# The Nature of Matter Waves

1. Sep 5, 2011

### mysearch

Hi,
I am trying to understand whether quantum theory provides any sort of physical interpretation of matter waves. So far, I have been attempting to work my through some of the developments, but possibly failing to grasp what appear to be some fairly fundamental issues. Therefore, I was hoping that by listing these issues, different members of this forum might be able to clarify some of the specific points raised. I will expand the details of each issue in subsequent posts.
• deBroglie’s assumptions
• the matter wave model
• matter wave dispersion
• Schrodinger’s wave derivation
I realise that this is a lot of issues within one thread, but they do all appear to be interrelated. While I have attempted to read into these issues, it seems that many references simply start by accepting deBroglie’s assumptions, without ever questioning them, and even fewer seem to attempt qualify the idea of a matter wave against any physical interpretation. Therefore, I was hoping that some members might have a better insight and be willing to lend a helping hand. Thanks

2. Sep 5, 2011

### mysearch

Issue-1: deBroglie’s Assumptions

I am confused by what appears to be a degree of circular logic that always seems to come back to deBroglie’s assumptions about matter waves, i.e. the relationship between wavelength and momentum. In the overall debate about wave-particle duality, it seems that Compton first made the logical argument for photons having a particle-like nature by equating Planck’s and Einstein’s energy equations.
$$E=mc^2=hf=\frac {hv}{\lambda}$$
However, the rationalisation to [h/p] only seems to work for a photon with [v=c] and momentum [p=mc]:
$$\lambda=\frac {hc}{mc^2}=\frac {h}{p}$$
As far as I can see, deBroglie’s wavelength for a particle appears to rest on the assumption that the relationship [h/p] can be extended from a particle-like photon to a wave-like particle, where [v<c]. However, if this is the case, then the first equation above doesn’t appear to reduce to the deBroglie assumption, unless we make a fairly radical assumption about the definition of momentum [p]
$$\lambda=\frac {hv}{mc^2}= \frac {h}{p}$$
$$only \ if \ p= m \frac {c^2}{v}$$
An alternative approach might be to assume that the energy associated with the deBroglie wavelength relates to the kinetic energy of the particle, not its total energy. However, following this logic still does not seem to explain the deBroglie formulation, e.g.
$$E= \frac {mv^2}{2} = \frac {p^2}{2m}=hf=\frac {hv}{\lambda}$$
$$\lambda=\frac {2h(mv)}{p^2}=2 \left( \frac {h}{p} \right)$$
While this now appears closer to the form of deBroglie’s equation, there is still an additional factor of 2. One suggestion that I have seen is that if the particle conforms to a wave packet, which is localised in space, its length would equal $x=\lambda/2$. If so, the formulation [h/p] would represent the length of wave packet, but only half the wavelength. While I know that deBroglie wavelength is said to have been verified by experiments, I would appreciate any clarification of the maths, which supports this basic assertion, which then appears to be used in so many other key derivations.

3. Sep 5, 2011

### mysearch

Issue-2: Matter Wave Model?

Given that a particle always moves with a velocity [v<c], there is the suggestion that a matter wave might be better described in terms of a wave packet, which can be mathematically constructed as a superposition of components waves, each with some given phase velocity [vp], which then results in some overall group velocity [vg] corresponding to the particle velocity [v]. However, if you predicate the derivation of the phase velocity on what appear to be deBroglie’s assumptions, you are led to a relationship between [E/p]
$$v_p=f \lambda= \frac {E}{h} \frac {h}{p} = \frac {E}{p}$$
What is not clear to me is the scope of the energy [E], i.e. total or kinetic?
$$v_p = \frac {mc^2}{mv} = \frac {c^2}{v} \ or$$
$$v_p = \frac {mv^2}{2mv} = \frac {v}{2} \ ????$$
Again, we appear to have another factor of 2, which doesn’t seem to fit. Equally, in the context of a superposition wave model, a finite matter wave packet would not appear to be a ‘real’ wave, but rather the superposition resulting from the constructive and destructive inference of the underlying waves. However, without being able to rationalise the nature of these underlying waves, it is difficult to reconcile the ‘construction’ of a matter wave packet, localised in space. There is also an issue as to whether the matter wave model would be dispersive, which is raised in the next post.

