Particle in the box eigenfunctions

[DrDu]

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?

 

[fog37]

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.

 

[fog37]

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...

 

[PeterDonis]

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.

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.

 

[fog37]

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 $\Psi(x,t)$?

 

[PeterDonis]

Is it always true that $[ \hat H, \hat p]=0$ , i.e. they commute?

AFAIK, yes.

$p= \hbar k$.

No, the momentum operator is $\hat{p} = - i \hbar \partial / \partial x$. Squaring that gives the kinetic energy operator, which you will see in the expression I wrote down for $\hat{H}$ in my last post.

The expression $p = \hbar k$ is just a way of defining a "wave number" $k$ for a momentum eigenstate with momentum $p$.

 

[DrDu]

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?

This depends on the space of functions you use. Is H defined on the Hilbert space of square integrable functions from minus infinity to plus infinity or is it defined on the square integrable functions on the range L, only? That's your choice.
In the first case, the commutator is nonzero, as the potential changes at 0 and L, and p contains a derivative.
In the second case, the momentum operator is not well defined, because $\exp (ipa) \psi(x)=\psi(x+a)$, i.e. the momentum operator will shift a function out of its range of definition.

 

[fog37]

Hello,

Ok, so I agree that $\hat H$ is the sum of the momentum^2 operator divide by $2m$ and the potential energy operator $V(x)$.

• I will replace $\hat H$ in the exponent of the evolution operator with the the expression $\frac{-\hbar^2}{2m} \frac {\partial ^2}{\partial x^2}$ and set $V(x)=0$ since we are in the region inside the potential well:

$$ \Psi(x,t) = e^{ - \frac{\hbar} {2m} \frac {\partial ^2}{\partial x^2} (t-t_0) } \int e^{ikx} dx$$

Would it be ok to replace $\frac {\partial ^2}{\partial x^2} (t-t_0)$ in the exponential by $\hbar ^2 k^2$? Otherwise, I feel like I am stuck with an operator in the exponential and unless I make a replacement I don't know how to use it...

 

[PeterDonis]

Would it be ok to replace $\frac {\partial ^2}{\partial x^2} (t-t_0)$ in the exponential by $\hbar ^2 k^2$?

The $\hbar$ factor comes from the $\hbar$ that is already present in the exponential. There is a way to get a factor of $k^2$ from $\partial^2 / \partial x^2$, by applying it to a particular function. Can you see which function, and where it appears in the expression you wrote down?

I feel like I am stuck with an operator in the exponential

You also have an exponential inside the integral. Can you see how to use that, in combination with the hint I gave above?

 

[fog37]

Ok. If we consider the function $e^{ikx}$, which is a momentum eigenfunction and also an energy eigenfunction, the operator $\frac {\partial^2} {\partial x^2}$ acting on that eigenfunction produces

$$\frac {\partial^2} {\partial x^2} e^{ikx} = - k^2 e^{ikx}$$

so $$\Psi(x,t) = e^{\frac {\hbar}{2m k^2} (t-t_0) } \int e^{ikx} dx$$

$$\Psi(x,t) = \int e^{ikx \frac {\hbar}{2m k^2} (t-t_0)} dx$$

This means that each momentum eigenfunction receives a different time dependent $(t-t_0)$ and $k^2$ dependent relative phase shift. If the phase shift associated to each momentum eigenfunction was the same, the position eigenstate $\delta(x)$ would be a stationary state. But since the relative phase terms are all different, this imply that the probability associated to the new state $\Psi(x,t)$ is different than the probability of the state $\Psi(x,t_0)$...

Did I succeed? I think so.

 

[PeterDonis]

If the phase shift associated to each momentum eigenfunction was the same, the position eigenstate $\delta(x)$ would be a stationary state. But since the relative phase terms are all different, this imply that the probability associated to the new state $\Psi(x,t)$ is different than the probability of the state $\Psi(x,t_0)$

Yes, exactly.

 

[jtbell]

$$\frac {\partial^2} {\partial x^2} e^{ikx} = - k^2 e^{ikx}$$

so $$\Psi(x,t) = e^{\frac {\hbar}{2m k^2} (t-t_0) } \int e^{ikx} dx$$

Um... do you really want that $k^2$ to be in the denominator of the exponent?

$$\Psi(x,t) = \int e^{ikx \frac {\hbar}{2m k^2} (t-t_0)} dx$$

And what do you do with the exponents when you multiply two exponentials that have the same base? e.g. $e^a e^b = \cdots$

Also, your integral needs to be with respect to $k$, not $x$. Your representation of the delta function is actually a definite integral: $$\delta(x)= \int^{+\infty}_{-\infty} \frac {1}{\sqrt {2\pi \hbar}} e^{ikx}dk$$ Taking a definite integral "integrates out" the variable of integration, in this case $k$, by evaluating the limits.

