, which is how do we go about finding these functions ##\Phi(x)##?
On what functions is this operator defined? Is it self-adjoint?

PeterDonis,

thanks. I will complete the derivation.

By the way, last night I was reading how the time evolution is a unitary transformation. In general, unitary transformations affect the space and time properties of the system and don't change the state of the system. If the time evolution acts on a function that is an eigenstate of an operator that does not share eigenstates with the generator of the evolution operator (which is the Hamiltonian), a relative phase is given to each eigenstates in the superposition but the total probability of that state is not altered.

However, if the state is an eigenstate or a superposition of eigenstates of an operator that does not share any eigenstate with the evolution operator, then the probability of the future state ##\Psi(t)## is different from the probability of the state ##\Psi(t_0)##. A relative phase is applied to the eigenstates in the superposition but this produce a change in the probability.

Ok,

you are right, the actual solution, only for a time-independent ##\hat H##, is given by:

$$\Psi(x,t) = e^{- \frac {i} {\hbar} \hat H (t-t_0)} \Psi(x,t_0)$$

where ##\Psi(x, t_0)= [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) ] e^{ - i \omega t_0}##

So by substitution, we get $$\Psi(x,t) = [e^{- \frac {i} {\hbar} \hat H (t-t_0)} ] [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) e^{- i \omega t_0} ]$$

Based on your hint, ##\frac {1} {\hbar} E \Psi = \omega \Psi##, we get
$$\Psi(x,t) = [e^{- \frac {i} {\hbar} E (t-t_0)}] [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) e^{- i \omega t_0} ]$$

$$\Psi(x,t) = [e^{- \omega (t-t_0)}] [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) e^{- i \omega t_0} ]$$

$$\Psi(x,t) = [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) e^{-i \omega t} ]$$

Finally! This is correct. What helped me was your hint and the fact that the evolution operator is an exponential operator that can be expressed as a Taylor series. By looking at the terms of the Taylor series I realized that we can replace the Hamiltonian operator ##\hat H## in the exponential by ##E##. This does not really mean that ##\hat H## and ##E## are the same thing but in the context of the operation on a wavefunction they produce the same result so the substitution is fair to do...

So we have shown that the energy eigenstates (expressed in position representation) evolve in time but their associated probability distribution does not change in time since the factor ##e^{-i \omega t} ## is just a phase factor that does not modify the squared modulus of a complex number.

Now, let's return to see how a position eigenstate, which is a non normalizable state, evolves in time. For example, in the case of a 1D particle in the box, the particle remains confined to the interval ##0 \le x \le L## where ##L## is the width of the box. That leads me to conclude that the uncertainty ##\Delta x## does not change with time.

Let's repeat the process above for the position eigenstate:

$$\Psi(x,t) = e^{- \frac {i} {\hbar} \hat H (t-t_0)} \Psi(x,t_0)$$

where I think the position eigenstate at time ##t_0## ##\Psi(x, t_0)= \delta (x-x_0) \delta(t-t_0)##.
Is ##\Psi(x,t_0)## the correct wavefunction to start with? I believe so since ##\delta (x-x_0) \delta(t-t_0)## seems to indicate a specific position ##x_0## at a specific instant of time ## t_0 ##.

##\Psi(x, t_0)= [\sqrt \frac {2}{L} sin (\frac {n \pi x} {L}) ]## is a linear superposition of position eigenstates...

fog37 said:
in the case of a 1D particle in the box, the particle remains confined to the interval ##0 \le x \le L## where ##L## is the width of the box. That leads me to conclude that the uncertainty ##\Delta x## does not change with time.

Yes, but that is not telling you that a particle that is in a position eigenstate at time ##t_0## will not change its state. It's telling you that if the particle is confined in the box, its state cannot be a position eigenstate.

fog37 said:
where I think the position eigenstate at time ##t_0## ##\Psi(x, t_0)= \delta (x-x_0) \delta(t-t_0)##.

No. A delta function in time would mean the particle only exists at ##t = t_0##, not at any other time! That's not what you want. Also, you are assuming that the ##x## dependence of ##\Psi## will be a delta function at all times, but that is not a valid assumption (as we'll see).

If you want to try to express the time dependence of the state as a function, i.e., the ##t## dependence of ##\Psi(x, t)##, you first have to find out what it is! You could do that by guessing: we know that for a free particle, a position eigenstate is a superposition of momentum eigenstates (and hence energy eigenstates), because, roughly speaking, ##\delta(x) = \int dp e^{ipx}## (with infinite limits of integration, and I've left out a constant in front that can be chosen based on your desired normalization). But momentum eigenstates are also energy eigenstates, so you can rewrite the superposition in terms of energy eigenstates, whose form we know. Putting the particle in the box then just makes the integral a sum since the energy spectrum is now discrete. Then you can apply the time evolution operator to the superposition, since you know how it acts on each energy eigenstate. What do you think you will find? (Hint: each energy eigenstate "oscillates" at a different frequency, so time evolution will change how they superpose.)

But there is another way to proceed: start with the function ##\delta(x - x_0)## without worrying about time dependence. This is equivalent to assuming that, if we evaluate the general function ##\Psi(x, t)## at ##t = t_0##, without specifying any value of ##x## (i.e., we take whatever the general formula is and plug ##t = t_0## into it), we get the function ##\delta(x - x_0)##--without assuming that the ##x## dependence will still be a delta function at other values of ##x##. The Hamiltonian operator is

$$\hat{H} = \frac{\hat{p}^2}{2m} + V(x) = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)$$

Now plug those two things into the time evolution equation and see what happens.

Well, I agree that the delta function, position eigenstate, is equal to

$$\delta(x)= \int \frac {1}{\sqrt {2\pi \hbar}} e^{ikx}dx$$

and that the Hamiltonian and momentum operator commute so the momentum eigenstates are also eigenstates for ##\hat H##. Question: the Hermitian depends on the problem and the potentials ##V(x)##. Is it always true that ##[ \hat H, \hat p]=0## , i.e. they commute?

Now, $$\Psi(x,t) = e^{-\frac {i}{\hbar} \hat H (t-t_0)} \int \frac {1}{\sqrt {2\pi \hbar}} e^{ikx}dx$$

I know that momentum and kinetic energy (not total energy) are related: ##KE =\frac {p^2}{2m}## and ##p= \hbar k##. Should I use this relationships in the expression for ##\Psi(x,t)##?

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