4. Sep 5, 2011

### mysearch

Issue-3: Dispersion of Matter Waves

$$\omega = \frac {\hbar \kappa^2}{2m}$$
In contrast, a photon propagating in vacuum is described by the non-dispersive relationship $\omega=c \kappa$, which appears to reflect the relationship $v=f \lambda$. Looking at the equation above, it is not obvious where the velocity has disappeared to in the particle equation. For example, is it meaningful to re-arrange as follows:
$$\omega = c \kappa$$
$$\omega = v \kappa$$
$$v = \frac{\hbar \kappa}{2m}$$
However, if you come at the same issue from deBroglie’s assumptions, we might start with the equation for kinetic energy
$$\frac {mv^2}{2} =\frac {p^2}{2m} = \frac { \hbar^2 \kappa^2}{2m}$$
$$v = \frac {\hbar \kappa}{m} \propto \frac {1}{\lambda}$$
Again, a factor of 2 seems to show up. However, I am more interested in how to interpret the idea of dispersion for a particle wave, e.g. electron, travelling through a vacuum. If a particle is modelled as a finite wave packet, localised in a vacuum, constructed as a superposition of a large number of harmonics, i.e. different wavelengths, how can the particle model maintain any local coherence in space, if the harmonics all travel at a different velocity?

Last edited by a moderator: May 5, 2017
5. Sep 5, 2011

### mysearch

Issue-4: Schrodinger’s Derivations

The derivation of Schrodinger wave equation appears in many references, http://www.physics.mq.edu.au/~jcresser/Phys201/LectureNotes/SchrodingerEqn.pdf" [Broken]; it would seem that Schrodinger adopted his approach due to the non-linear relationship between [w] and [k] as described in terms of dispersion. As such, Schrodinger appears to have adopted the complex form of the wave equation so that he could equate the first derivative with respect to time with the second derivative with respect to space. However, this approach never loses the complex number see equation 6.7 on page 43 of the link above. In contrast, the following approach starts with the classical solution of the wave equation in terms of the second order derivatives only, which has no complex component:
$$\frac {\partial^2 A}{\partial t^2} = \frac {\omega^2}{\kappa^2} \frac {\partial^2 A}{\partial x^2}$$
I have switched back to using a generic variable [A] simply to highlight that this seems to be a non-standard approach. However, deBroglie’s assumptions allows us to substitute for [w] and [k] based on the following associations, first [k]
$$\lambda = \frac {h}{p}$$
$$p= \frac {h}{\lambda}= h \frac {\kappa}{2 \pi}= \hbar \kappa$$
$$\kappa = \frac {p}{h}$$
Now [w]:
$$E=hf= \frac {h \omega}{2 \pi} = \hbar \omega$$
$$\omega = \frac {E}{\hbar}$$
As such, we appear to be able to jump to some sort of wave equation based on substituting the previous equations into the original wave equation:
$$\frac {\partial^2 A}{\partial t^2} = \frac {E^2}{p^2} \frac {\partial^2 A}{\partial x^2}$$
So my question relates to whether this approach, which appears to based on deBroglie assumptions, is equally valid?

Last edited by a moderator: May 5, 2017
6. Sep 5, 2011

### kith

I have not read de Broglie's thesis, but his assumptions can be justified along the following line of reasoning.

First, Einstein introduced the photon with energy E=hw. From this, one can deduce p=hk when taking into account E=cp. So light can be described as a particle with energy E and impulse p. De Broglie now asked: if electromagnetic waves show particle-like behaviour, can particles also show wave-like behaviour? Particles are well-localized, so if such a description is possible, the particles (moving with velocity v) will be represented by wavepackets (propagating with group velocity vg).

If we assume that energy and impulse obey the same relations as for the photon -like de Broglie-, we can derive vg=v in the non-relativistic as well as in the relativistic case (see wikipedia for example). This is what justifies this hypothesis. Like other parts of the 'old' quantum theory, it was just a plausible assumption at this time, and was not derived from fundamental principles.

The question is not whether to use kinetic or total energy, but whether to use the non-relativistic or the relativistic energy-impulse relation, which depends on the physical situation. However, both should yield the same results. And you always use the total energy, which happens to be the same as the kinetic one in the non-relativistic case.

I think you're too sloppy with your formulas. For example, what's 'm' supposed to mean, when you're talking about the photon in your second post?

