# I The Debroglie Relation and SR?

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1. Apr 6, 2016

### referframe

The function ei(p.r – Et) is the central player in non-relativistic QM. Yet the expression (p.r – Et) is the Minkowski inner product of the space-time four-vector, (t,r), and the four-momentum (E,p) and as such is Lorentz invariant. According to Feynman, De Broglie realized the relativistic significance of the above expression.

Is there any physical significance to the above in SR? Should the units be regarded as action or angular momentum?

Thanks in advance.

2. Apr 6, 2016

### rubi

Your wave function is valid in both relativistic and non-relativistic physics. Depending on which dispersion relation you specify, it satisfies a relativistic wave equation or a non-relativistic one. For example, given $E=\frac{p^2}{2m}$, your wave function satisfies $i\partial_t\Psi = -\frac{1}{2m}\Delta\Psi$. However, for the relativistic dispersion relation $E^2=p^2+m^2$, the function satisfies $(\Box+m^2)\Psi =0$.

3. Apr 8, 2016

### referframe

Ok, I can see that. But even before one considers which dispersion relation to use, the general expression (p.r - Et) just happens to be the Minkowski inner product of two four-vectors. Do you think that has any physical significance?

4. Apr 9, 2016

### vanhees71

No, it hasn't a physical significance. It's just that math that's similar, because you have a wave function of a rotation-invariant theory in both Newtonian (e.g., sound waves in a gas) as well as in special-relativistic field theory (in an arbitrary but fixed inertial frame).

5. Apr 9, 2016

### rubi

What kind of physical significance do you have in mind? I'm not sure, whether I understand the question.

6. Apr 9, 2016

### referframe

It does seem like just a mathematical coincidence. Feynman (Lectures, Vol III, paragraph 7-2) does take note of the fact that the expression happens to be the product of two four-vectors and is Lorentz invariant and hints that De Broglie was aware of that fact. He only spends a couple of paragraphs on the subject.

7. Apr 9, 2016

### referframe

Actually, I am not sure. Maybe a different kind of relativistic Action? Just wanted to see to what extent this was common knowledge.

8. Apr 9, 2016

### stevendaryl

Staff Emeritus
Well, the classical action can be written in a similar way:

$A = \int L(x, v) dt$

where $L$ is the Lagrangian, and $v = \frac{dx}{dt}$.

If we define $p = \frac{\partial L}{\partial v}$ and define $H = p v - L$, then

$L = p v - H$

So the action can be written as:

$A = \int (p v - H) dt = \int (p dx - H dt)$

where I used $v = \frac{dx}{dt}$

So in the case where $p$ and $H$ are both constants, we can write the action as:

$A = p x - E t$

So it is an action. I'm sure there is some deep connection here, but I'm not sure what.

9. Apr 12, 2016

### Ilja

I would like to object that, not, it is not the product of two four-vectors, but of a 4-vector and a 4-covector.

10. Apr 12, 2016

### referframe

In the 1st and 3rd posts above, I referred to the inner product as a "Minkowski inner product". The definition of that type of inner product assumes that one of the 4-vectors be treated as a co-vector to insure that the sign of the time component comes out correct.

11. Apr 13, 2016

### secur

Surely there is a deep connection, and it's action. I think.

action is the fundamental invariant. All the terms in a Lagrangian represent the action of those components. One must be careful here because "Lagrangian" is used a little loosely. The Lagrangian density is the action at each 4-point. Non-relativistically Lagrangian means the density integrated only over 3-space, and the integral through time is the term whose variation is 0 on the solution path. (The "Action principle"). But in relativity the entire integral over 4-space is termed the Lagrangian. Anyway it can all be thought of as "action" - at a point, throughout 3-space, or throughout 4-space.

The Lagrangian can be a little vague also because only its derivatives (usually partial) are physically relevant, so there's considerable freedom in how you express it. A negative sign more or less is almost a matter of taste :-)

In the simple case action is just Et - pq, or pr - Et, as you say. Schroedinger's equation basically says the partial time derivative is energy. And the standard procedure substitutes partial space derivative for momentum. Both of these can be read directly off that simple expression. The same things are found to be true with the relativistic Lagrangian. Noether's theorem expresses (among other symmetries, which work essentially the same way) the fact that when the Lagrangian is symmetric in t (or x) - meaning essentially the terms don't appear in the L - then those partials are 0 so Energy (or momentum) is a constant of the motion. Leaving out details.

When we solve the wave equation the solution is essentially exp (i * action). Sometimes this is seen as exp (i * integral (Lagrangian)). That's when we're calling the 3-space integral "Lagrangian" and must integrate over time to get the total, sometimes called "S". As mentioned calling this "action" is a bit loose, the reason being that all that counts physically is derivatives.

Anyway when expressed as psi = exp (i * action) the partial derivatives now work as operators. Applied to psi they result in the partial times psi. For instance partial time derivative of psi is E * psi (otherwise expressed as H * psi), when we're using the simple Et - pq. And partial space derivative gives p * psi. Just another way of manipulating action.

