Understanding Momentum Eigenfunctions in Quantum Mechanics

[snoopies622]

I'm enjoying this introductory essay about quantum mechanics found here

http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/psi.html

and I have a question. About five-eighths of the way into it a wave function is given "at time t=0",

<br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br /> <br />

and some questions and answers follow. If I am understanding the authors, the answers imply that this wavefunction is a (normalized) linear combination of two momentum eigenfunctions, where the momenta are h/2L and 5h/2L.

My question is, shouldn't

<br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} (cos (\pi x / L) + i sin (\pi x / L)) + \frac { \sqrt {3} }{2} (cos (5 \pi x / L)<br /> + i sin (5 \pi x / L))]<br /> <br />

or - more succinctly -

<br /> <br /> <br /> \psi = \sqrt {\frac {2} {L} } [ \frac {1}{2} e ^ {i \pi x / L } + \frac { \sqrt {3} } {2} e ^ {i 5 \pi x / L } ]<br /> <br />

?

 

[snoopies622]

Oops - I just looked backward and noticed that this is one of those infinite potential well situations. That explains why both the real and imaginary components of \psi have to be sine waves here.

And yet, how do I square this fact with the postulate that

<br /> e ^ { ipx/ \hbar }<br />

- not sin (px / \hbar ) - are the momentum basis states?

 

[sonoluminated]

Write the sin as exp(ipx) - exp(-ipx), so isn't the sine just a different basis?

 

[sweet springs]

Hi.

If I am understanding the authors, the answers imply that this wavefunction is a (normalized) linear combination of two momentum eigenfunctions, where the momenta are h/2L and 5h/2L.

No, the answers imply that this wavefunction is a (normalized) linear combination of two ENERGY eigenfunctions, not momentum.
Regards.

 

[snoopies622]

Write the sin as exp(ipx) - exp(-ipx), so isn't the sine just a different basis?

Ah, yes - that did it. Thanks sonoluminated.

No, the answers imply that this wavefunction is a (normalized) linear combination of two ENERGY eigenfunctions, not momentum.

When the potential energy is zero, don't those amount to the same thing?

 

[sweet springs]

When the potential energy is zero, don't those amount to the same thing?

Hi.
When the potential energy is not zero but infinite square well or particle in a box, that is I thought your case is ( Am I wrong? ) consideration of momentum is shown in the text of Schubert.
http://www.ecse.rpi.edu/~schubert/C...um mechanics/Ch03 Position&momentum space.pdf
Regards.

 

[snoopies622]

That's a nice source, thanks sweet springs.

What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.

 

[sweet springs]

Hi, snoopies622.

What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.

Shubert (3.22),(3.23) and Figure 3.3 show us that the discrete energy eigenstate is superposition of continuous momentum eigenstates.
Energy at somewhere e.g. inside the well means simultaneous consideration of energy E or Hamiltonian H and position X. Commutation relation [H,X] ≠0 does not allow such consideration.
Regards.

 

[snoopies622]

...the discrete energy eigenstate is superposition of continuous momentum eigenstates.

I haven't studied the math yet but conceptually I am confused. In a place where there is no force acting on a particle and the potential energy is zero, the total energy must be the kinetic energy, which is a function of the momentum. How then can discrete energy states not imply discrete momentum states, and vice versa?

 

[sweet springs]

Hi, snoopies622.
I say here not using mathematics so do not take it so strict. See ψ0 in Shubert Figure 3.2. that is the ground energy state. You see near the well walls particle rarely exists and around center of the well particle is most probably found. This behavior is different from classic particle whose probability density is uniform in the well. The particle seems to be knowing the distances from the walls though local information e.g. potential energy does not change as you stated;

In a place where there is no force acting on a particle and the potential energy is zero, the total energy must be the kinetic energy, which is a function of the momentum.

In quantum physics such a global behavior replaces classical localism. In quantum physics amount of energy cannot be distinguished to kinetic and potential parts according to where the particle is.
Regards.

 

[snoopies622]

By "in a place" I didn't mean at a particular location inside the well, just inside the well. That is, at any location inside the well, the potential energy is zero and therefore all the energy is kinetic energy. I think we agree on this.

 

[sweet springs]

Hi, snoopies622

By "in a place" I didn't mean at a particular location inside the well, just inside the well. That is, at any location inside the well, the potential energy is zero and therefore all the energy is kinetic energy. I think we agree on this.

