Heisenberg algebra Isomorphic to Galilean algebra

I admit my notation ##\begin{bmatrix}0&K&P\\0&0&H\\0&0&0\end{bmatrix}## has been a bit sloppy (lazy). It should have been:$$K=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\; , \; H=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\; , \;P=\begin{bmatrix}0&0&-i\\0&0&0\\0&0&0\end
  • #1
Azure Ace

Homework Statement


Given for one-dimensional Galilean symmetry the generators ##K, P,## and ##H##, with the following commutation relations: $$[K, H] = iP$$ $$[H,P] = 0$$ $$[P,K] = 0$$

Homework Equations


Show that the Lie algebra for the generators ##K, P,## and ##H## is isomorphic to the Heisenberg algebra $$[X, P] = i \hbar I$$.

The Attempt at a Solution


The Heisenberg algebra is a non-trivial central extension of the Galilean algebra. I don't know how to prove how they are isomorphic.
 
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  • #2
What is a Galilean algebra? The multiplication rules you gave under section 1. is directly how the three dimensional Heisenberg algebra is defined. If you switch the names, then what is the Heisenberg algebra to you?
 
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  • #3
fresh_42 said:
What is a Galilean algebra? The multiplication rules you gave under section 1. is directly how the three dimensional Heisenberg algebra is defined. If you switch the names, then what is the Heisenberg algebra to you?
I think by isomorphic, the commutation relations I have for my so-called "Galilean" algebra really are for the Heisenberg algebra. By Galilean algebra, I mean the Lie algebra for the Galilean group, which is the set of all Galilean transformations. I think Heisenberg algebra is the same as Bargmann Lie algebra, which as I understood it, is a non-trivial central extension of the Galilean algebra.
 
  • #4
Azure Ace said:
I think by isomorphic, the commutation relations I have for my so-called "Galilean" algebra really are for the Heisenberg algebra. By Galilean algebra, I mean the Lie algebra for the Galilean group, which is the set of all Galilean transformations. I think Heisenberg algebra is the same as Bargmann Lie algebra, which as I understood it, is a non-trivial central extension of the Galilean algebra.
Can you write down the multiplications within a given vector space basis of the Galilean algebra?

One possible representation of the three dimensional Heisenberg algebra by matrices is the algebra of matrices of the form ##\begin{bmatrix}0&*&*\\0&0&*\\0&0&0\end{bmatrix}## with Lie multiplication ##[X,Y]=XY-YX##. It is nilpotent with a one-dimensional center. Can you do similar for the Galilean algebra, whatever this is? I don't know the term Galilean group either. Of course I can make some guesses what I think you mean, but it's probably better if you define it.
 
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  • #5
fresh_42 said:
Can you write down the multiplications within a given vector space basis of the Galilean algebra?

One possible representation of the three dimensional Heisenberg algebra by matrices is the algebra of matrices of the form ##\begin{bmatrix}0&*&*\\0&0&*\\0&0&0\end{bmatrix}## with Lie multiplication ##[X,Y]=XY-YX##. It is nilpotent with a one-dimensional center. Can you do similar for the Galilean algebra, whatever this is? I don't know the term Galilean group either. Of course I can make some guesses what I think you mean, but it's probably better if you define it.

The Galilean group, as what was discussed to us, is the set of Galilean transformations given in matrix form as:
a3e57643c70bb66bcf57936968f4435d1a8c3ba1

where R is for rotations, v is for Galilean boosts, a is spatial translations and s is for time translations. The commutation relations are the following.
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98fc7c4a468ce7fa72d4ce1a447632324337ff52

01b7aa61bc2856d4e3cfc6ce5f3d72564535f1f5

ade86ef409f8f767be149b605b2a76127684a3dc

285a4d16555cfca4c80d963f36f2925c64d7037d

7337d08fee8d9be08a2969bb16b578504d9182b1

61e8488de4717b3e616094f5b8b3d8ab464aa502

ca5972e950d30691dba2fafc2e58977300b6ad69

2c57b7e93beefa09f215b05f16f8f2a309ba7775

I just took these equations from wikipedia. For the Bargmann algebra,
c5de574832089c79e6f579c52ef4de1f1f950c13

8e56cbe7fc5ad8aaa4b5e99bd208e961edcbb242

334b98de717fad3a286c171163b52e65222e72a5

7c93f26d28dda58f1c60e3938f66753530cd0fec

627522438a338d0623fe653177fbf778892aa7d3

bc33032edd56c7309ae548c711c86071e157dc0b

8f774a5b5348649053a05ecb9349301f2f659c5e

eaa99e5eac46bb60b94276b49313baea2e8dabc3

235c0adc062f85c03c8d1263d159233b1c483f23

I think this one is the same as the Heisenberg algebra my professor is talking about with ##[C'_{i}, P'_{j}] = iM\delta_{ij}## equivalent to ##[X,P] = i\hbar I \delta_{ij}##. But I'm not really sure about this. A hint given to me is that I shouldn't really take the notations seriously and that ##P## in the Galilean algebra is the central generator. Does this mean that perhaps if I compare the two, maybe the isomorphism is ##K \to X, H \to P, P \to \hbar I##?

