I am taking an introductory course to quantum field theory. The lecturer goes on saying that some transforms as (represented by $\to$) . I tried to ask the lecturer, and he said that he means some regular transformation consisting of a domain, a codomain and a rule for assigning each element of the domain to an element of the codomain. So can you explain to me what is the domain, the codomain and the rule in the following expression, for example? I dont understand how to read expressions like these.

$$ a \to A a $$

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    $\begingroup$ Having some more context might make it easier to answer this question. Transformations occur in many different situations in QFT. $\endgroup$ – David Z Sep 14 '16 at 9:22
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    $\begingroup$ Transforms means it is subject to a function. Which function is context dependent. The function might be a linear transformation (like in Special Relativity), a conformal or a gauge transformation (like in QFT), etc. $\endgroup$ – QuantumBrick Sep 14 '16 at 10:32
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    $\begingroup$ Saying something 'tranforms as' means that there is a group acting on the class of somethings. $\endgroup$ – user1504 Sep 14 '16 at 14:10

I don't know this particular expression, but the usual picture is the following:

  • On the one hand, you have a system and you assign quantities to it that describe the system such as the total energy, the momentum four-vector, the spin, etc.

  • On the other hand, you have symmetries of this system, for instance Lorenz transformations, parity, time reversal and transformations like time evolution operators (if those are not considered symmetries).

Obviously, symmetries behave differently for different objects. If I do a time reversal, the total energy stays the same while the momentum changes signs because the object travels in the other direction. Obviously, you need to know how everything behaves.

The mathematics

Each object lives in some sort of space or manifold. For instance, the energy might live in $\mathbb{R}$, the four vector might live in $\mathbb{R}^4$ or on some four-dimensional manifold. Now the symmetries are usually a group and for every group, you can study group representations. "Transforms as" just refers to the particular representation of the symmetry.

To illustrate the point, let's stick to the Lorentz symmetry: The Lorentz symmetries form a group that consists of rotations and "boosts" and is usually denoted $SO(3,1)$. You can consider a representation of the Lorentz group on $\mathbb{R}$ or $\mathbb{R}^4$ (or any vector space; the representations are not necessarily unique, but let's just suppose they are). Call these representations $\pi_1$ and $\pi_2$ then we have:

The object $\phi$ transforms as a vector and the object $\psi$ transforms as a scalar


A Lorentz group element $L\in SO(3,1)$ acts on $\phi\in \mathbb{R}^4$ as $\pi_2(L)\phi$ while it acts on $\psi\in \mathbb{R}$ as $\pi_1(L)\psi$.

In other words: You have to classify the different irreducible representation of the symmetry group. Saying that an object transforms as, for instance, a tensor, vector or a scalar (or whatever) just means that it transforms according to the representation that you labeled as "tensor", "vector" or "scalar" (or whatever). Note that this also specifies the mathematical space that the objects actually live in.

  • $\begingroup$ my answer i believe partially share your views about his question as well. $\endgroup$ – AMS Sep 14 '16 at 11:13
  • $\begingroup$ Thanks Martin. Suppose that $\phi \to \phi e^{i \alpha}$ $(Eq.1)$ is a symmetry (I suppose we read it as $\phi$ transform as $\phi e^{i \alpha}$). Then there is a $g$ in some group $G$ and a representation $\pi$ such that $g$ acts on $\phi$ as $\pi(g)\phi$. But that means that $\pi(g) = e^{i \alpha}$. Correct? Now, how do I determine only using Eq.1 that $g$ acts on $\phi^*$ as $\pi(g) = e^{-i \alpha}$ and not $e^{i \alpha}$? $\endgroup$ – Mikkel Rev Sep 14 '16 at 11:50
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    $\begingroup$ @mas: Yes. I think the two answers (and also the gist of the others) complement each other very well. I try to give a mathematician's "meta" perspective, while your approach is a bit more hands-on and technical. $\endgroup$ – Martin Sep 14 '16 at 13:22
  • $\begingroup$ @MariusJonsson: I'm not completely sure about your question. In principle, there could also be different kinds of group actions (left vs. right) and corresponding representations. In your case however, it looks like you are interested in the dual represenation (en.wikipedia.org/wiki/Dual_representation) - but that's an entirely different question. $\endgroup$ – Martin Sep 14 '16 at 13:26

As David Z mentioned that transformation occurred in different situation in QFT. Instead of being formal, we can try to have a example, where this "transform as" can be realized.

The representations of the Lorentz algebra can be labeled by two half-integers: $(j_{-},j_{+})$. The dimension of the representation $(j_{-},j_{+})$ is given by $(2j_{-}+1)(2j_{+}+1)$. In particular

  • $\left(\frac{1}{2},0\right)$ known as left-handed spinor ($\psi_{L}$) as $\psi_{L} \in \left(\frac{1}{2},0\right)$.
  • and $\left(0,\frac{1}{2}\right)$ known as right-handed spinor ($\psi_{R}$) as $\psi_{R} \in \left(0,\frac{1}{2}\right)$.