 

[fog37]

Thanks jtbel,

Lots of mistakes indeed:

so $$\Psi(x,t) = e^{\frac {\hbar k^2}{2m} (t-t_0) } \int e^{ikx} dk$$

$$\Psi(x,t) = \int e^{ikx+ k^2 (t-t_0) \frac {\hbar}{2m}} dk$$

This should be correct. The conclusion is the same: each momentum eigenstates in the integral is multiplied by a different phase factor altering the entire probability density function...

 

[fog37]

Great.
Thanks everyone, especially PeterDonis, for the fruitful and steady discussion and for making me work through the conclusions myself.

I enjoyed expressing the position eigenfunction $\delta(x)$ as a summation of momentum eigenstates $e^{ikx}$ to prove that the $\delta(x)$ is not a stationary state for the particle in the 1D infinite box. Could we dare to say that the $\delta(x)$ is never a stationary state for any other physical situation and Hamiltonian, time dependent or not that it may be?
Also, I am still not clear on the other method that PeterDonis suggested to prove the same result and I would be interested in understanding that method too.

I hope there are other members following this thread and making as much progress as I am. The saga continues:

• As we know, a stationary state is a state whose associated probability density function $|\Psi|^2$ does not change with time. Eigenstates of a certain time-independent operator, for a certain physical system, can be stationary states but not all eigenstates of all time-independent operators are stationary states. A state that is not an eigenstate can also be stationary. For example, a superposition of stationary energy eigenstates is not an eigenstate but should also be a time stationary state.

• Spatial confinement of the system determines the discreteness of the spectra of certain observables (but not all). For the particle in the 1D infinite box we saw that the energy operator and the momentum operator have a discrete spectrum while the position operator does not. The time stationary (of certain eigenstates0 is instead determined by the Hamiltonian of the system being time independent.

• For the particle in the 1D infinite box, the operators of interest are mainly the energy operator and the momentum operator. There is also the position operator but that does not seem to receive the same attention. Is there any other operator (observable) that I am forgetting and should be included in the list? I know that the angular momentum operator does not apply to 1D situations but only to 2D and 3D systems.

• For the particle in the 1D infinite box, the energy eigenvalue equation came up easily and naturally from Schrödinger equation once we set the potential $V(x)=0$ in the Hamiltonian. What about the eigenvalue equation for the linear momentum and the eigenvalue equation for position operator? Are the functions $$\delta(x-x_0)$$ always the position eigenfunctions for all physical systems? Same goes for the momentum eigenfuctions $e^{ikx}$? The energy eigenfunctions, instead, change depending on the form of the Hamiltonian which changes depending on the potential $V(x,t)$ that we have to deal with.

• Both the operators and the states of a physical system can have different representations according to the basis of other operators. For instance, the momentum operator can have a position representation, the position operator can have a momentum representation, the energy operator commonly has a position representation but nothing forbids it from having a momentum or an angular momentum representation. Is that correct? I guess we run into trouble when we try to have the mentioned operator have a spin representation. Certain operators (almost all of them) are vector operators and have multiple operator components. For example, the momentum operator is $\hat p = (\hat p_x, \hat p_y, \hat p_z)$. Does that mean we could three momentum representations, one for each momentum component, i.e. $\Psi(p_x)$, $\Psi(p_y)$, $\Psi(p_z)$ ?

Thanks,
Fog 37

 

[PeterDonis]

Could we dare to say that the $\delta(x)$ is never a stationary state for any other physical situation and Hamiltonian, time dependent or not

As far as I know this is the case.

a superposition of stationary energy eigenstates is not an eigenstate but should also be a time stationary state.

No, it isn't; you've already proven this false. Can you see why? (Hint: momentum eigenstates are also energy eigenstates.)

the momentum operator can have a position representation, the position operator can have a momentum representation, the energy operator commonly has a position representation but nothing forbids it from having a momentum or an angular momentum representation. Is that correct?

Almost. Any set of mutually orthogonal states that span the Hilbert space can be used to construct a representation. The set of all position eigenstates, the set of all momentum eigenstates, and the set of all energy eigenstates all meet this requirement. So does (AFAIK) the set of all eigenstates of orbital angular momentum for particles with zero spin. But I'm not sure about angular momentum when we include particles with nonzero spin.

