Here's where I make a fool of myself in Physics

[Qfwfq]

Corollary: a scalar is a rank zero tensor - well we knew that already, right?

Yup! Just like a vector is a rank 1 tensor.

 

But it illuminates (but doesn't wholly excuse!) the annoying (to me) habit of physicists to define tensors in terms of how they transform. I prefer to define them in terms of what they actually are, but I can now see where you guys are coming from.

:hyper: Well I kinda see what you mean.

 

alibi or alias? ;)

[Pyrotex]

I have a theory why all of this Action stuff sounds so unfamiliar. Last night I remembered the grad school class where I took advance mechanics. At mid-term, we were told to come up with a simple mechanical system, construct its Lagrangian, L = V - U, perform the Hamiltonian thingy on it, and derive the equations of motion of the system.

 

I used my class ring as my simple mechanical system. Actually, I simplified it to a hemisphere with a rod attached to the flat side of the hemisphere at the center of the sphere and normal to that surface, of length = radius of the (hemi)sphere.

 

You remember class rings? You put them stone down on a surface and spin them, right? And the heavy stone winds up on TOP! Well, that's what I was hoping to get out of my equations of motion.

 

To make a long story short, and less pathetic, even with assistance from my professor, we were not able to perform the Hamiltonian on L of the system. :thumbs_up I got a B in the class for serious effort, but the whole experience was disappointing and somewhat humiliating. And so I forgot about Hamiltonians and Action. [sniff]

[Ben]

First arkain 101 I believe "uniform motion" is the standard term for constant speed i.e. no acceleration

To make a long story short, and less pathetic, even with assistance from my professor, we were not able to perform the Hamiltonian on L of the system.

This is no great surprise to me! How can one "perform" a Hamiltonian on a Lagrangian?

 

I have been thinking deeply about the Legendre transformation which maps the Lagrangian [math] \mathcal{L} = T -U[/math] onto the Hamiltonian [math]\mathcal{H} = T+U[/math]. It looks like a simple parity argument, right? But this seems to make no sense, to me at least. Something else has to give, I think, though I am not sure what.

 

Aaargh: phucking physics.........

 

Leave it with me a while, see if I can come good (which I seriously doubt).

 

Anyhoo, today's dumb question.

 

We have agreed that a rank 0 tensor is a scalar, and a rank 1 tensor is a vector. Is the converse true in general?

 

That is, may we think of an element in any vector space as a rank 1 tensor? I think perhaps not; consider the space of functions [math]\mathbb{C} \to \mathbb{C}[/math], closed under addition and scalar multiplication. Does it make any sense to think of these as rank 1 tensors?

 

I doubt it, but I'm not sure, hence my question

[Qfwfq]

That is, may we think of an element in any vector space as a rank 1 tensor?

Actually, not such a dumb question! :)

 

It's a kinda subtle matter. Strictly, the answer is no. The briefest reason is that these two terms are used in linear algebra and aren't necessarily geometric entities.

 

The geometric entities (apart from scalars), can however hardly be concretely handled except in terms of the corresponding algebraic ones... which represent them for a given choice of coordinates (or of basis). Now that is where the matter of "how they transform" comes into play. The same transformation, applied to the vector, may be viewed as a map between different elements (alibi) of the linear space, or alternatively as the map between the different components (alias) of the same geometric entity for two coordinate choices.

 

The use for terms alibi and alias describes the use in ordinary geometry:

 

• alibi - other place (point of space)
  
• alias - other name (labelling)
  

 

P. S. Even a linear space such as the functions [math]f:\,\mathbb{C} \to \mathbb{C}[/math] can constitute a geometric space providing a metric is chosen, after which a given one of the functions may be regarded as the components of a rank 1 tensor (it has one variable rather than one index).

[Pyrotex]

First arkain 101 I believe "uniform motion" is the standard term for constant speed i.e. no accelerationThis is no great surprise to me! How can one "perform" a Hamiltonian on a Lagrangian?...

My bad! Mea Culpa! :shrug:

 

I meant to say, that we failed at taking the Action Integral over the Lagrangian:

 

S = Integral[over path P] { U - V } dt

 

so we could set that = 0 (the principal of least action)

 

and from this, derive the equations of motion for the system.

 

Eight weeks of furious calculus, trig, algebra, transforms... nothing. :)

[Ben]

Yeah Pyro, I have been feeling bad about that flip comment ever since I made it. I apologize, truly.

