# A question for Sakurai's advanced qm

1. Oct 30, 2011

### kof9595995

On page 96, he defined $\psi'(x')=S\psi(x)$, but it seems what he later derived for S only transforms the spinor part not the space-time coordinate of the 4-component wave function, then how is the space-time coordinate primed after acted by S? I'm pretty sure it's not a typo according to what he did on page101, eqns (3.158)~(3.161).

2. Oct 30, 2011

### ytuab

I had Sakurai's advanced qm a few month ago.
But now I don't. Sorry, I imagine from your sentence.

First, Dirac's solution includes exponential function

$$\psi = \int_{-\infty}^{\infty} \frac{d^3 p}{\sqrt{2E_p}} a_p u(p) e^{-ipx} \cdots$$

Here, both the p (momentum, energy) and x (time , spece) are 4 vectors.
So "px" means scalar (= vector x vector). (px doesn't change under Lorentz transformation.)
And the integration of p is from -infinity to +infinity, so the change from p to p' is meaningless.

But of course, the differentiation by time and space coordinate in Dirac equation changes under Lorentz transformation ( x' = a x ).

$$\frac{\partial }{\partial x'_{\mu}} = \sum_{\nu} \frac{\partial x_{\nu}}{\partial x'_{\mu}} \frac{\partial}{\partial x_{\nu}} = \sum_{\nu} a_{\mu \nu} \frac{\partial}{\partial x_{\nu}}$$

So under Lorentz transformation, Dirac equation becomes

$$(-i\hbar \gamma^{\mu} \partial_{\mu}' + mc) \psi' (x') = (-i\hbar \gamma^{\mu} a_{\mu\nu} \partial_{\nu} + mc) S \psi (x) = (-i\hbar \gamma^{\mu} \partial_{\mu} + mc) \psi (x) =0$$

Is this what is discussed here ? (If not, I transfer your question to someone with advanced qm.)

3. Oct 30, 2011

### George Jones

Staff Emeritus
I won't have access to my copy of Sakarai until tomorrow. Is $S$ the 4-component spinor versionof a Lorentz transformation $L$? If so, then $x' = Lx$ and
\begin{align} \psi \left( x' \right) &= S \psi \left( x \right) \\ \psi \left( Lx \right) &= S \psi \left( x \right) \\ \psi \left( x \right) &= S \psi \left( L^{-1} x \right) \end{align}

4. Oct 30, 2011

### kof9595995

Yes, S is purely written in terms of gamma matrices and parameters like angle and rapidity, and the expression came into my mind first was $\psi'(x')=S\psi({\Lambda}^{-1}x)$

5. Oct 30, 2011

### ytuab

George Jones, I want to confirm this equation.
According to p.60 of an introduction to quantum field theorey by Peskin, Dirac field change under Lorentz transformation,

$$\psi (x) \quad \to \quad \ \psi' (x) = \Lambda_{1/2} \psi (\Lambda^{-1} x) \quad (S = \Lambda_{1/2}) \quad ( x' = \Lambda x )$$

Changing x to x', this meaning is the same as

$$\psi' (x') = \Lambda_{1/2} \psi (x) = S \psi (x) \qquad ( x = \Lambda^{-1} x')$$

S includes gamma matrices, so I think the form of psi changes under Lorentz trandformation.

Sorry. the next equation is what you mean ?

$$\psi' (x') = S \psi (\Lambda^{-1} x') = S \psi (x) \quad or \quad \psi' (x) = S \psi (\Lambda^{-1} x) \qquad x' = \Lambda x \quad ( x = \Lambda^{-1} x')$$

6. Oct 31, 2011

### kof9595995

Ah..I see where my problem is, I was thinking $$\psi' (x) = S \psi (\Lambda^{-1} x)$$, which means an active transformation acting on the wavefunction not the reference frame, so $$\psi' (x') = S \psi (\Lambda^{-1} x') = S \psi (x)$$ means Sakurai transformed both the reference frame and the wavefunction? That's weird, why did he do this?

7. Oct 31, 2011

### ytuab

I think it is easier to imagine "vector" ( ex, 4-vector potential A(x) ) instead of spinor, first.

$$A^{\mu} (x) = ( A^0 (x), A^1 (x), A^2 (x), A^3 (x) ) \qquad x^{\mu} = (x^0, x^1, x^2, x^3)$$
This 4-vector A (x) means that there is a thing " vector A " at the position of x ( from the viewpoint of reference frame K ).
Here we can consider 4-vector A as an thing such as "arrow".
So there is one "arrow" ( which vector direction is (A0, A1, A2, A3) ) at the coordinate of ( x0, x1, x2, x3 ) in the reference frame of K.

From the viewpoint of a different reference frame K', the position of the "arrow" changes (under Lorentz transformation),

$$x'^{\mu} = \Lambda^{\mu}_{\nu} x^{\nu} \quad ( x' = \Lambda x ) \qquad x'^{\mu} = ( x'^0, x'^1, x'^2, x'^3 )$$
This means that from this reference frame K', the "arrow" exists at x'.
( x' from K' originally exists at x from K. )
How do we see the direction of this arrow from this reference frame K' ?

A is 4-vector, so this direction changes as 4-vector,

$$A'^{\mu} = \Lambda^{\mu}_{\nu} A^{\nu} \quad ( A' = \Lambda A )$$
As a result, from the reference frame K', there is an arrow which direction is A' at the coordinate of x'. So,

$$A'^{\mu} (x') = \Lambda^{\mu}_{\nu} A^{\nu} ( \Lambda^{-1} x' ) \qquad x = \Lambda^{-1} x'$$
( x' from K' originally exists at x from K. )
In the case of spinor, the position (= x ) of spinor changes as 4-vector, which is the same as that of vector A above.
But the change of "spinor direction" is different from the vector A.
( As you said, we use "S" instead of "Lambda" )

$$\psi' (x') = S \psi (\Lambda^{-1} x')$$

Last edited: Oct 31, 2011
8. Nov 1, 2011

### kof9595995

Thanks, your explanation is very clear. I just get confused every now and then on this issue, now i get the moment of clarity, but probably someday I'll get confused again: (