In quantum mechanics, even if potential energy is zero anywhere inside the well, energy eigenstate is under the influence of the potential beyond the walls. I called it "global" in my previous post. Tunnel effect is on the same footing.
Regards.

 

[snoopies622]

...the discrete energy eigenstate is superposition of continuous momentum eigenstates.

The problem is that \psi as described above

<br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br /> <br />

can be expressed equivalently as

<br /> <br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {i}{4} (e ^ {-i \pi x / L} - e ^ {i \pi x / L}) + \frac { \sqrt {3} }{4} (e ^ {5i \pi x / L} - e ^ {-5i \pi x / L})]<br /> <br /> <br />

which is a linear combination of four momentum eigenfunctions,

<br /> <br /> e ^ {-i \pi x / L} \quad<br /> e ^ {i \pi x / L } \quad<br /> e ^ {5i \pi x / L} \quad<br /> e ^ {-5i \pi x / L} <br />

not a continuum of them.

 

[sweet springs]

Hi. Snoopoies622
You are right in case the formula stands from -infinity x to +infinity x. I am afraid you are not right in case the formula stands only within the well and turns to zero for the outside. Which one is your case?
Regards.

 

[Dickfore]

Oops - I just looked backward and noticed that this is one of those infinite potential well situations. That explains why both the real and imaginary components of \psi have to be sine waves here.

And yet, how do I square this fact with the postulate that

<br /> e ^ { ipx/ \hbar }<br />

- not sin (px / \hbar ) - are the momentum basis states?

Because when the particle is in an external potential, the momentum does not commute with the total Hamiltonian of the particle and, hence, it does not have a definite value in a stationary state.

 

[snoopies622]

You are right in case the formula stands from -infinity x to -infinity x. I am afraid you are not right in case the formula stands only within the well and turns to zero for the outside

I was wondering about that...

Because when the particle is in an external potential, the momentum does not commute with the total Hamiltonian of the particle and, hence, it does not have a definite value in a stationary state.

So when the energy of the particle (inside the well) is measured to be E _n we cannot assume that its momentum is therefore <br /> <br /> \pm \sqrt {2m E_n }<br /> <br /> ?

 

[sweet springs]

Hi. snoopies622

So when the energy of the particle (inside the well) is measured to be E _n we cannot assume that its momentum is therefore <br /> <br /> \pm \sqrt {2m E_n }<br /> <br /> ?

No, in case the system has potential energy so there is a well.
Yes, in case the system has no potential energy so there is no well.
Regards.

 

[snoopies622]

This surprises me. I always looked at the particle-in-a-one-dimensional-box situation like this:

The wave must fit neatly inside the box, so the wavelength must be 2L/n where n is integer. Therefore, the particle's momentum must be

<br /> <br /> p = \frac {h}{\lambda} = \frac {hn}{2L}<br /> <br />

and its energy

<br /> <br /> E_n = \frac {p^2}{2m} = \frac {h^2 n^2}{4 L^2} \frac {1}{2m} = \frac {\hbar ^2 \pi ^2 }{2 m L^2} n^2<br /> <br />

 

[sweet springs]

Hi, snoopies622.
Fig.1 is wave function of momentum eigenstates p=h'/λ and p=-h'/λ superposed in equal weight.
Fig.1
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Fig.2 is wave function of an energy eigenstate in infinite square well potential.
Fig.2
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Though Fig.1 and Fig.2 coincide within the well region, They are different states. Fourier transform of Fig.2 gives continuous momentum spectrum. See Schubert.
You can rewrite En as pn^2/2m where pn is a quantity whose dimension is momentum, but pn is not momentum of the system. However as classical limit taking n large, the state momentum tends to +-pn. See Schubert Fig3.2.
Regards.

 

[snoopies622]

I think I understand now. I have one more question: what does the Schrodinger equation look like in the momentum basis (instead of the position basis)?

 

[sweet springs]

Hi, snoopies622.
Shrodinger equations of stationary state for Hamiltonian H(p,x) are H(h'/i d/dx, x) ψ(x)=Eψ(x) in coordinate space and H(p, ih'd/dp) φ(p)=Eφ(p) in momentum space where φ(p) is Fourier transform of ψ(x).
Regards.

 

[snoopies622]

Thanks for that, sweet springs.