Does that mean we could three momentum representations

No, because in 3 dimensions, the eigenstates of momentum in only one direction do not span the Hilbert space (think of all the possible states that have momentum in the other two directions but zero momentum in the chosen direction--they are all mutually orthogonal but cannot be expressed in terms of momentum eigenstates in the chosen direction, since they all look the same in terms of those states).

 

[fog37]

Ok,

My (bad) mistake on the stationarity point. What I really meant, after re-reading my words, is that it is possible for a state that is not an eigenstate of a certain operator to be a stationary. Of course, if a state is not an eigenstate then it is a superposition of eigenstates for that specific operator. That same state may be an eigenstate for a another operator though...

I will have to read your comment more than once since I am not fully clear on it:

As far as I know this is the case.
No, because in 3 dimensions, the eigenstates of momentum in only one direction do not span the Hilbert space (think of all the possible states that have momentum in the other two directions but zero momentum in the chosen direction--they are all mutually orthogonal but cannot be expressed in terms of momentum eigenstates in the chosen direction, since they all look the same in terms of those states).

In 1D, where momentum is one-dimensional, we only have one component, say $\hat p_x$, and there is no problem representing the state as $\Psi(p_x)$.
In 2D or 3D we know that the momentum components do not commute. If I understand your answer correctly, you are saying that the eigenstates of the momentum component operator #\hat p_x# do not form a complete set and cannot represent any of the momentum eigenstates of the other two momentum operators, either $\hat p_y$ or $\hat p_z$... Does that mean that in 3D we need to choose one among the operator components if we want to express the state $|\Psi>$ in the momentum representation as a wavefunction?

 

[PeterDonis]

it is possible for a state that is not an eigenstate of a certain operator to be a stationary.

For some operators, yes; for example, any stationary state is not an eigenstate of the position operator. But you have to figure that out for every specific operator.

In 2D or 3D we know that the momentum components do not commute.

That's incorrect. The linear momentum operators in different directions do commute. The angular momentum operators don't.

 

[houlahound]

Just wanted to say what a great honest thread.

 

[DrDu]

I will replace $\hat H$ in the exponent of the evolution operator with the the expression $\frac{-\hbar^2}{2m} \frac {\partial ^2}{\partial x^2}$ and set $V(x)=0$ since we are in the region inside the potential well:

You cannot do this: the eigenfuntions of tje p operator extend over all the space, not only the region inside the box.

 

[fog37]

Hello DrDu.

So, if that is a mistake, does it mean that the rest that follows also has mistakes?

 

[DrDu]

Yes, your hamiltonian is not defined on momentum eigenstates. You basically discussed the evolution under the free particle hamiltonian, not the box hamiltonian. You could try to repeat your analysis expressing the delta function in terms of the eigenstates of the box hamiltonian.

 

[fog37]

I see. But isn't the Hamiltonian where the particle exists, inside the well, essentially equal to the free-particle Hamiltonian since $V(x)=0$ in that region? That seems to be what is done to find the energy eigenstates. Setting $V(x)=0$ is how the energy eigenvalue equation arises to find those energy eigenstates...

 

[PeterDonis]

where the particle exists, inside the well

Are you assuming an infinitely high potential outside the well? In other words, $V(x) = \infty$ outside $-\frac{L}{2} \le x \le \frac{L}{2}$?

 

[fog37]

Yes, that I have been thinking/assuming that infinite potential at the edges of the well.

 

[PeterDonis]

I have been thinking/assuming that infinite potential at the edges of the well.

Then that won't affect the energy eigenvalues, but it will affect the relationship between energy eigenstates and other eigenstates. For example, a momentum eigenstate won't just be $e^{ikx}$, since the wave function in any representation has to vanish outside the box, and that function doesn't. I don't think that affects the general arguments in this thread (for example, the general argument for why position eigenstates can't be stationary), but it does affect specific details like the form of the wave functions.

 

[fog37]

Well, if that is the case, even the accepted energy eigenfunctions in the form $$\sqrt \frac {2}{L} sin(\frac {\pi n x}{L})$$ are not mathematically acceptable since a sine is not zero outside the walls of the well. The function should be multiplied by another function that renders it zero (step function)...

 

[DrDu]

Either this or you consider from the very beginning only functions on 0<x<L.

 

[DrDu]

You could express a delta function as $\delta(x-a)=\sum_{n=1}^\infty \psi_n(a)\psi_n(x)$ where $\psi_n(x)=\sqrt{2/L} \sin(\pi n x/L)$. Then it is easy to write down its time dependence.

 

[houlahound]

I can read all the equations in this thread, the last post is unreadable on my screen?

 

[DrDu]

I can read all the equations in this thread, the last post is unreadable on my screen?

Sorry, I did correct the Latex tags.