I Energy and momentum in quantum mechanics

  • I
  • Thread starter Thread starter Vyurok
  • Start date Start date
Vyurok
Messages
7
Reaction score
1
Here is an excerpt from a lecture by my teacher Emil Akhmedov (MIPT)
from AkhmedovEmil's lecture.webp

And I have the following question.
It turns out that the probability wave describing a free particle is determined by its energy and momentum, right? But what do these two words—energy and momentum—actually mean in quantum mechanics?
In classical mechanics, we had Noether’s theorem, which gave us a quantity conserved along the trajectory due to time homogeneity—that’s what we called energy—and a quantity conserved due to space homogeneity—that’s what we called momentum.
But how are these concepts defined in quantum mechanics? And how can we show that they determine the probability wave describing a free particle in the way presented here?
 
Physics news on Phys.org
Energy and momentum are still defined as the generators of the infinitesimal symmetry transformations of the states of the system. The difference is that, while in classical mechanics the state is given in terms of (generalized) coordinates and momenta, in non-relativistic quantum mechanics it is given by a complex wave function ##\psi(\mathbf{r})## depending on coordinates.

Now, the infinitesimal spatial translation ##\mathbf{r} \rightarrow \mathbf{r} + \boldsymbol{\varepsilon}## of coordinates must be represented by some operator which acts on the wave function ##\psi##. Wigner's theorem states that such operator, representing a symmetry operation on the system, must be either unitary or anti-unitary. In particular, if it is a unitary operator, then the appropriate generator of the infinitesimal symmetry must be Hermitian.

If, say, ##\widehat{T}_{\varepsilon}## is an operator of spatial translation by an infinitesimal vector ##\boldsymbol{\varepsilon}##, then the wave function changes appropriately as
$$
\widehat{T}_{\varepsilon}\psi(\mathbf{r}) = \psi(\mathbf{r} + \boldsymbol{\varepsilon}) \approx \psi(\mathbf{r}) + \boldsymbol{\varepsilon}\cdot\nabla\psi(\mathbf{r}) = \left[1 + \boldsymbol{\varepsilon}\cdot\nabla \right]\psi(\mathbf{r})
$$
so you could say that ##\nabla## is the generator of this symmetry transformation. Since, however, ##\widehat{T}_{\varepsilon}## must be a unitary operator, the generator must be Hermitian and ##\nabla## is not. You make it Hermitian by additional imaginary unit ##\mathrm{i}##, and according to historical convention (and the choice of appropriate physical units) this ##\mathrm{i}## is introduced alongside ##\hbar## in the following way
$$
\widehat{T}_{\varepsilon} \approx 1 + \frac{\mathrm{i}}{\hbar}\boldsymbol{\varepsilon}\cdot\frac{\hbar}{\mathrm{i}}\nabla \rm{.}
$$
The generator of the infinitesimal spatial translation is therefore a Hermitian operator ##-\mathrm{i}\hbar\nabla## (I changed ##\hbar/\mathrm{i}## into ##-\mathrm{i}\hbar##). On the other hand, in classical mechanics the appropriate generator of spatial translations is the momentum. So in quantum mechanics we say that ##\widehat{\mathbf{p}} = -\mathrm{i}\hbar\nabla## is the momentum operator. But it is still associated with the spatial-translation symmetry. Similar arguments go for the Hamiltonian.
 
  • Like
Likes nasu, Dale, Spinnor and 1 other person
Also, this is more for the purpose of your "scientific entertainment", but I see that in your lecture notes there is mention of "wave function of a photon". This is a tricky concept, because you need relativistic version of quantum mechanics, i.e., quantum field theory, to speak properly in the language of photons. It is no coincidence that Dirac did not specify explicitly what he meant by "photon wave function" in his texbook on QM, despite starting from the example of light interference. If you want to look more into the problems (and their potential resolutions) regarding the concept of wave function for photons, take a look at the review "Photon Wave Function" by Iwo Białynicki-Birula (it is available on arXiv here: https://arxiv.org/abs/quant-ph/0508202) and references therein.
 
Is the following understanding correct?

We are considering such spacetime transformations — symmetries — under which the laws of physics remain unchanged. For example, if space is homogeneous, time is homogeneous, and space is isotropic, then the corresponding symmetry transformations are spatial translation, time translation, and rotation, respectively — is that right?

The way the system's state changes under such transformations can be described mathematically using certain operators: ##U(\varepsilon)=\exp(-i\varepsilon G)##

Here, ##G## is called the generator of the given symmetry transformation — correct?

You claim that the generator of spatial translation is momentum, both in classical and quantum mechanics. In your example, you consider how the wave function (which describes the system's state in quantum mechanics) changes under an infinitesimal spatial translation and conclude that the generator of such a symmetry transformation should look like: ##-i\hbar\nabla##

And you identify this with the momentum.

I know Noether's theorem in the following form (see attached images). This theorem connects homogeneity or isotropy with a conserved quantity along a trajectory, as follows:

Spatial homogeneity gives the conserved quantity:
$$ p = \frac{\partial L}{\partial \dot{r}} $$
which we call momentum;

Temporal homogeneity gives:
$$ E\overset{\mathrm{def}}=\frac{\partial L}{\partial v}v-L $$
which we call energy;

And spatial isotropy gives:
$$ l_{z}\overset{\mathrm{def}}=\frac{\partial L}{\partial \dot{x}}(-y)+\frac{\partial L}{\partial \dot{y}}=xp_{y}-yp_{x} $$
which we call angular momentum.

This is how I’m used to defining momentum, energy, and angular momentum.

But you are saying that these quantities can be defined as generators of the corresponding symmetry transformations. So, there must be some relationship between these generators and the conserved quantities along trajectories — but what is it exactly?