I'm still having a problem with the mathematical equivalence I mentioned in entry 13. I see how neither representation takes into account the boundaires of x=0 and x=L, and that this is why the momentum representation (the second equation) doesn't work. But why then does the first one? I assume that a function that is sinusoidal between two x values and then suddenly zero outside of that domain is much more complicated than the sum of either two or four sinusoidal terms.

 

[Dickfore]

Schroedinger Equation in momentum basis is:

<br /> \frac{p^{2}}{2 m} \, a(\mathbf{p}) + \int{\frac{d^{3}p}{(2 \pi \hbar)^{3}} \, \tilde{U}(\mathbf{p} - \mathbf{p}&#039;) \, a(\mathbf{p}&#039;) = E \, a(\mathbf{p})<br />

where

<br /> a(\mathbf{p}) \equiv \int{d^3{r} \, \psi(\mathbf{r}) e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}<br />

is the wave function in momentum space. It's physical significance is that:

<br /> |a(\mathbf{p})|^{2} \frac{d^{3}p}{(2 \pi \hbar)^{3}}<br />

is the probability that the particle will have a value of momentum inside an elementary cube of volume d^{3}p around the value \mathbf{p}, and

<br /> \tilde{U}(\mathbf{p}) = \int{d^{3}r \, U(\mathbf{r}) \, e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}<br />

is the spatial Fourier transform of the potential energy.

If the potential depends only on one Cartesian coordinate (denoted by x), then:
<br /> \tilde{U}(\mathbf{p}) = \tilde{U}(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y}) \, \delta(p_{z})<br />

and we can choose a wave-function with a definite value of the components of the momenta in the y and z direction:

<br /> a(\mathbf{p}) = a(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y} - p_{y0}) \, \delta(p_{z} - p_{z0})<br />

Then,

<br /> |a(p_{x})|^{2} \frac{dp_{x}}{2 \pi \hbar}<br />

is the probability that p_{x} has a value in the interval (p_{x}, p_{x} + dp_{x}). The Schroedinger equation becomes one dimensional:

<br /> \frac{p_{x}^{2}}{2 m} \, a(p_{x}) + \int{\frac{dp_{x}}{2 \pi \hbar} \, \tilde{U}(p_{x} - p&#039;_{x}) \, a(p&#039;_{x})} = \epsilon \, a(p_{x}), \ \epsilon = E - \frac{p^{2}_{y} + p^{2}_{z}}{2 m}<br />

 

[sweet springs]

Hi. snoopies622

I see how neither representation takes into account the boundaires of x=0 and x=L, and that this is why the momentum representation (the second equation) doesn't work. But why then does the first one?

Both the representation take into account the potential energy generating boundaries at x=0 and x=L. The coordinate representation is easier to solve because equation includes up to (d/dx)^2. In the momentum representation equation (d/dp)^n of any large n appear corresponding to power series expression of V(x). Dickfore discussed it in another way using Fourier transform of the potential energy.
Regards.

 

[snoopies622]

Wow, thanks Dickfore. I'll have to give all that some time.

The coordinate representation is easier to solve...

Actually, the two representations I had in mind in my previous question were not position vs. momentum. They were both in position representation. (See entry #13.) The first was a linear combination of energy eigenfunctions, and the second was a linear combination of momentum eigenfunctions. I was wondering why one could use the first one to find the particle's allowed energy values, but not the second one to find its allowed momentum values.

 

[sweet springs]

Hi. snoopies622 From your post #13

The problem is that \psi as described above

<br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br /> <br />

inside the well and zero outside, I remind you again. This can be expressed equivalently as 1/h＾1/2∫a(p)e^ipx/h' dp where a(p)=1/h^1/2∫ψ(x)e^-ipx/h' dx where h=2πh', Fourier transformation.

not a continuum of them.

So continuum of momentum eigenfunctions, i.e. 1/h＾1/2∫a(p)e^ipx/h' dp.
You don't have to solve two different equations of different representation of an energy eigenstate independently. Once you get a solution in one representation, Fourier transform of it becomes the solution in other representation.

I was wondering why one could use the first one to find the particle's allowed energy values, but not the second one to find its allowed momentum values.

On the state ψ(x), when you want to know about energy, energy representation is convenient. When you want to know about momentum, momentum representation is convenient. When you want to know about position, coordinate representation is convenient, so on.
Regards.

