1. Dec 18, 2009

### 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?

2. Dec 18, 2009

### Fredrik

Staff Emeritus
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: Dec 18, 2009
3. Dec 18, 2009

### fermi

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.

4. Dec 18, 2009

### meopemuk

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.

5. Dec 18, 2009

### ginda770

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
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.

6. Dec 18, 2009

### ginda770

7. Dec 18, 2009

### diazona

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.

8. Dec 18, 2009

### Fredrik

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

9. Dec 18, 2009

### meopemuk

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.

10. Dec 18, 2009

### ginda770

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.

11. Dec 18, 2009

### meopemuk

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.

12. Dec 20, 2009

### dextercioby

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.