Thank you very much for your interest in helping me!
 
  • #6
Yes, that's what Wikipedia says. But what dimension do we have here? Certainly a significantly higher number than three. So either we must consider generalized Heisenberg algebras (with increased dimension) or a low dimensional version of the Galilean group.

We have as Heisenberg algebra ##\mathcal{H} = \left\{ \begin{bmatrix} 0&K&P\\0&0&H\\0&0&0\end{bmatrix} \right\}##. The same as we get by ##\{C_1,H,P_1\}## with ##C_1 \leftrightarrow K## and only one index, no ##L## and eventually a scalar correction for ##P=p\cdot P_1##. This means: no rotation and one-dimensional boost and translations, i.e. a different algebra than the large algebra you just described.
 
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  • #7
fresh_42 said:
Yes, that's what Wikipedia says. But what dimension do we have here? Certainly a significantly higher number than three. So either we must consider generalized Heisenberg algebras (with increased dimension) or a low dimensional version of the Galilean group.

We have as Heisenberg algebra ##\mathcal{H} = \left\{ \begin{bmatrix} 0&K&P\\0&0&H\\0&0&0\end{bmatrix} \right\}##. The same as we get by ##\{C_1,H,P_1\}## with ##C_1 \leftrightarrow K## and only one index, no ##L## and eventually a scalar correction for ##P=p\cdot P_1##. This means: no rotation and one-dimensional boost and translations, i.e. a different algebra than the large algebra you just described.

Yes, I would only like to consider a smaller Galilean group. For 1D Galilean symmetry we have no rotations, right? So does that mean for a 1D physical space the two algebras are isomorphic (which now I think they are)? How about for 2D space with SO(2) rotational symmetry? I'm sorry I'm a little bit of a beginner here. Also thank you very much for your reply! I really appreciate it and I actually learned something new. ^_^
 
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  • #8
Azure Ace said:
Yes, I would only like to consider a smaller Galilean group. For 1D Galilean symmetry we have no rotations, right?
Yes.
So does that mean for a 1D physical space the two algebras are isomorphic (which now I think they are)?
Yes. You don't even need a complicated basis transformation. The basis vectors only have different letters (in the versions above) and maybe one has to deal with a factor ##\pm i## or so for the central vector ##P##. I haven't checked.
How about for 2D space? I'm sorry I'm a little bit of a beginner here. Also thank you very much for your reply! I really appreciate it and I actually learned something new. ^_^
I admit my notation ##\begin{bmatrix}0&K&P\\0&0&H\\0&0&0\end{bmatrix}## has been a bit sloppy (lazy). It should have been:
$$
K=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\; , \; H=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\; , \;P=\begin{bmatrix}0&0&-i\\0&0&0\\0&0&0\end{bmatrix}
$$
I haven't checked the two-dimensional case. You could do it as an exercise.
 
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  • #9
fresh_42 said:
Yes.Yes. You don't even need a complicated basis transformation. The basis vectors only have different letters (in the versions above) and maybe one has to deal with a factor ##\pm i## or so for the central vector ##P##. I haven't checked.
I admit my notation ##\begin{bmatrix}0&K&P\\0&0&H\\0&0&0\end{bmatrix}## has been a bit sloppy (lazy). It should have been:
$$
K=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\; , \; H=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\; , \;P=\begin{bmatrix}0&0&-i\\0&0&0\\0&0&0\end{bmatrix}
$$
I haven't checked the two-dimensional case. You could do it as an exercise.
Wow! Thank you very much! I learned a lot of new things from your replies. Thank you again! ^_^
 

FAQ: Heisenberg algebra Isomorphic to Galilean algebra

1. What is the Heisenberg algebra and how is it related to the Galilean algebra?

The Heisenberg algebra is a mathematical structure that describes the fundamental commutation relations between position, momentum, and time in quantum mechanics. It is isomorphic, or structurally equivalent, to the Galilean algebra, which describes the same relationships in classical mechanics.

2. How is the Heisenberg algebra used in quantum mechanics?

In quantum mechanics, the Heisenberg algebra is used to describe the uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. This algebra helps us to understand and calculate these uncertainties.

3. What are the main differences between the Heisenberg algebra and the Galilean algebra?

The Heisenberg algebra takes into account the effects of quantum mechanics, such as the uncertainty principle, while the Galilean algebra only applies to classical mechanics. Additionally, the Heisenberg algebra uses operators to represent physical quantities, while the Galilean algebra uses classical variables.

4. Can the Heisenberg algebra be extended to other physical systems?

Yes, the Heisenberg algebra has been extended to other areas of physics, such as quantum field theory and general relativity. These extensions help us to understand the behavior of particles at the smallest scales and in extreme environments, such as near black holes.

5. What are the practical applications of the Heisenberg algebra?

The Heisenberg algebra is used extensively in quantum mechanics and has led to many practical applications, including the development of transistors and lasers in electronics, as well as advancements in medical imaging technologies. It also plays a crucial role in the understanding and development of quantum computing.

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