 

[snoopies622]

...when you want to know about energy, energy representation is convenient. When you want to know about momentum, momentum representation is convenient. When you want to know about position, coordinate representation is convenient, so on.

I'm not sure I'm completely following you here. Both equations I mentioned in entry 13 are in the position basis, and both are linear combinations of eigenfunctions. The first one uses energy eigenfunctions, the second one momentum eigenfunctions. One can look at the first equation and see what the allowable energy values will be, but not the second equation and see what the allowable momentum values will be. Why the difference? I assume that they both take the boundary conditions into account.

 

[sweet springs]

Hi, snoopies622

Both equations I mentioned in entry 13 are in the position basis, The first one uses energy eigenfunctions,

Yes. You see energy eigenfunctions has non zero value only within the well region. This feature is kept even if we rewrite it in another form, so,

the second one momentum eigenfunctions.

No, the second one does not use momentum eigenfunctions. Eigenfunction of momentum is e^ikx for all the regions of x.　
Function that is e^ikx for some region and zero outside of it is not eigenfunction of momentum that is the case of your second one. If you want to rewrite the formula with momentum eigenfunctions, you should apply Fourier transform.
Regards.

 

[snoopies622]

Eigenfunction of momentum is e^ikx for all the regions of x. Function that is e^ikx for some region and zero outside of it is not eigenfunction of momentum...

Why does this restriction not apply to energy eigenfunctions as well? They are both sine waves after all, which by definition extend forever in both directions.

 

[sweet springs]

Hi, snoopies622

Why does this restriction not apply to energy eigenfunctions as well? They are both sine waves after all, which by definition extend forever in both directions.

Why not? Read remind to you in my post #26, please. Energy eigenfunctions are sinusoidal waves in the well and are zero outside.
Do not be bored to see the same Figures again. They are NOT like Fig.1
Fig.1 (superposition of two) momentum eigenstate(s)
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but like Fig.2.
Fig.2 an energy eigenstate
...○○○○○○○○○○○○●●○○○○○○○○○...
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Regards.

 

[hmm.max]

I'm not sure if this is what you're looking for, but here's a simple answer-

Even in situations with NO potential energy anywhere, energy eigenstates are not generally momentum eigenstates. A particle's kinetic energy depends on the magnitude, but not the direction of its momentum. So most energy eigenstates are linear combinations of many different momentum eigenstates with the same magnitude but different direction.

In a one-dimensional situation, there are only two directions, positive and negative. So a standing wave like

psi = A sin( k x ) = A/2i ( e^(i k x) - e^(-i k x)

is a superposition of TWO momentum eigenstates, one in each direction. The energy states for the infinite square well have this form. They aren't eigenstates of momentum.

 

[hmm.max]

What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.

You may have resolved this already, but to elaborate on what I just said, there isn't a one to one correspondence, even for zero potential energy. There are many different momentum states with the same energy. So in general, energy states are superpositions of different momentum states. For square potential wells, the wave functions must be standing waves, which means they must always be linear superpositions of at least two momentum states with opposite directions.

Hope this is relevant to something.

-max

 

[snoopies622]

In a one-dimensional situation, there are only two directions, positive and negative. So a standing wave like

psi = A sin( k x ) = A/2i ( e^(i k x) - e^(-i k x)

is a superposition of TWO momentum eigenstates, one in each direction.

Yes, this makes complete sense to me. But two momentum eigenstates is not a continuum of them.

 

[snoopies622]

Energy eigenfunctions are sinusoidal waves in the well and are zero outside.

So you're saying that a simple sine wave like

\psi (x) = sin (kx) = sin ( \frac { \sqrt {2mE} } {\hbar} x )

that goes on forever in both directions - that is, is not bounded at two x values like the ones in the examples - is not an energy eigenfunction?

Edit: Perhaps I should have made

\psi (x) = e ^ {ikx} = e ^{ i( \frac { \sqrt {2mE} } {\hbar} x )}

 

[sweet springs]

Hi, hmm.max

There are many different momentum states with the same energy.

In general case that there exists potential energy, momentum and energy do not commute. Momentum state cannot have definite energy value. Thus it is not appropriate to refer same energy or not.

For square potential wells, the wave functions must be standing waves, which means they must always be linear superpositions of at least two momentum states with opposite directions.