[Ben]

The same transformation, applied to the vector, may be viewed as a map between different elements (alibi) of the linear space, or alternatively as the map between the different components (alias) of the same geometric entity for two coordinate choices.

Sure, I am well aware of this (though I never heard them called alibi and alias before).

 

I am also well aware that, in ye olde tymes, if the coordinate transformation was signed oppositely to the component transformation, then a tensor was called contravariant, covariant otherwise.

 

That is not quite what I meant, however. One frequently encounters texts, good ones too (Lovelock & Rund, for e.g.) where the existence theorem for a tensor runs something like "if it transforms like {this} it is a tensor". No matter, it's a point of preference I guess.

P. S. Even a linear space such as the functions [math]f:\,\mathbb{C} \to \mathbb{C}[/math] can constitute a geometric space providing a metric is chosen, after which a given one of the functions may be regarded as the components of a rank 1 tensor (it has one variable rather than one index).

This is much more interesting. I take it that by "geometric space" you mean one where some abstract concept of length and angle have some sort of meaning?

 

OK, let [math]F[/math] be the space of all functions [math]\mathbb{R} \to \mathbb{C}[/math], continuous on the closed real interval [math][a,b] \ni x [/math]. Then I may define the sesquilinear form [imath]\int^b_a \overline{f}(x)g(x) \,dx = (f,g)[/imath] as an inner product, by analogy with the more familiar form; i.e. for fixed basis, [imath] \sum_i \overline{\alpha}^i\beta^i = (v,w)[/imath], where the continuity of our functions sends the sum over to the integral.

 

This suggests the norm [imath]\int^b_a\overline{f}(x)f(x)\,dx = \int^b_a|f^2|\,dx \equiv ||f||[/imath]. Except for two things: how do I know this integral exists, and, if it does, it is unique?

 

First is easy: simply define [math]F[/math] to be the space of square-integrable functions [imath]L^2[/imath]!

 

Second is less so, for the following reason: the uniqueness condition implies that [imath]||f|| \equiv (f,f) =0 \Rightarrow f=0[/imath], that is that [math]f(x) = 0[/math] for all [math]x \in [a,b][/math].

 

The sad fact is that [math]f(x)[/math] may be non-zero for finitely many points in [a,b] and the Riemann integral won't notice this - it will still return an integral of zero.

 

So we need a "new" sort of integral called the Lebesgue integral, and this is where I start to unravel slightly. Any guidance on this guy would be appreciated.

 

Moreover: every vector space deserves a set of basis vectors. So, for the sake of example, if [math]\vec{f} \in F[/math] is a vector, and [math]f^x \equiv f(x)[/math] are the components of this vector on some basis, what is the basis?

 

This is no trick - it's a genuine question.

 

Sorry to be so long-winded; I had a coupla beers after work.

[Pyrotex]

Yeah Pyro, I have been feeling bad about that flip comment ever since I made it. I apologize, truly.

No, you're cool.

I totally misused the word "Hamiltonian". I demand that you flip comments at me when I do that! :D

 

Ever since I corrected myself, the memories keep flooding in and bugging me.

Like, WHAT is the Langrangian for a spinning top consisting of one hemisphere with mass 1 , and a spindle of length r and mass 2; where r is the radius of the hemisphere?

 

It's the Kinetic Energy - the Potential Energy. V-U

 

KE would be replaced by angular something... :shrug: angular momentum * radial velocity? Does that need a cross product? :) Where are my vectors? Where are my damn vectors?!?! :confused: :confused:

 

PE would be... :confused: oh no... er, mass 1 * height above table + ... :confused:

Is there an angular potential? oh no... the whole thing can rotate in TWO axes... how do I... what is the... :confused:

 

 

help...

[Qfwfq]

Tee hee! Just out of curiosity i did a google on the two words and, out of almost half a million results, some are actually relevant! :D

 

A neat depiction: File:Alias and alibi rotations.png - Wikimedia Commons

 

So we need a "new" sort of integral called the Lebesgue integral, and this is where I start to unravel slightly. Any guidance on this guy would be appreciated.

I must admit! It isn't easy to choose a non-degenerate metric in the case we considered...

but that's nitpicking!!!!!! :rant:

 

Of course it's something to account for in a proper definition of states in QM for continuous spectra, no point using a fancier metric than the one you chose, but this doesn't imply it being impossible. ;) A good treatment, sure enough, considers Lesbegue integrability but all this means is that a difference on any subset of zero measure goes unnoticed. In order to make the metric good, one may use a suitable singular measure.