Last edited: Sep 5, 2011
7. Sep 5, 2011

### kith

Re: Issue-1: deBroglie’s Assumptions

Your expressions for the phase velocities seem to be right. But I don't think, there's much physics in them. As you noted, the group velocity is what counts.

I'm not sure, what you're troubled with here. With the use of E=hw and p=hk, the energy-impulse relations E(p) translate to frequency-wavenumber relations w(k).

Broadening of wavepackets should not be confused with loss of coherence. For every potential, there is a characteristic revival time for which the wavepacket is restored to it's initial shape.

In free space, the wavepacket doesn't stop broadening, but that's okay, since there is infinite space. ;-) In real space, there are also interactions which localize the particle and therefor cut off parts of the wavepacket.

Last edited: Sep 5, 2011
8. Sep 5, 2011

### KWillets

Re: Issue-1: deBroglie’s Assumptions

I can't tell for sure, but you seem to be confusing phase velocity and group velocity. I labelled a few of the v's in the quote below.

(And more importantly, $$v_{phase} = v_{group} = c$$ . For photons, the two factors are indistinguishable, but for massive particles, $$v_{phase}v_{group} = c^2$$ but the group velocity is < c.)
(It does)

9. Sep 6, 2011

### mysearch

Thank you for the responses in #6. #7 and #8 to the issues raised in posts #1-#5. Here are some comments in response.
Addressing sloppiness, did you mean h-bar in the equations above? I also noticed a mistake in one of my own references, so for general reference, I will table the following equations, which will hopefully explain my use of ‘m’ in the context of a photon.
$$E= \hbar \omega ; \ \ \ \ \ \ p= \hbar \kappa$$
$$E^2 = m_0^2 c^4 + p^2 c^2$$
This, as you state, leads to E=pc for a photon having no rest mass. However, if you equate and re-arrange Planck’s and Einstein’s energy equation, you get some ‘notional’ idea of a kinetic mass associated with the photon:
$$E=mc^2=hf; \ \ m_k=\frac {hf}{c^2}$$
$$mc= \frac {hf}{c}=p$$
If we accept ‘h’ and ‘c’ as constants in this context, then there is some equivalence between ‘m’ and ‘f’, which must also be subject to the same relativistic effects. I raise this point in passing because I believe that deBroglie thesis discusses 3 frequencies associated with the particle, 1) the internal frequency in the rest system 2) the internal frequency as measured by an external observer who sees the system moving with velocity ‘v’ and 3) the frequency the external observer would associate with the particle’s total energy. However, I have to admit that I have not work through the implications of these 3 perspectives. Maybe somebody might be able to clarify this aspect?
I have no problem with this, as a general assumption, but the issues raised in post #2 were specific to the formulation of deBroglie’s wavelength.
Sorry, this is too generalised for me to be sure about what you are trying to clarify. If you reference the total relativistic energy equation above, which does not account for any potential energy, then the total energy of a particle includes its rest mass. Again, post #2 tried to resolve the deBroglie wavelength in terms of both total and kinetic energy without success. If you know a way to resolve the issue that I was trying to highlight, I would be most grateful if you could detail any sort of derivation that substantiates deBroglie’s wavelength equation. Thanks.
I wasn’t trying to present the physics, but rather establish the definition of the phase velocity, when predicated on what appears to be deBroglie’s assumptions, linked to either total energy or kinetic energy. Again, I will present the ‘physics’ purely by way of reference before returning to the main issue being raised in post #3.
$$v_p= \frac {E}{p} = \frac {mc^2}{mv}= \frac {c^2}{v}$$
$$v_g= \frac {\partial E}{\partial p} = \frac {p}{m} = \frac {mv}{m} = v$$
The energy associated with the phase velocity appears to be the total energy, while the energy associated with the group velocity appears to be kinetic. However, I am unsure what you mean by ‘the group velocity is what counts’ as the issue I was attempting to raise was linked to the physical meaning of both the phase and group velocity. Please note I am not trying to forward any interpretation of my own, but the description of a group wave packet seemed to suggest some process of wave superposition. If so, I don’t understand how you get a group wave packet without describing the underlying waves associated with the phase velocity. The fact the equation above seems to suggest these wave travelling in excess of ‘c’ appears confusing, at least, to me.
Again, if you could clarify some of your points I would be grateful; especially in the context of the following Wikipedia statement that you referenced in post #6.
Are the caveats being raised here applicable to a dispersive particle wave packet, which appears to be constructed from a superposition of harmonic waves of different wavelengths, at least, mathematically?
While I understand where you are coming from, can you justify the phase velocity within any physical description of the wave packet?

