# Why the name Second Quantization ?

1. Dec 2, 2005

### DaTario

Why the name "Second Quantization"?

Hi all,

The title said it all. My question is:

How is one to interpret the name second quantization ?

Which specifically is quantized twice ?

Best Wishes

Sincerely,

DaTario

2. Dec 2, 2005

### Meir Achuz

Quantum mechanics takes the classical variables p and x that describe the state of a classical system into non-commuting operators that act on a wave function that describes the state of the quantum mechanical system.

For quantum field theory, the old QM wave function is now considered a quantum mechanical operator. This is a second quantization, hence the name. The old QM wave function is an operator that can create and destroy particles. It acts on a new type of state function tht includes the details about the particles.

3. Dec 2, 2005

### ZapperZ

Staff Emeritus
Why the name Second Quantization? Because First Quantization was already taken. [A,B] is roughly the First Quantization. Second Quantization doesn't mean something is done twice (or else we would have called it Twice Quantized).

Zz.

4. Dec 2, 2005

### marlon

marlon

ps : do not make "strange" quotes on a photon's energy
That is a big no no if you read the forum's guidelines

5. Dec 5, 2005

### DaTario

what do you mean by strange quotes on photon's energy ?

By the way, thank you for the indication of the thread.

Zapperz, thank you also for your comments. But I am not sure I have understood well your understanding of the term second quantization.

Anyway, thank you all.

creation and anilillation operators is one of the main products of second quantization formalism while the wave function is one of the main products of First quantization.

6. Dec 6, 2005

### dextercioby

Well, we don't have "first quantization" and "second quantization". We just have QUANTIZATION, or "canonical formalism" if you prefer this terminology. Quantum mechanics is QUANTIZATION applied to classical dynamical systems with FINITELY MANY degrees of freedom, while Quantum field theory is QUANTIZATION applied to classical dynamical systems with INFINITELY MANY degrees of freedom, the latter a.k.a. "classical fields", or "finite dimensional nonunitary irreducible linear representations of the universal covering group of the restricted Poincaré group"...

That's all there is to know...

Daniel.

7. Dec 6, 2005

### marlon

This is not accurate. Not only do the degrees of freedom need to be finite in the case of QM, they also need to be FIXED !!!

regards
marlon

8. Dec 6, 2005

### dextercioby

What do you mean by fixed ? If you mean certain requirements upon the rank of the hessian matrix

$$W_{ij}=:\left(\frac{\partial^{2}L}{\partial \dot{q}^{i}\partial \dot{q}^{j}} \right)$$

to be constant in time when evaluated on the surface of all primary constraints, then you're absolutely right and i appologized not to have mentioned this fact in my first posting.

Daniel.

9. Dec 6, 2005

### marlon

This is indeed my point.

You directly gave the techical version but there is a more intuitive explanation on this on to the Wikipedia website (if you want, just look for second quantization)

regards
marlon

10. Dec 6, 2005

### George Jones

Staff Emeritus
Actually, infinite-dimensional unitary irreducible linear representations of the universal covering group of the restricted Poincaré group.

Regards,
George

11. Dec 6, 2005

### marlon

To Dexter and George,

I do not know who of you is right but shouldn't we keep the discussion onto a more intuitive level ? I mean, i assure you that 99,9% of the people (including science advisors and mentors) will not consider such an answer to be very clear. Ofcourse this does NOT mean that it is wrong.

Just my opinion.

regards
marlon

12. Dec 6, 2005

### Physics Monkey

George,

I'm a bit confused by your reply. Are not the quantum fields one usually makes use of finite dimensional representations (vector, spinor, etc) of a non-compact group and hence non-unitary?

DaTario,

Part of the problem with the name second quantization is that it is something of a historical misnomer. Early quantum field theorists thought that they were somehow "quantizing" the wavefunction. For example, Dirac proposed his famous equation for the electron as a single particle relativistic wave equation. However, the relativistic quantum field describing the electron obeys the same equation. Confusion arose. I tend to think of it as something analogous to the way we still use the term "electromotive force" for something that isn't a force.

13. Dec 6, 2005

### George Jones

Staff Emeritus
Ok, I will try and be a bit more clear at least on the infinite-dimensional representation part. To this end, I'm going to forget about untarity, covering spaces, Poincare etc., and I'm going to consider some toy examples.

Take ordinary physical space to be $$\mathbb{R}^3$$, and consider the action of rotations on fields defined in space.