Could you please explain this in more detail?

For instance, in classical mechanics, the state of the system is defined by generalized coordinates and momenta: ##(q, \; p)##, right?

Let’s consider a spatial translation by s. The system's state transforms as: ## (q, \; p)\rightarrow(q-s, \; p) ##

This can be represented using operators as:

$$
\begin{pmatrix}
q-s\\
p
\end{pmatrix}
= \left(
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+(-i)\cdot(-i)s
\begin{pmatrix}
\frac{1}{q} & 0\\
0 & 0
\end{pmatrix} \right)
\begin{pmatrix}
q \\
p
\end{pmatrix}
\Rightarrow G = -i
\begin{pmatrix}
\frac{1}{q} & 0\\
0 & 0
\end{pmatrix}
$$

How exactly is this resulting matrix connected to momentum as the conserved quantity: ## p = \frac{\partial L}{\partial \dot{r}} \; ? ##

And what if we consider a time translation?

In this case, the state transforms as: ## (q, \; p)\rightarrow(q, \; p) ##

Which can be expressed using operators as:

$$
\begin{pmatrix}
q\\
p
\end{pmatrix}
= \left(
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+(-i)s
\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix} \right)
\begin{pmatrix}
q \\
p
\end{pmatrix}
\Rightarrow G =
\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}
$$

This seems odd. Could it be that I’m misunderstanding something?

If so, could you please explain where exactly the mistake in my reasoning is?

But if it is correct, then how is the resulting matrix

$$
\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}
$$

related to energy as the conserved quantity:

$$ E\overset{\mathrm{def}}=\frac{\partial L}{\partial v}v-L \; ? $$

As for Wigner’s theorem, I roughly know what it is, but could you recommend a source where I can read the best explanation and proof?

Also, could you recommend good textbooks on quantum mechanics in general?
 

Attachments

  • Киселёв_page-0001.webp
    Киселёв_page-0001.webp
    37 KB · Views: 16
  • Киселёв_page-0002.webp
    Киселёв_page-0002.webp
    35.8 KB · Views: 18
  • Киселёв_page-0003.webp
    Киселёв_page-0003.webp
    30.2 KB · Views: 8
  • Киселёв_page-0004.webp
    Киселёв_page-0004.webp
    30.4 KB · Views: 22
  • Киселёв_page-0005.webp
    Киселёв_page-0005.webp
    21.5 KB · Views: 23
  • Киселёв_page-0006.webp
    Киселёв_page-0006.webp
    22.6 KB · Views: 22
  • Киселёв_page-0007.webp
    Киселёв_page-0007.webp
    32.1 KB · Views: 12
Last edited:
@Vyurok your question doesn't look like a question about QM interpretations, just about basic QM. Is that correct?
 
PeterDonis said:
@Vyurok your question doesn't look like a question about QM interpretations, just about basic QM. Is that correct?
Yes, that's right. Should I write to another forum topic?
 
Vyurok said:
Yes, that's right. Should I write to another forum topic?
No, I'll just move this one to the regular QM forum.
 
Moderator's note: Thread moved to the regular QM forum since it is not about interpretations but standard QM.
 
Vyurok said:
Is the following understanding correct? (...) you consider how the wave function (which describes the system's state in quantum mechanics) changes under an infinitesimal spatial translation and conclude that the generator of such a symmetry transformation should look like: ##-i\hbar\nabla##. And you identify this with the momentum.
Yes, this understanding is essentially correct. The operator ##-i\hbar\nabla## can be called the momentum operator, because - just like in classical mechanics - it was obtained by considering the effect of infinitesimal spatial translation on the system (the technical difference, of course, is that in quantum mechanics the state of the system is not given by coordinates and momenta, but by a complex wave function ##\psi##).

Vyurok said:
I know Noether's theorem in the following form (...). This is how I’m used to defining momentum, energy, and angular momentum.
Classical mechanics can be formulated neatly using the approach based on the Lagrangian or on the Hamiltonian. In the Lagrangian approach you have an explicit recipe for how to calculate the conserved quantities - the integrals of motion - using Noether's theorem. In the approach based on the Hamiltonian, you have to "guess", in some sense, the form of the conserved quantities - namely, the integrals of motion in the canonical formulation of classical mechanics are such phase-space functions that their Poisson bracket with the Hamiltonian is zero. In practice, however, you would not "guess" but rather calculate the conserved quantities using Noether's theorem, then go over from ##q##'s and ##\dot{q}##'s into the ##q##'s and ##p##'s, and substitute them into the expression for the calculated integral of motion.

Vyurok said:
But you are saying that these quantities can be defined as generators of the corresponding symmetry transformations. So, there must be some relationship between these generators and the conserved quantities along trajectories — but what is it exactly?
"Generators of symmetry transformations" is just the group-theoretic language for the conserved quantities.

I'm not sure at the moment about the second part of your question @Vyurok, where you play around with matrices, but essentially there is also a way of obtaining the integrals of motion in the canonical formulation of classical mechanics by considering the generating functions and the canonical transformations (you can look up these "keywords" in the meantime :smile: ). Perhaps someone competent will jump in and explain this point clearly here.

Vyurok said:
Also, could you recommend good textbooks on quantum mechanics in general?
There is an entire thematic sub-forum here on PhysicsForums that deals with textbook recommendations, check it out:
https://www.physicsforums.com/forums/science-and-math-textbooks.21/

R. Shankar's "Principles of Quantum Mechanics" seems like one of the good choices - the book is not too thick, contains interesting topics (in addition to the standard ones) and includes a self-contained mathematical preliminary on vector spaces. It is also good to ask your teachers about their recommendations.
 
Back
Top