# Why is wavefunction a lorentz scalar? So confused

• ginda770
In summary: Schrodinger wave function would transform like a volume, and that's not what he's saying.Anyway, thanks for clearing it up for me!
ginda770
I hope someone can explain this to me:

In multiple textbooks I've seen it said that a single particle wave function (no spin) transforms as a Lorentz scalar. I.e. if we have a Lorentz transformation from an old frame to a new frame

$$\overline{x}=\Lambda x$$

(x is short for (t,x,y,z)) then the wave function in the new frame and the old frame are related by

$$\overline{\psi}(\overline{x})=\psi (x)$$

My question is this: The square of the wave function is a probability density, so why doesn't it transform as the zeroth (time) component of a 4-vector as other densities do?

Because it's just a function from Minkowski space into $\mathbb C$. Your last equation is just the value of the function at a point in spacetime, expressed using two different coordinate systems. Click me.

Last edited:
There is some confusion here on two counts. They are related to the difference between the wave functions and the fields. (1) It is true that spin=0 Klein-Gordon field transforms as a Lorentz scalar. However this field is not the same as a wave function. In particular the absolute square of this field is not the probability density of any kind. And if you integrate the absolute square of this field over the entire space you don't get a constant value; the result could even be infinite. (2) On the reverse side of the coin, if you talk about the Schrodinger wave function without spin, it is not a Lorentz scalar. It transforms like a volume (which is again not Lorentz invariant.)

I hope this clarification helps.

ginda,

I think that fermi is right. Your transformation formula is valid for quantum field, and it is not valid for the wave function of a particle. If you want to find the wave function transformation you need to follow general Rules of Quantum Mechanics. First, set times t=t'=0, because you are looking for a position-space wave function $$\psi'(x',0)$$ in the moving reference frame at time (measured in that frame) t'=0 assuming that the position-space wave function $$\psi(x,0)$$ in the frame at rest is known at time t=0. Next note that the wave function is the inner product of the state vector $$|\psi \rangle$$ with eigenvectors $$|x \rangle$$ of the position operator $$X$$

$$X |x \rangle = x|x \rangle$$
$$\psi(x) = \langle x |\psi \rangle$$

In the Heisenberg picture you can assume that the state vector does not depend on the frame, while operators (and their eigenvectors) change. So, you need to know how the position operator transforms under boosts. For each quantum system you must know the boost generator $$\mathbf{K} = (K_x, K_y, K_z)$$. Then, the transformation of the position operator with respect to boosts along the x-axis is

$$X' = \exp(-iK_x \theta) X \exp(iK_x \theta)$$......(1)

where $$\theta$$ is the boost rapidity that is connected to the boost velocity $$v$$ by formula $$v/c = \tanh \theta$$. In relativistic QM, position operator is given by the Newton-Wigner formula. The transformation law (1) is known, see for example

A. H. Monahan and M. McMillan, "Lorentz boost of the Newton-Wigner-Pryce position operator", Phys. Rev. D, 56 (1997), 2563.

To find the wave function at non-zero times you should add time translations (unitary operator $$\exp(iHt)$$, where H is the Hamiltonian) to the boosts.

If you do the math you'll see that the position-space wave function transformation will be different from your formula.

Eugene.

Thanks so much for your replies Fermi and Eugene, I think you are both correct. When I look back at the texts, most of them are indeed talking about the Klein Gordon field, not the Schrodinger wave function. I think the reason I got confused was that I was reading Srednicki's online QFT notes (I'm taking a class using Peskin and Schroeder, but Srednicki's book was recommended to me) and he has a confusing remark in there which says
When we say that physics looks the same, we mean that two observers (Alice and Bob, say) using two different sets of coordinates (representing two different inertial frames) should agree on the predicted results of all possible experiments. In the case of quantum mechanics, this requires Alice and Bob to agree on the value of the wave function at a particular spacetime point, a point that is called x by Alice and x' by Bob. Thus if Alice’s predicted wave function is ψ(x), and Bob’s is ψ'(x'), then we should have ψ(x) = ψ'(x'). Furthermore, in order to maintain ψ(x) = ψ'(x') throughout spacetime, ψ(x) and ψ'(x') should obey identical equations of motion. Thus a candidate wave equation should take the same form in any inertial frame.

Then he goes on to explore if the K-G equation is relativistically invariant...