(Which you may already familiar with from your QFT course)

Now, both $\psi_{L}$ and $\psi_{R}$ transform differently under Lorentz transformations. Whereas \begin{eqnarray} \psi_{L} \to \psi_{L}' = \Lambda_{L}\psi_{L} & = &\exp\left[(-i\theta-\eta)\cdot\frac{\sigma}{2}\right]\psi_{L}\\ \psi_{R} \to \psi_{R}' = \Lambda_{R}\psi_{R} & = &\exp\left[(-i\theta+\eta)\cdot\frac{\sigma}{2}\right]\psi_{R} \end{eqnarray} where $\sigma's$ are Pauli matrices and $\theta,\psi$ are rotation and boost parameters, respectively. Now by using the basic properties of Pauli matrices one can show that $\sigma^{2}\psi_{L}^{*}\sigma^{2}=\psi_{R}$. From this it can be shown that, \begin{equation} \sigma^{2}\psi_{L}^{*} \to \sigma^{2}(\Lambda_{L}\psi_{L})^{*}=(\sigma^{2}\Lambda^{*}_{L}\sigma^{2})\sigma^{2}\psi^{*}_{L} = \Lambda_{R}(\sigma^{2}\psi_{L}^{*}) \end{equation} Bottom line is though $\psi_{L}$ transform as left handed spinor which belongs to $(1/2,0)$, the $\sigma^{2}\psi_{L}^{*}$ transform as right handed spinor, which is $(0,1/2)$.

So its merely how a state is transforming under a given transformation is the matter of question, without going to deep into mathematical terminology.


What your teacher means is just a map, say $f$, which takes elements from a set (domain) $\mathcal{S}$ and assigns them to an element of another set $\mathcal{S}'$ (the codomain). In mathematical form this is usually written as

$$f: \mathcal{S} \to \mathcal{S}'$$

If $x\in \mathcal{S}$, then $f(x)\in \mathcal{S}'$. For instance, the map $f(x)=e^x$ takes an element $x\in\mathbb{R}$ (domain) and transforms it to the element $e^x\in\mathbb{R}^+$ (codomain).

In your example, I take that $\psi$ is the wavefunction and $A$ some linear operator acting on it, which is just a linear map which takes an element $\psi$ of the space of wavefunctions $\mathcal{H}$ (domain) and transforms it to another element $A\psi$ of some other (or same) space $\mathcal{H'}$ (codomain).


I will start with an example from linear algebra. Suppose we are given a finite-dimensional vector space $V$. We choose a basis $e_i$, $i=1...n$ and expand all vectors in terms of basis. For $x \in V$ we have $$ x = x^i e_i, $$ where summation over repeated indices is implicit (Einstein's summation convention). From now I will identify vector $x$ with its components $x^i$. Linear operator $O$ is represented by a matrix $O^i_{\ \ j}$ etc. Now suppose I want to change basis to $f_i = \beta^j_{\ \ i}$, where $\beta$ is some nonsingular matrix. Then the components of $x$ also change, but $x$ as a geometric object does not. Therefore I write $$ e_i \mapsto f_i = e_j \left( \beta ^{-1} \right)^j_{\ \ i}, $$ $$ x^i \mapsto \beta^i_{\ \ k} x^k, $$ $$ x \mapsto x .$$ Of course linear operators need to change accordingly. For example, $$ O^{i}_{\ \ j} \mapsto \beta^i_{\ \ k} O^k_{\ \ m} \left( \beta^{-1} \right)^m_{\ \ j}.$$ This transformation by similarity matrix assures that $(Ox)^i=O^i_{\ \ j}x^j$ transforms as components of a vector, as it should be: $$ (Ox)^i \mapsto \beta^i_{\ \ j} x^j. $$ This is only a specific example of a general scheme: there is some group of transformations acting on objects in the theory. In this case it was group of all possible changes of basis. We don't give a name or special symbol to the transformation, we just write explicitly what it does to specific objects.

What I presented above is the so called passive interpretation of transformations. I regarded vectors as invariant objects. Change of basis is thought of as "change of point of view". In many physical situations there are physically equivalent ways of describing the system. You can use plenty of coordinates to describe the same physical system. Or you can use various reference frames. Of course some objects used in the description, such as numerical values of velocities or functional form of the Hamiltonian will change.

There is also active point of view. The idea is that you think of basis $e_i$ as fixed. You change coordinates of $x$ according to $$ x^i \mapsto \beta^i_{\ \ j} x^j. $$ Note that this is genuine transformation of $x$ vector: $$ x \mapsto \beta x, $$ where $\beta$ is now though of as linear operator (rather than transition matrix). Now what we are usually after is covariance: since $Ox$ is also a vector, we would like it to transform accordingly, $$ Ox \mapsto \beta Ox .$$ This demand forces us to transform $O$ accordingly, $$ O \mapsto \beta O \beta^{-1} .$$

Note that these two ways of looking at transformations laws are dual to each other. In one case set of all possible transformations is set of all nonsingular matrices and in the other, set of all nonsingular linear operators. Of course matrices are just components in some basis of linear operators. These theme is rather universal in mathemathics and profoundly important.


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