 

Moreover: every vector space deserves a set of basis vectors. So, for the sake of example, if [math]\vec{f} \in F[/math] is a vector, and [math]f^x \equiv f(x)[/math] are the components of this vector on some basis, what is the basis?

 

This is no trick - it's a genuine question.

To be honest and truthful, this is what I dearly hoped wouldn't be asked!!!!!!

 

Actually a countable basis exists in the case of the finite interval (Fourier series) but with the non-degenerate metric one can even count in the functions everywhere 0 except at one point.

 

Sorry to be so long-winded; I had a coupla beers after work.

Back in my uni days, there was a professor in the math dep't (who I fortunately escaped), quite a character. During one exam, he was posing a rather tricky topology problem and the poor student was reeling with it, the professor asked him: "Don't you drink?" The student gaped and replied no and the professor said: "Well then how the heck can you hope to understand topology, if you don't!?!?!?!?!"

[Ben]

Just back from a weeks vacation, so here is another dumb question. And when I say dumb, I mean really, really dumb.

 

At some point in my reading I encountered the the terms "momentum operator" and "position operator". This is baffling me! Ordinarily, I would like to think of momentum and position as vectors (given the appropriate choice of coordinates).

 

Equally ordinarily, I would like to think of an "operator" as a linear transformation on a vector space.

 

Now, I have the equation for the momentum operator scribbled in front of me. [math]\vec{p} = -i \hbar \nabla = -i \frac{h}{2 \pi} \frac{\partial}{\partial x^i}[/math]. I notice there is no mass term!

 

So what gives? Obviously the "momentum operator" is not the same as "momentum". What is it doing, in simplistic terms? Likewise the

position operator?

 

Anyhoo....Looking in my yellowing college Phys. Chem. text, and almost pulling my beard out, I think I see this is true......

 

Start with the time-independent Schroedinger wave function (written in longhand):

 

[math](-\frac{\hbar ^2}{2m} \nabla ^2 + U)\psi = \epsilon \psi[/math] where the [math]\epsilon [/math] is interpreted as the eigenvalue for the operator [math]\mathcal{H}=-\frac{\hbar ^2}{2 m} \nabla ^2 + U [/math] ([math]U[/math] is the potential). This is, of course, the Hamiltonian operator.

 

Then [math]\mathcal{H} \psi = \epsilon \psi[/math], a very familiar identity from linear algebra.

 

As is customary, since the Hamiltonian is defined as [math]\mathcal{H} =[/math] kinetic + potential, one thinks of the quadratic term, i.e. the first, as kinetic energy of a particle with mass = m.

 

Write [math]\text{k.e}=\frac{1}{2}mv^2 = \frac{(mv)^2}{2m}[/math]. Then comparing this to the kinetic (i.e. first) term in the Hamiltonian, I deduce that [math]\frac{(mv)^2}{2m} = -\frac{\hbar ^2}{2 m} \nabla ^2 [/math] and so that

 

[math]mv = \sqrt{(-1)\hbar ^2 \nabla ^2}= i \hbar \nabla \equiv \vec{p} [/math], which is NOT what I wrote above.

 

I have a sign wrong somewhere, where have I gone wrong? Moreover, I feel decidedly queasy taking the square root of a second-order differential operator

[Qfwfq]

But, what exactly is the problem Ben? That's kinda the way it works.

[Ben]

OK, granted that I may have made a freshman arithmetic error somewhere, I have the desired equality "up to a sign".

 

The question is really, which I failed to ask fully.... how can position and momentum be operators? What are the vectors they are acting upon? Or is the term "operator" being used in a sense different from that used in linear algebra?

[Qfwfq]

OK, granted that I may have made a freshman arithmetic error somewhere, I have the desired equality "up to a sign".

Back in my student days, a lecturer who was to do the exercises of theoretical physics, taking the folks from the solutions of the free Dirac & KG equations through propagotors, Wick's theorem, LSZ formalism and Feynman diagrams, was just starting the first computations of the course on the blackboard when he paused and turned toward the audience with a wide and fiendish grin. "Of course, we shall adopt natural units, such as to have a value of 1 for all things like [imath]c[/imath], [imath]\cancel{h}[/imath], [imath]\pi[/imath], [imath]i[/imath], 1, 2, 3 &c..."