Instead let me say, for energy eigenstates in square potential well of infinite depth, the wave functions must be standing waves WITHIN THE WELL AND ZERO OUTSIDE, which means they must always be linear superpositions OF CONTINUOUS INTEGRALS OF momentum states THAT EACH ONE IS NOT BOUNDED AT TWO VALUES.　Think of Fourier transform please.

Hi, snoopies622.

So you're saying that a simple sine wave like
\psi (x) = sin (kx) = sin ( \frac { \sqrt {2mE} } {\hbar} x )

that goes on forever in both directions - that is, is not bounded at two x values like the ones in the examples - is not an energy eigenfunction?

No, it is not an energy eigenfunction of infinite depth square well potential. Why such an idea is haunting on you? Imagine particle can be outside the well as you say. There potential energy is infinite thus the eigenvalue of this eigenstate cannot be finite. Is this acceptable?

PS
In order to avoid misunderstanding it would be helpful to write wave function of energy eigenstate in the infinite square well potential system as ψ (x) = sin (kx) Θ(x) = 1/2i e^ikx Θ(x) - 1/2i e^-ikx Θ(x) where support Θ(x)=1 within the well region, 0 for outside.
sin (kx) Θ(x), not sin (kx), is energy eigenfunction.
e^ikx, not e^ikxΘ(x), is momentum eigenfunction.

Regards.

 

[hmm.max]

of course, i completely agree, sweet springs. in a non-zero potential, momentum eigenstates don't have definite energy. i was first talking specifically about the case of zero potential, in order to address the difference between a sinusoidal wave function and a momentum eigenstate, in concrete terms.

and yes, for a square potential well, many (infinite) more than two momentum components are required to represent the actual wave function, because it goes to zero outside the well. this is true. i should have focused that comment on standing waves in general, rather than on the energy states of the square well.

 

[hmm.max]

So you're saying that a simple sine wave like

\psi (x) = sin (kx) = sin ( \frac { \sqrt {2mE} } {\hbar} x )

that goes on forever in both directions - that is, is not bounded at two x values like the ones in the examples - is not an energy eigenfunction?

Edit: Perhaps I should have made

\psi (x) = e ^ {ikx} = e ^{ i( \frac { \sqrt {2mE} } {\hbar} x )}

Yeah those are eigenfunctions of the "free particle" hamiltonian (where V = 0 everywhere), but not of the infinite square well hamiltonian.

 

[snoopies622]

Since V(x)=0 inside the well, all this time I've been using

<br /> <br /> - \frac { \hbar ^2} {2m} \frac {\partial ^2 }{ \partial x ^2 }<br /> <br />

as the energy operator instead of

<br /> <br /> - \frac { \hbar ^2} {2m} \frac {\partial ^2 }{ \partial x ^2 } + V (x)<br /> <br /> \vspace {5 mm}.

 

[hmm.max]

In order to avoid misunderstanding it would be helpful to write wave function of energy eigenstate in the infinite square well potential system as ψ (x) = sin (kx) Θ(x) = 1/2i e^ikx Θ(x) - 1/2i e^-ikx Θ(x) where support Θ(x)=1 within the well region, 0 for outside.
sin (kx) Θ(x), not sin (kx), is energy eigenfunction.
e^ikx, not e^ikxΘ(x), is momentum eigenfunction.

Regards.

That's a good way of putting it.

 

[sweet springs]

Hi, snoopies622

Since V(x)=0 inside the well,

How about wave function outside the well? Does it keep to be a sinusoidal wave?
Regards.

 

[snoopies622]

No, ψ (x) = 0 outside the well.

 

[snoopies622]

No, it is not an energy eigenfunction of infinite depth square well potential. Why such an idea is haunting on you?

I understand now, sweet springs. Ψ(x) can be defined in a piecewise manner to solve the Schrodinger equation using the piecewise-defined potential, and this yields energy values, not momentum values. Thank you for all of your efforts.

Here's a follow-up question for you, and it's been at the root of all my confusion in this particle-in-a-box matter. Since inside the box there isn't a one-to-one (or two-to-one) relation between momentum states and energy states, isn't it odd that the method I described in entry #18 for finding the discrete energy values works? It assumes discrete momentum values, then squares them and divides by 2m.