When I read this I thought he was saying that all wave functions (at least spinless ones) transform this way. But now I think you're right that this does not apply to the Schrodinger wave function, although I'm still a little confused by the beginning of chapter 3 in Peskin & Schroeder where they seem to say that it's obvious that any scalar function should transform this way (they make a short argument about considering the scalar function as a "measuring the local value of some quantity distributed throughout space"), but it's not so obvious to me since functions like the Schrodinger wave function and charge (or mass) density don't transform this way (I guess this means those functions are not TRUE scalars, but don't they also "measure the local value of some quantity distributed throughout space"?).

Maybe this is just a case of confusing textbooks, but I'd appreciate to hear what you guys think.

Thanks again for responding to me, I really appreciate it.

ginda770 said:
(I'm taking a class using Peskin and Schroeder, but Srednicki's book was recommended to me)
Interesting... I'm in the opposite situation of taking a class that uses Srednicki's book, but Peskin & Schroeder was recommended...

...honestly, I'm not too thrilled with either of them.

My reply is valid for a scalar field, not for a wavefunction, but I think you already knew that.

ginda770 said:
Maybe this is just a case of confusing textbooks...

That's exactly the problem. Unfortunately, most QFT textbooks are very confusing. They often pretend that wave functions and quantum fields are the same things or, at least, that they are somehow related. They also create an impression that Klein-Gordon/Dirac equations for quantum fields are analogs (or generalizations) of the Scroedinger equation for wave functions. In fact, these things have nothing in common.

The most clear and logical no-nonsense presentation of QFT can be found in Weinberg's "The quantum theory of fields" vol. 1.

Eugene.

Thanks Eugene, I'll try to track down the Weinberg book.

diazona: I only just started reading Srednicki, but I agree with your assessment of P&S. I need to find an introduction to QFT that is actually an introduction, P&S leave out so many steps in their arguments, it reminds me of Jackson's E&M. I wish there was a Griffiths-like QFT book to get me started.

After Weinberg's book my second choice is

S. S. Schweber, "An introduction to relativistic quantum field theory"

50 years ago (when this book was written) people were more inclined to understand and explain the physical meaning of what they're doing.

Eugene.

It actually turns out that a wavefunction and a quantum field are two totally different things. A wavefunction is in the end a vector in a certain space, while a quantum field is an operator. Operators act on vectors to <produce> other vectors. You might say that a quantum field acts on a <wavefunction> to produce another one.

## 1. Why is the wavefunction a Lorentz scalar?

The wavefunction is a mathematical representation of a quantum system, describing the probability amplitude of finding a particle in a particular state. It is a scalar quantity because it does not change under Lorentz transformations, which are mathematical transformations that describe how the laws of physics appear the same for all observers moving at constant speeds. This is a fundamental principle of special relativity, and the wavefunction must follow this principle in order to accurately describe the behavior of particles in different reference frames.

## 2. How does the Lorentz scalar property of the wavefunction affect its behavior?

The Lorentz scalar property of the wavefunction means that it is not affected by changes in the frame of reference. This is important because it allows for consistent predictions of particle behavior, regardless of the observer's frame of reference. This is crucial in quantum mechanics, where the behavior of particles is probabilistic and can change depending on the observer's perspective.

## 3. Can the wavefunction be a Lorentz vector instead of a scalar?

No, the wavefunction must be a scalar in order to accurately describe the behavior of particles in different reference frames. If it were a vector, it would change under Lorentz transformations and could lead to inconsistent predictions of particle behavior. Additionally, the wavefunction is a complex-valued function and does not have the necessary properties to be a Lorentz vector.

## 4. How does the Lorentz scalar property of the wavefunction relate to the principle of relativity?

The Lorentz scalar property of the wavefunction is a direct consequence of the principle of relativity, which states that the laws of physics should appear the same for all observers moving at constant speeds. The wavefunction must also follow this principle in order to accurately describe the behavior of particles in different reference frames. Without this property, the predictions of quantum mechanics would not be consistent with the principle of relativity.

## 5. Is the Lorentz scalar property of the wavefunction a fundamental aspect of quantum mechanics?

Yes, the Lorentz scalar property of the wavefunction is a fundamental principle of quantum mechanics. It is necessary for the consistency of predictions and for obeying the principle of relativity. Without this property, the wavefunction would not accurately describe the behavior of particles in different reference frames and the foundations of quantum mechanics would be compromised.

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