Again, just by way of clarification, being new to quantum theory, I decided to start from an historical aspect and work forward. As such, I have only just reached deBroglie’s ideas about particle wave and the influence it appears to have had on Schrodinger’s derivation of his wave equation. This derivation seems to be predicated on the idea of a dispersive wave relationship based on deBroglie’s ‘assumptions’. Equally, it was unclear to me why Schrodinger elected to take the complex path within his derivation – see post #5. From a learning perspective, it would be nice if there was also some physical interpretation underpinning some of the particle wave packet ideas. Thanks

10. Sep 6, 2011

### kith

Hehe, good response. ;-) Yes, I do. So also in this post h=hbar.

I think your problems can be reduced to two issues:
1) Before writing out any equations, you have to decide whether you want to work in the non-relativistic regime or in the relativistic one. This decides which energy-impulse relation E(p) you will use, either E=p²/2m or E²=m²c^4+c²p². You should not switch between these viewpoints.
2) The dispersion relation w(k) is determined by the wave equation. For electromagnetic waves it is w(k)=ck, which is equivalent to lambda*f=c. This is not true for matter waves! So you can't just substitute c by v and expect v to be related with the particle velocity.

In fact, what you try to do in your second post, can't be done. For the photon, you start with E=hw and E=cp. Using the known dispersion relation of the electromagnetic wave equation, w(k)=ck, you derive p=hk=h/lambda. If you start with with E=hw and E=p²/2m you can't derive p=h/lambda, since you do not know the dispersion relation. Instead, de Broglie postulated p=hk in analogy to the photon. From this postulate, he got vg=v(particle), so it seems reasonable. And from this postulate, the dispersion relation w(k) is derived (to 'guess' the wave equation from it).

Before we proceed, you have to understand this line of reasoning.

11. Sep 7, 2011

### mysearch

Kith,
Thank you for the points raised in post #10, if possible, I would like to clarify some of the issues.
Not really sure what you mean by energy-impulse relation in this context or the scope of the restrictions you are implying by not switching between the classical and relativistic equations. Many of the references I have found on the internet seem to do this by invoking the caveat [v<<c]. By way of reference, the hyperphysics site seems to provide a reasonable derivation of ‘http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html" [Broken]provides 2 sub-headings that clarify the scope of ‘newtonian kinetic energy’ and ‘relativistic kinetic energy of rigid bodies’. However, my following comments on dispersion may serve as an example for clarification.
I agree. I thought my outline of the dispersion issue in post #4 accepted this point, but then goes on to highlight the issue of reconciling the equation given on the Wikipedia page under the sub-heading ‘http://en.wikipedia.org/wiki/Dispersion_relation" [Broken]’. In post #9, I also gave what I understood to be the definition of the phase and group velocities associated matter waves following on from your statement ‘de Broglie postulated p=hk in analogy to the photon’. I will repeat these equations below with some modification in order to clarify the energy I am assuming to be associated with each:
$$v_p= \frac {E_{total}}{p} = \frac {\gamma m_0 c^2}{\gamma m_0 v}= \frac {c^2}{v}$$
$$v_g= \frac {\partial E_{kinetic}}{\partial p} = \frac { \partial (p^2/2m) }{\partial p} = \frac {p}{m} = \frac {mv}{m} = v$$
The caveat applying to the second equation is v<<c, such that the classical approximation of kinetic energy can be applied. Is this unreasonable?
I agree with you and in some respect post #2 was simply trying to highlight what appear to be the inconsistencies in this approach.
As also stated in post #2, Compton’s logic of describing a photon as a particle seems to be unambiguous. The questions only arise when extrapolating this logic to matter waves.
Ok, let us try to approach this issue from the perspective of the group velocity of a matter wave. I am assuming that we can use the following generalised equation as our starting point:
$$v_g= \frac {\partial \omega}{\partial \kappa} = \frac {\partial ( \hbar \kappa^2/2m)}{\partial \kappa} = \frac {\hbar \kappa}{m}$$
This substitutes the accepted dispersion relationship for a matter wave, but seems to come to the same conclusion as outlined in post #4, i.e. it doesn’t align to deBroglie’s assumption by the factor 2, as also outlined in post #2.
Following on from the issue raised in post #5, the general form of the 2nd order wave equation for a plane wave is believed to conform to:
$$\frac {\partial^2 A}{\partial t^2} = \frac {\omega^2}{\kappa^2} \frac {\partial^2 A}{\partial x^2}$$
In this context, the ratio of [w/k] is interpreted as a phase velocity, which in the case of a photon is unambiguous, e.g. ‘c’. Whether this equation can be modified to reflect a group velocity is unclear, but Schrodinger wave solution suggests that it can; although I assume we have to modify the ratio [w/k] to [dw/dk]. If so, we could use the equation for group velocity above such that:
$$\left( \frac {\partial \omega}{\partial \kappa} \right)^2 = \left( \frac {\hbar \kappa}{m} \right)^2$$
Using the assumption $p=\hbar \kappa$ allows the equation above to collapse back to the expected group velocity that aligns to the kinetic velocity of particle, when v<<c, but we appear to have to accept the factor 2 discrepancy as discussed above. So, to summarise posts 1-5:
• Can the formulation of deBroglie’s wavelength can be mathematically derived and explained without the factor 2?
• Can the particle wave packet be described in terms of a superposition of waves and, if so, what physical description can be given to these waves?
Thanks