 

The question is really, which I failed to ask fully.... how can position and momentum be operators? What are the vectors they are acting upon? Or is the term "operator" being used in a sense different from that used in linear algebra?

The operators act on vectors of a Hilbert state (for discrete spectra at least). A state of the system is represented by a normalized vector, an observable by a self-adjoint operator (or Hermitian in simple cases) and symmetries by unitary ones. Then there are creation and destruction operators which are neither,they are the adjoint of each other. Well, time reversal is actually anti-unitary but who cares.

 

The so-called "time independent Schrödinger equation" is really just the Hamiltonian's eigenvalue equation; an observable's eigenvalues are the possible outcomes of a measurement of it.

[Ben]

......a wide and fiendish grin. "Of course, we shall adopt natural units, such as to have a value of 1 for all things like [imath]c[/imath], [imath]\cancel{h}[/imath], [imath]\pi[/imath], [imath]i[/imath], 1, 2, 3 &c..."

Yeah, I have heard these called "God's units". I like that - good old God

 

The operators act on vectors of a Hilbert state

Not quite sure I follow this. Maybe you mean they act on state vectors in an Hilbert space?

 

Anyway, I must have been having another of my "senior moments". It is easy to show that, given a vector space, then the set of all operators acting on this space is itself a vector space. The obvious implication is that, if an operator/vector in this set can be written as the sum of elements in the same set, then they themselves are, by definition essentially, operators.

 

Thus for the Hamiltonian. Anyway, recall I was a little unhappy with my last extraction, mainly because I'm not sure it's legal (at any age) to take square roots of differential operators. So.....

 

I remember a chum (well, chummess, actually) saying I could do this via the plane wave equation. I was rather drunk at the time and forgot the details, but I wrote down the outline on a beer mat.

 

Recall I want to show that [math] \vec{p} = -i \hbar \nabla [/math]

 

I start with the plane wave solution to the general wave equation, which I shamelessly copy from the Wiki (let's leave aside the rather embarrassing fact I'm not completely sure what a plane wave is);

 

[math]\psi(\vec{r},t) = Ae^{i(\vec{k} \cdot \vec{r} - \omega t)}[/math] where [math]\vec{r} [/math] the position vector, [math] \vec{k} [/math] the wave propagation vector, [math]A[/math] is amplitude and [math] \omega[/math] is angular speed.

 

For simplicity (and I think it makes sense in context) I will assume that the norm [math]||\vec{k}|| = \frac{2\pi}{\lambda}[/math] where [math]\lambda [/math] is wavelength.

 

Looking again in my chemistry text, I follow the de Broglie conjecture that [math]E = h \nu = mc^2 \Rightarrow \lambda = \frac{h}{\vec{p}}[/math].

 

Then I will have, by the above, that [math]\lambda = \frac{2\pi}{||\vec{k}||} = \frac{h}{\vec{p}} \Rightarrow ||\vec{k}|| = \frac {2 \pi \vec{p}}{h} = \frac{\vec{p}}{\hbar}[/math].

 

Plugging this into my plane wave solution above, I get that [math]\psi({\vec{r},t}) = Ae^{\frac{i}{\hbar}(\vec{p} \cdot \vec{r}- \omega t)}[/math].

 

I now assume that [math]\vec{r}[/math] and [math]\vec{p}[/math] have components in the set of coordinate functions [math]\{x,y,z\}[/math] and partially differentiate w.r.t these. I find that [math]\frac{\partial }{\partial x}\psi = (\frac{i}{\hbar})p_x \psi \Rightarrow p_x \psi = -i\hbar \frac{\partial}{\partial x}[/math] etc for the other coordinates.

 

Then I may have that [math]\vec{p}[/math] is the vector sum [math]\vec{p} = -i \hbar \sum \frac{\partial}{\partial x^{\mu}} =- i \hbar \nabla[/math], and we're done.

 

So, what's for dessert?

 

Well, noticing that the second derivatives take the form [math]\frac{\partial ^2}{\partial x^2}\psi = - (\frac{p_x^2}{\partial \hbar^2})\psi[/math], after some rearrangements, taking the same sum, and popping in the factor [math]\frac{1}{2m}[/math], I arrive at [math] -\frac{\hbar^2}{2m}\nabla^2 \psi[/math] for the kinetic term in Schroedinger's Hamiltonian.

 

Yay!