Minkowski metric expresses the action used by the object, and it must be invariant - like all action terms. Thus if energy (or mass) increases by gamma (1 / sqrt (1 - v^2/c^2) then time must be dilated by the same amount. And if p increases by gamma, x must be contracted by gamma. The product, action, is always constant.

I'm not confident I'm expressing these thoughts strictly correctly. But I am rather confident the key is action. It's the "only" real invariant. All the terms of a Lagrangian express it. And the wave function is, essentially, action. It's expressed as exp (i * action) for various good reasons - that's how operators work, and that's the solution of the wave function - but essentially it's action.

Not surprisingly Planck's quantum is of action. It's the fundamental granularity of the physical world, and the fundamental invariant.

Yes you're noticing, with your question, a very deep connection: action.

Last edited: Apr 13, 2016
12. Apr 13, 2016

### Ilja

You can, of course, use the Minkowski metric to transfer a vector into a covector. But my point is that even without a metric, coordinate differences will be vectors, and the momentum will be a covector. And their product will be well-defined and have to correct signs. See stevendaryl's derivation of pdx-Hdt, it does not use any Minkowski metric at all.

Feel free to use the Minkowski metric to transform the covector into a vector, and, then, again to define the scalar product between the two vectors. But this does not make the result depending on the Minkowski metric, it gives the same product of a vector and a covector which is well-defined without any metric.

13. Apr 14, 2016

### muscaria

Although I wasn't thinking about it from a QM perspective, this same thought struck me just over a month ago as I was trying to think about action in general and get more of an intuition of it as a physical quantity!
Yes I felt there seemed to be a very deep connection indeed and it is no coincidence.. I even wondered whether there was some way pre Lorentz-Minkowski-Einstein of anticipating the geometry of spacetime and formulating special relativity via a variational principle of the type: requiring the invariance of the differential one form of the action $$dS = \sum_{i=1}^n\left(p_idq_i - Hdt\right)$$
under certain point transformations in extended phase space of $(2n + 2)$ dimensions, where time is placed on the same level as the coordinates and varied as such, in other words taken to be a dynamical variable and obtain Lorentz invariant solution trajectories for the $q_i$ and $q_t\equiv q_{n+1}$ in terms of a unspecified parameter, via some form of constraint condition out of which generators of Lorentz Transformations could be shown to magically emerge :p and combine with the generating function of infinitesimal canonical transformations! But "real" work got in the way I guess.. :p, and the idea seemed obviously too vague in my own mind. But in any case this emerged from thinking about what the differential form of the action represented physically, which I feel may contribute to this discussion. See many of us think of the action as a $\textit{functional}$ to be used in order to obtain the equations of motion and consider it as such, often not appreciating that $S(q,t)$ can be expressed as a $\textit{function}$ along solution trajectories, taking on a physical meaning of a measure of how much flow has been carried over space and time (not absolute, but relative: it's $dS(q,t)$ that is unique along solution trajectories. What do I mean by flow..? Well, in classical field theory, a conserved linear/angular momentum of a field (Noether current) results from specific configurations which follow from the form of the mutual interactions between the field degrees of freedom and represents linear/circulating flow of energy in the field. In quantum theory, momentum serves as the generator of space translations and we have the same principle taking place here, where momentum density is the description of energy density being translated through space, a steady flow of energy. The differential form of the action $dS$ from above, to me seems to be a measuring how much energy flow is being carried through space $(pdq)$. Complementing this flow of energy in space, or , we have the notion of a Hamiltonian serving as a generator of time translations in QM. From the invariance of the action functional under Lorentz transformations Relativistic field theory is clearly saying that a contribution in the differential of the action over parametric curves in space time, must enter, corresponding to a measure of energy flow which is being carried over time. However, the differential form appearing above by virtue of the $-Hdt$ term, does have the required difference in sign but yet at first sight seems to be appearing on a somewhat different footing than its spatial counterpart, given that $H$ represents the total energy and not a flow of energy and more fundamentally that time plays the role of the independent variable for the generalised coordinates.

I have presented things in this fashion as this was the thought process I went through, but it is in fact the following part which is at the heart of it all and seems to reveal a rich structure where the invariant differential 1-form $dS$ with its relativistic counterpart. It was either Jacobi or Lagrange (or both) who realised that if you considered time as a mechanical variable taking part in the variation process, the single particle configuration space becomes 4-dimensional. So in extended phase space, the action functional in parametric form for n +1 mechanical degrees of freedom reads: $$S = \int_{\lambda_1}^{\lambda_2}d\lambda L\left(q_1,\cdots,q_n,q_{n+1},\frac{q_1'}{q'_{n+1}},\cdots, \frac{q'_n}{q'_{n+1}}\right)q'_{n+1}$$
where $q_{n+1}\equiv q_t\equiv t$ and prime denotes total derivative with respect to this as of yet unspecified parameter. if you now consider the momentum associated with time: $$p_{n+1}\equiv p_t = \frac {\partial \left(Lq'_{n+1}\right)}{\partial q'_{n+1}}$$ you find that it comes out as negative the Legendre transform of the Lagrangian $L$ with respect to the $\dot{q}_i$, in other words $-H$.