 

[Dickfore]

The nth stationary state is:

<br /> \psi_{n}(x) = \left\{\begin{array}{cl}<br /> \sqrt{\frac{2}{L}} \, \sin\left(\frac{n \pi x}{L}\right) &amp;, \ 0 \le x \le L \\<br /> <br /> 0 &amp;, x &lt; 0 \vee x &gt; L<br /> \end{array}\right., \ n = 1, 2, \ldots<br />

The momentum eigenfuniction is:

<br /> \phi_{p}(x) = \frac{1}{(2 \pi \hbar)^{1/2}} \, e^{\frac{i \, p \, x }{\hbar}}, \ -\infty &lt; x &lt; \infty<br />

These momentum eigenfunctions (plane waves) form a complete orthonormal basis of functions and we can expand any function in them. Let us expand the nth stationary state from above. The expansion is actually an integral and the coefficient is a function of the continuous parameter p:

<br /> a(p) = \int_{-\infty}^{\infty}{\phi^{\ast}_{p}(x) \, \psi_{n}(x) \, dx}<br />

<br /> a(p) = \frac{1}{(2 \pi \, \hbar)^{1/2}} \, \left(\frac{2}{L}\right)^{1/2} \, \frac{1}{2 i} \left[ \int_{0}^{L}{\exp(i (\frac{n \pi}{L} - \frac{p}{\hbar}) \, x) \, dx} - \int_{0}^{L}{\exp(-i (\frac{n \pi}{L} + \frac{p}{\hbar}) \, x) \, dx}\right]<br />

<br /> a(p) = \frac{1}{2 i} \left( \frac{1}{\pi \hbar L} \right)^{1/2} \left[\frac{1}{i (\frac{n \pi}{L} - \frac{p}{\hbar})} - \frac{1}{-i (\frac{n \pi}{L} + \frac{p}{\hbar})}\right] \, ((-1)^{n} exp(-i \frac{p \, L}{\hbar}) - 1)<br />

<br /> a(p) = \frac{n \, \pi}{L} \left(\frac{1}{\pi \, \hbar \, L} \right)^{1/2} \, \frac{1 - (-1)^{n} \exp(-i \, \frac{p \, L}{\hbar})}{(\frac{n \, \pi}{L})^{2} - (\frac{p}{\hbar})^{2}}<br />

The square of the absolute value of these complex numbers gives the probability density for dfifferent values of p. The only complex number is the expression in the numerator. We evaluate its square absolute value as:
<br /> \begin{array}{l}<br /> \left| 1 - (-1)^{n} \exp(-i \, \frac{p \, L}{\hbar}) \right|^{2} = \left( 1 - (-1)^{n} \exp(-i \, \frac{p \, L}{\hbar})\right) \left( 1 - (-1)^{n} \exp(i \, \frac{p \, L}{\hbar})\right) \\<br /> <br /> =1 - (-1)^{n} \exp(-i \, \frac{p \, L}{\hbar}) - (-1)^{n} \exp(i \, \frac{p \, L}{\hbar}) + 1 \\<br /> <br /> = 2 \left[ 1- (-1)^{n} \, \cos\left(\frac{p \, L}{\hbar}\right)\right]<br /> \end{array}<br />

<br /> |a(p)|^{2} = \frac{\frac{n^{2} \, \pi}{\hbar \, L{3}}}{\left(\frac{\pi}{L}\right)^{4}} \, \frac{2 \left[ 1- (-1)^{n} \, \cos\left(\frac{p \, L}{\hbar}\right)\right]}{\left[n^{2} - \left(\frac{p \, L}{\hbar \pi}\right)^{2}\right]^{2}}<br />

<br /> |a(p)|^{2} = \frac{L}{\hbar \, \pi} \, \frac{2 \left(\frac{n}{\pi}\right)^{2} \, \left[1 - (-1)^{n} \, \cos\left( \pi \, \frac{p \, L}{\hbar \, \pi}\right) \right]}{\left[n^{2} - \left(\frac{p \, L}{\hbar \pi}\right)^{2}\right]^{2}}<br />

The denominator has double zeros for:

<br /> \tilde{p} \equiv \frac{p \, L}{\hbar \, \pi} = \pm n<br />

These values correspond to the values for momentum that give "standing De Broglie's waves" with nodes at the walls of the potential well. However, the numerator is also zero for these values and the limit 0/0 can be evaluated by using the L'Hospital's Rule:

<br /> |a(\tilde{p}_{n})|^{2} = \frac{1}{4} \, \frac{L}{\hbar \, \pi}<br />

The plot of this probability distribution is given on the graph below (for n = 1, 2, 3):

As you can see, the distribution has a non-zero, but finite probability density any other (continuous) value for \tilde{p} as well.