An example of a scalar field defined on space is temperature $$T$$, which associates a temperature $$T \left( \vec{r} \right)$$ with every position $$\vec{r}$$ in space. In other words, $$T$$ is a function, with $$T: \mathbb{R}^3 \rightarrow \mathbb{R}$$, so $$T$$ is a member of the infinte-dimensional function space $$\{f: \mathbb{R}^3 \rightarrow \mathbb{R} \}$$.

Now let a rotation $$R$$ act on the space of scalar fields. Since $$R$$ operates (rotates) on (3-dim) vectors, this action is defined through a representative operator as follows. The representative of $$R$$ acts on a scalar field $$T$$ to give a new scalar field (i.e., another function) $$T'$$ such that the new temperature at the rotated positon $$\vec{r}' = R \vec{r}$$ is the same as the old temperature at the unrotated position $$\vec{r}$$:

$$T' \left( \vec{r}' \right) = T \left( \vec{r} \right).$$

Thus, the function $$T'$$ is defined by

$$T' \left( \vec{r} \right) = T \left(R^{-1} \vec{r} \right).$$

Now consider a field that assigns a vector $$\vec{E} \left( \vec{r} \right)$$ to each position $$\vec{r}$$ in space. $$\vec{E} \left( \vec{r} \right)$$ is an element of a 3-dimensional space (say) $$V$$ (to distinguish $$V$$ from the space of positions), but $$\vec{E}$$ itself is an element of the infinte-dimensional space $$\{ \vec{A}: \mathbb{R}^3 \rightarrow V \}$$ of vector-valued functions of position.

A rotation $$R$$ acts on the space as vector fields as follows. A rotation $$R$$ acts on the vector field $$\vec{E}$$ to give a new vector field $$\vec{E}'$$ such that the new field evaluated at the rotated position $$\vec{r}'$$ field is the same as the old field evaluated at the unrotated position and then rotated $$\vec{r}$$.
This makes sense because $$R$$ can act directly on $$\vec{E} \left(\vec{r} \right)$$, since $$\vec{E} \left(\vec{r} \right)$$ is an element of the 3-dimensional space $$V$$.

$$\vec{E}' \left( \vec{r}' \right) = R \left( \vec{E} \left( \vec{r} \right) \right).$$

Thus, the vector-valued function $$T'$$ is defined by

$$\vec{E}' \left( \vec{r} \right) = R \left( \vec{E} \left(R^{-1} \vec{r} \right) \right).$$

For both scalar and vector fields, finite-dimensional (1-dim for scalar fields, 3-dim for vector fields) have been used to get at the actual required infinite-dimensional representations.

It is an interesting mathematical exercise to show that $$R^{-1}$$, not $$R^$$, is need in the arguments in order to define a representation, i.e., a homomorphism of groups.

It is also interesting to go through this for the action of the Poincare group on "classical" Dirac spinor-valued fields on spacetime. I may try and give a pedagogical exposition of this, including unitarity, in another thread.

Regards,
George

Last edited: Dec 6, 2005
14. Dec 6, 2005

### marlon

:) George :)

Now this is what we call : "a piece of theoretical art"

regards
marlon

ps : i have been thinking to set up some kind of "general introduction to theoretical physics"-thread covering the intro of QFT all together with basic gauge symmetry and the implementation of group theory. I would like to invite you to participate (along with all others that like this idea) and check out some of the texts i already have written in my journal. What do you think ?

15. Dec 6, 2005

### marlon

ps George, we can also cover the difference between QM and QFT , and "why we use fields" (see my journal). I have written a first attempt on this. I provided a link to that text in my first post of this thread. Let me know what you think of it, please. I would really appreciate it

regards
marlon

16. Dec 6, 2005

### George Jones

Staff Emeritus
I didn't notice your post until after I made my second post. My second post might shed some light, or it might just roil the waters! My examples are somewhat poor because I have used a compact group, but as I said in that post, I may start a thread and go through this (except for proving the irreducibility!) in some detail for "classical" Dirac fields

Regards,
George

17. Dec 6, 2005

### George Jones

Staff Emeritus
Sounds very interesting.

I would like to participate, but I think and write *very* slowly, so I don't know how much this will happen.

Regards,
George

18. Dec 6, 2005

### marlon

No problem, me too.

For example, you already have written a nice text in this thread that can serve our goals.

regards
marlon

19. Dec 6, 2005

### George Jones

Staff Emeritus
Last edited by a moderator: Apr 21, 2017
20. Dec 6, 2005

### Physics Monkey

Thanks for the reply, George, I can see it was just some confusion over terminology.