14. Apr 14, 2016

### muscaria

Apologies I just realised i didn't really motivate the differential form $dS$ which appears at the start of my previous post, but you can find this in most analytical mechanics textbooks.. Landau&L mechanics volume has it towards the end, in a terrific a la Landau manner in fact. You can obtain this expression by considering the usual action in canonical integral form, i.e Legendre transform under the integral and then consider the change in the integral due to infinitesimal boundary variation of a special kind (while the equations of motion hold-so just boundary contributions), namely those which corresponds to the actual motion undergone by the system and that way you can infinitesimally evaluate the change in action along solution trajectories. Same spirit as for evaluating Conserved currents (Noether) or Stress tensors (Current injection at boundary).

15. Apr 14, 2016

### stevendaryl

Staff Emeritus
There is a way to think about non-relativistic physics in terms of 4-D spacetime, as well. The four differentials $(dx,dy,dz,dt)$ form a 4-vector and the four momenta $(p_x, p_y, p_z, -E)$ form a 4-dimensional covector. Just like in SR. The difference with SR is that Galilean spacetime doesn't have a metric, so there is no way to talk about the "length" of a 4-vector or a covector. There is no way to take the "dot product" of two 4-vectors, or two covectors, but you can still form a kind of dot product of one 4-vector and one covector: $dX^\mu P_\mu = p_x dx + p_y dy + p_z dz - E dt$

16. Apr 14, 2016

### Demystifier

That gave my an idea. We know that in relativistic physics Lagrangian is (Lorentz) invariant and Hamiltonian is not. But suppose that we have a theory in which time is a fourth dimension in a 4-dimensional space with Euclidean (rather than Minkowski) geometry. Then Hamiltonian would be invariant (with respect to 4-dimensional rotations) and Lagrangian would not.

17. Apr 14, 2016

### stevendaryl

Staff Emeritus
The most obvious use of 4-vectors and 4-covectors is gradients. If $\phi(x,y,z,t)$ is some scalar field function of space and time (for instance, the temperature), then you can form a 4-gradient $\nabla \phi = (\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}, \frac{\partial \phi}{\partial t})$. Then if you consider $\phi$ at a different point, at a different time, then you can define a 4-vector

$\delta X = (\delta x, \delta y, \delta z, \delta t)$

and you can use 4-dimensional dot product: $(\nabla \phi) \cdot \delta X = \frac{\partial \phi}{\partial x} \delta x + \frac{\partial \phi}{\partial y} \delta y + \frac{\partial \phi}{\partial z} \delta z + \frac{\partial \phi}{\partial t} \delta t$ to compute the change in $\phi$:

$\phi(x+\delta x, y + \delta y, z + \delta z, t + \delta t) \approx \phi(x,y,z,t) + (\nabla \phi) \cdot \delta X$

You don't need special relativity to justify this type of use of 4-vectors.

18. Apr 14, 2016

### muscaria

Some Hamiltonians are not Lorentz invariant sure, but if that is the case the Lagrangian won't be either.. I don't see how one could be Lorentz invariant and not the other, that seems to make no sense given their content is the same and follows simply from the duality of the Legendre transform.. Is it that you mean usually you can see that the Lagrangian is invariant explicitly by just looking at it but it's encoded in a implicit fashion for the Hamiltonian?

19. Apr 14, 2016

### muscaria

The differential 1-form dS isn't a scalar though, and is not invariant under general point transformations (in phase space), I'd be careful ;) For canonical transformations which preserve the structure of Hamilton's equations that is the case, but arbitrary transformations would destroy that structure.. That's not the case for the E-L equations in configuration space which has a definite geometric metric structure and the structure of the E-L eom are always preserved underarbitrary coordinate transformations of configuration space, because of this. Phase space doesn't have a metric, unlike configuration space which is generally of Riemannian and definitely of that type when constraints enter amongst coordinates, there's no definite geometry or metrical structure to phase space, any geometry will do and no reason for introducing a metric in the first place. In my first post where I talked about Lagrange including time into configuration space itself and consider time as a mechanical variable, the point is configuration space really becomes 4 dimensional with definite metrical structure! Not merely 3 +1.. Hence the depth in all of this!! :p

20. Apr 14, 2016

### Demystifier

The Legendre transform is indeed a duality, but Legendre transform is not Lorentz invariant. The Legendre transform in one Lorentz frame is not the same thing as Legendre transform in another Lorentz frame. In other words, for given a Lagrangian, the Legendre transform is not unique: each Lorentz frame defines another Legendre transform. Consequently, each Lorentz frame defines another Hamiltonian. Indeed, Hamiltonian is related to energy, and you should know that energy depends on the Lorentz frame.

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