 

[snoopies622]

Hey thanks, Dickfore! I had no idea how to calculate those coefficients.

 

[sweet springs]

Hi, snoopies622.

It assumes discrete momentum values, then squares them and divides by 2m.

To adjust discrete wavelength according to the size of well or box should not be regarded as momentum fitting. de Broglie relation h'/λ=p applies to waves of infinite length, not to waves of finite length or wave train. In order to know what momentum wave train has, we have to decompose it to momentum eigenstates by Fourier transform. However even wave train, square of h'/λ gives energy multiplied by 2m in the well where V=0.
Regards.

 

[snoopies622]

The momentum eigenfunction is:

<br /> \phi_{p}(x) = \frac{1}{(2 \pi \hbar)^{1/2}} \, e^{\frac{i \, p \, x }{\hbar}}, \ -\infty &lt; x &lt; \infty<br />

These momentum eigenfunctions (plane waves) form a complete orthonormal basis of functions and we can expand any function in them.

I have a couple questions about this:

1. Why are these vectors not dimensionless? (or are you using a system of units such that Planck's constant is just a number?)

2. What does it mean to say that they are orthonormal? I understand what "orthonormal" means when it refers to a finite-dimensional vector space like \bold {R}^3, and I understand how these vectors are "ortho" at least in the sense that are linearly independent. But what about the "normal" part? Is there a sense in which a sine wave can have unit length?

Thanks.

 

[Dickfore]

1. Why are these vectors not dimensionless? (or are you using a system of units such that Planck's constant is just a number?)

2. What does it mean to say that they are orthonormal? I understand what "orthonormal" means when it refers to a finite-dimensional vector space like \bold {R}^3, and I understand how these vectors are "ortho" at least in the sense that are linearly independent. But what about the "normal" part? Is there a sense in which a sine wave can have unit length?

Eigenfunctions are never dimensionless. In accordance with the probabilistic interpretation of
the wave function, when a particle is in a state described by the wave function \Phi(\mathbf{r}), then |\Phi(\mathbf{r})|^{2} \, d^{d}r is the probability of finding the particle in the infinitesimal volume dV = d^{d}r (d is the dimensionality of space) around the point with position vector \mathbf{r}. SInce probabilities are dimensionless quantities, from here it immediately follows that the dimension of the wavefunction is:

<br /> [\Phi]^{2} \, \mathrm{L}^{d} = 1 \Rightarrow [\Phi] = \mathrm{L}^{-\frac{d}{2}}<br />

Since the total probability of finding the particle somewhere in space must be one, the wave functions ought to be normalized:

<br /> \int{|\Phi(\mathbf{r})|^{2} \, d^{d}r} = 1<br />

Orthonormal is a shorthand amalgamation of two terms coined by lazy physicists :) It stands for orthogonal and normalized.

For observables with continuous spectra (such as momentum), one has to be careful what is meant by "normalization", since the eigenfunctions are not square-integrable. When we say the momemtnum eigenfunctions are orthonormal, we mean they satisfy the following identity:

<br /> \int{\psi^{\ast}_{\mathbf{p}}(\mathbf{r}) \, \psi_{\mathbf{p}&#039;}(\mathbf{r}) \, d^{d}r} = \delta^{(d)}(\mathbf{p} - \mathbf{p}&#039;)<br />

where

<br /> \delta^{(d)}(\mathbf{p} - \mathbf{p}&#039;) = \prod_{i = 1}^{d} {\delta(p_{i} - p&#039;_{i})}<br />

is the d-dimensional Dirac delta-function, with \delta(p_{i} - p&#039;_{i}) being the continuum generalization of the Kronecker delta \delta_{m, n}. A Dirac delta-function has the following fundamental property:

<br /> \int_{-\infty}^{\infty}{f(x&#039;) \, \delta(x -x&#039;) \, dx&#039;} = f(x)<br />

as a direct generalization of the discrete case:

<br /> \sum_{m = -\infty}^{\infty}{f_{m} \, \delta_{n, m}} = f_{n}<br />

(the index n goes to the continuous label x, a sequence f_{n} goes to a function f(x) and summation over m goes to integration over x&#039;:

<br /> \begin{array}{rcl}<br /> n &amp; \rightarrow &amp; x \\<br /> <br /> f_{n} &amp; \rightarrow &amp; f(x) \\<br /> <br /> \sum_{m = -\infty}^{\infty}{(\ldots)_{m}} &amp; \rightarrow &amp; \int_{-\infty}^{\infty}{dx&#039; \, (\ldots)(x&#039;)}<br /> \end{array}<br />

From the fundamental property of the delta function, one can specficially see that its dimension is the reciprocal of that of its variable x:

<br /> [\delta(x)] = [x]^{-1}<br />

Specifically, for [\delta^{(d)}(\mathbf{p} - \mathbf{p}&#039;)] = [\delta(p)]^{d} = [p]^{-d}, and, from the orthonormality condition of the momentum eigenfunctions from above, we get:

<br /> [\psi_{\mathbf{p}}(\mathbf{r})]^{2} \, \mathrm{L}^{d} = [p]^{-d} \Rightarrow<br /> <br /> [\psi_{\mathbf{p}}(\mathbf{r})] = ([p] \, \mathrm{L})^{-\frac{d}{2}}<br />

Using the fact that:

<br /> [p] \, \mathrm{L} = \mathrm{M} \, \mathrm{L}^{2} \, \mathrm{T}^{-1} = [\hbar]<br />

we see that:

<br /> [\psi_{\mathbf{p}}(\mathbf{r})] = [\hbar]^{-\frac{d}{2}}<br />

and the above formula for the momentum eigenfunctions is dimensionally correct (d = 1 there). The factor of 2 \pi comes from the formulas:

<br /> \int_{-\infty}^{\infty}{e^{i \, k \, x} dx} = 2\pi \, \delta(k)<br />

<br /> \delta(a \, k) = \frac{1}{|a|} \, \delta(k)<br />

 

[snoopies622]

Thanks again, Dickfore. I'm going to have to look into Fourier transforms a little more.

 

[snoopies622]

When we say the momentum eigenfunctions are orthonormal, we mean they satisfy the following identity:

<br /> \int{\psi^{\ast}_{\mathbf{p}}(\mathbf{r}) \, \psi_{\mathbf{p}&#039;}(\mathbf{r}) \, d^{d}r} = \delta^{(d)}(\mathbf{p} - \mathbf{p}&#039;)<br />

I take it then that

<br /> <br /> \int ^ {\infty} _ {-\infty} sin (x) dx = 0 \hspace {5 mm} ?<br /> <br />

(And likewise for cosine.)

 

[Dickfore]

I take it then that

<br /> <br /> \int ^ {\infty} _ {-\infty} sin (x) dx = 0 \hspace {5 mm} ?<br /> <br />

(And likewise for cosine.)

Strictly speaking, these intergrals are not defined in the sense of Riemann integrable. However, if you consider the Gaussian integral:

<br /> \int_{-\infty}^{\infty}{\exp\left[i \beta x - \frac{a x^{2}}{2}\right] \, dx} = \sqrt{\frac{2 \pi}{a}} \, \exp\left(-\frac{\beta^{2}}{2 a}\right), \ a &gt; 0<br />

and is well behaved for the stated values of a.

Furthermore, taking a \rightarrow 0^{+} recovers an integrand on the left hand side whose real and imaginary parts give the integrals you are interested in. However, the right hand side does not have a limit. You need the definition of Dirac delta-function through limits to express it:

<br /> \delta(x) = \lim_{\lambda \rightarrow 0}{\frac{1}{\sqrt{2 \pi} \lambda} \, e^{-\frac{x^{2}}{2 \lambda^{2}}}}<br />

Taking \lambda = \sqrt{a}, we see that the right hand side of the above integral tends to:

<br /> 2\pi \, \delta(\beta), \ a \rightarrow 0^{+}<br />

So, we can write:

<br /> \int_{-\infty}^{\infty}{e^{i \beta x} \, dx} = 2\pi \, \delta(\beta)<br />

Taking the real and imaginary part of this equality, we have:

<br /> \begin{array}{l}<br /> \int_{-\infty}^{\infty}{\sin(\beta x) \, dx} = 0 \\<br /> <br /> \int_{-\infty}^{\infty}{\cos(\beta x) \, dx} = 2\pi \, \delta(\beta)<br /> \end{array}<br />

Taking \beta = 1, you get that even the cosine integral is zero (the sine is an integral of an odd function on a symmetric interval, so it should always be zero).