A question for Sakurai's advanced qm

• kof9595995
In summary: K' )So the vector A(x) is now A(x',K').But this A(x', K') is same thing as A(x, K).Here the point is that the thing "arrow" is fixed ( not changed ) in any reference frame.But the position of the "arrow" changes depending on the reference frame.So the view point of K, the arrow is at the position of x, and the viewpoint of K', the arrow is at the position of x'.But the "arrow" itself is not changed.If we apply this idea to spinor, it is the same.The spinor itself does not change under Lorentz transformation.But
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).

kof9595995 said:
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).

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.)

kof9595995 said:
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).

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}

George Jones said:
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}
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 $\psi'(x')=S\psi({\Lambda}^{-1}x)$

George Jones said:
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}

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.

kof9595995 said:
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 $\psi'(x')=S\psi({\Lambda}^{-1}x)$

Sorry. the next equation is what you mean ?

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

ytuab said:
Sorry. the next equation is what you mean ?

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

kof9595995 said:
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?

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:
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: (

1. What is "A question for Sakurai's advanced qm"?

"A question for Sakurai's advanced qm" is a theoretical physics problem that was posed by physicist Jun John Sakurai in his influential textbook on quantum mechanics.

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