Last edited by a moderator: May 5, 2017
12. Sep 7, 2011

### kith

mysearch, it seems to me that you want to derive de Broglie's hypothesis, which can't be done. You also keep mixing the relativistic and the non-relativistic viewpoint (in your last post, you define vp by the relativistic and vg by the non-relativistic energy). As a whole, your posts are concerned far too much with details. So before I answer your questions, please state explicitly what you want to derive from what and describe the steps that should be taken - without using formulas!

For simplicity, lets disregard all relativistic issues for massive particles and work strictly in the realm of Newtonian mechanics. This means, the energy of a free particle is given by E=p²/2m which constitutes the energy-impulse relation E(p).

13. Sep 7, 2011

### KWillets

Just to clarify, I had stated that kinetic E might be correct in an earlier thread, but I was wrong, and my later posts had more correct information.

Also, my QM textbook mentions that dispersion of the wavepacket is not significant in the lifetime of the universe or some such. I can look it up if you want me to confirm.

As to this question:

I had posted this question earlier, and I have since resolved most of my concerns by reading De Broglie's thesis. In my assessment the key point is that a fairly simple waveform in the rest frame (albeit with infinite phase velocity) Lorentz-transforms into a waveform with spatial and temporal periodicity in a moving frame, and the frequency, wave number, and dispersion relation all emerge from this transformation.

14. Sep 8, 2011

### PhilDSP

Yes, Kith's last post describes why.

For the most part de Broglie described the situation for a monochromatic photon having only a single characteristic frequency and assumed massive particles would have a single characteristic frequency as such

$$f = \frac{E}{h} = \frac{mc^2}{h \sqrt{1 - \beta^2}}$$

Where f is related to the wavelength through the dispersion relation. In this case the wave is always a simple sinusoidal signal.

If the signal has a more complex shape the wave must be become a wave packet composed of more than one sinusoidal component, each of which has a different frequency (Lookup Fourier analysis if you're unfamiliar with this concept) The amplitude of all of these components add linearly to produce the wavefront, the frequency effectively related to a portion of the energy

$$f_n = \frac{E_n}{h}$$

Here the frequency of each component of the signal $f_n$ obeys the same relationship to its associated wavelength that holds from the dispersion relationship for a monochromatic wave.

Last edited: Sep 8, 2011
15. Sep 10, 2011

### PhilDSP

On page 76 of "Introduction to the Study of Wave Mechanics" de Broglie gives this expression (where I've re-labeled some variables to comply with our usage)

$$E_{total} = \frac{mc^2}{\sqrt{1 - \beta^2}} + U(x, y, z, t) = mc^2 + K + U = mc^2 + E_{newton}$$

where U is potential energy, K is kinetic energy, $E_{newton}$ is the classical Newtonian energy and $mc^2$ is the internal energy of the particle

He goes on to derive and explain the Schrödinger equation for one particle as well as giving what seems to me to be an especially lucid description of configuration space and why it was required by Schrödinger in applying his equation. He anticipates the need for the Dirac equation though it hadn't been discovered at that point in time.

Last edited: Sep 10, 2011