# On the definition of Active Transformation

1. Nov 11, 2009

### neelakash

In Franz Schwabl's QM book the idea of active transformation has been put in this way:

"transformation of a state Z to Z' and view Z' from the same reference frame".The statement follows by the argument that the state which arises through the transformation $$\Lambda^{-1}$$ is given as $$\psi\ '(\ x) =\psi\ (\Lambda^{-1}\ x)$$

$$\psi\ (\ x)$$ has been actively moved from $$\ x$$ to $$\ x'$$ where $$\ x'=\Lambda^{-1}\ x$$.

That is, here x' denotes a point in the same reference gotten from $$\ x'=\Lambda^{-1}\ x$$

In the same reference frame $$\psi\rightarrow\psi\ '$$; But how can that $$\psi\ '(\ x)$$ be equal to $$\psi\ (\Lambda^{-1}\ x)$$?

I know that active transformation is expressed in literature most commonly as $$\psi\ '(\ x) =\psi(\ x')$$,[as opposed to passive transformation $$\psi\ '(\ x') =\psi\ (\ x)$$---here primed co-ordinate means a new primed co-ordinate system],however,I am struggling a bit with the definition...

Last edited: Nov 12, 2009
2. Nov 12, 2009

### Fredrik

Staff Emeritus
Suppose that you use a coordinate system with the x axis to your right, the y axis in the forward direction, and the z axis up. Suppose also that you're describing a particle with a wavefunction $\psi$ that has a peak somewhere on the positive y axis (in front of you), and that you rotate yourself (and your coordinate system) 90° counterclockwise around the z axis. This is equivalent to rotating all vectors 90° clockwise. Let $\Lambda$ be the matrix of that clockwise rotation. After the rotation, the wavefunction $\psi'$ that you must use to describe the same particle has a peak on the positive x axis (to your right).

Let $\vec x$ be the position of the peak in your new coordinate system. It's on the positive x axis. You know that if you write $\psi'(\vec x)=\psi(\vec y)$, then $\vec y$ must be on the positive y axis (because that's where $\psi$ has its peak). That means that $\vec y$ can be obtained from $\vec x$ by applying a 90° counterclockwise rotation. That rotation is represented by $\Lambda^{-1}$, so $\vec y=\Lambda^{-1}\vec x$.

I would call this an active transformation by $\Lambda$, or a passive transformation by $\Lambda^{-1}$. This appears to be the opposite of the terminology you're using. I don't know if I'm the one who has misunderstood the terminology. I thought "active" was supposed to mean that the transformation is applied to the points in the vector space you're considering (in this case $\mathbb R^4$).

Last edited: Nov 12, 2009
3. Nov 12, 2009

### Fredrik

Staff Emeritus
When you see an equation like that, you should interpret it like this

$$f(p)=f\circ y^{-1}(y(p))=f\circ z^{-1}(z(p))$$

where

$$f:M\rightarrow\mathbb R$$ (a scalar field on M)
$$y,z:M\rightarrow\mathbb R^4$$ (two coordinate systems on M)

4. Nov 12, 2009

### neelakash

Let us forget about the passive transformation,for the time being.We do not invoke any new co-ordinate system rotated w.r.t. the original...I wish to focus on the meaning of Active transformation in this (x) system.
Active transformation should well be done without ever rotating my frame...operating the wave function by some operator in my unrotated frame.Such that $$\psi\ (\ x)\rightarrow\psi\ (\Lambda^{-1}\ x)$$...this is the point where I am failing to convice myself...

5. Nov 12, 2009

### neelakash

The scenario is clearer if the following viewpoint is taken: You take the wavefunction $$\psi$$.Give some spatial translation to $$\psi$$,w.r.t. the co-ordinate system where you are.This retains the form (dependence on the variable) of the wave function the same.$$\psi$$ remains $$\psi$$,but the location of its peak etc. is shifted up to a constant distance.Thus,

$$\psi(\ x) =\psi\ (\ x\ +\ a)$$
It is a bit obscure to see if you first shift the point,and according to that,shift your wavefunction. From the book I saw,they depicted the transformation as rotation in that frame.Obviously,that is a bit difficult to see;because,it looks like $$\psi\rightarrow\psi'$$

Last edited: Nov 12, 2009
6. Nov 12, 2009

### Fredrik

Staff Emeritus
I don't see why my explanation didn't help, and I'm a bit confused now. As I said, a rotation of $\mathbb R^3$ by a matrix R is equivalent to rotating yourself as described by R-1. If you want to express the transformation of the wavefunction in the form $\psi'=T\psi$ where T is an operator, that T is defined by $T\psi(\vec x)=\psi(\Lambda^{-1}\vec x)$.

Edit: You're saying some really strange things here, for example "$$\psi$$ remains $$\psi$$,but the location of its peak etc. is shifted". This is obviously not possible. If the location of the peak has shifted, it's not the same function. $\psi$ has been replaced by $\psi'=\psi\circ T_a{}^{-1}$, where $T_a$ is the translation function defined below.

It's a bit easier to visualize translations than rotations. Suppose that you move a distance a to the left or that you move the universe the same distance to the right. Either way you'll end up describing everything as being further to your right. Define $T_a(x)=x+a$. If you were using the wavefunction $\psi$ before, and $\psi'$ is the one you're going to use now, we must have $\psi'(x)=\psi(x-a)$. The new wavefunction must (for example) take the same value at 0 as the old one did at -a. So $\psi'(x)=\psi(T_a{}^{-1}(x))$, where $T_a$ is the function that moves every point in space a distance a to the right.

This is the same thing I said in #2, but with translations instead of rotations.

Last edited: Nov 12, 2009
7. Nov 12, 2009

### neelakash

What you said was perfectly fine...The only thing is that I was trying to avoid bringing another frame in the picture...

8. Nov 12, 2009

### Fredrik

Staff Emeritus
I think I've been making this more complicated than it needs to be. The value of the wavefunction at a specific (but arbitrary) point p in space is a coordinate independent quantity that's related to the probability that the particle will be detected there. Let's call this value f(p). You should think of "space" as just an abstract set of points here. A coordinate system is a function that takes points in "space" into $\mathbb R^3$. Let's express f(p) using two different coordinate systems, y and z

$$f(p)=f\circ y^{-1}(y(p))=\psi(x)$$ (where I have defined $x=y(p)$ and $\psi=f\circ y$)

$$f(p)=f\circ z^{-1}(z(p))=\psi'(x')$$ (where I have defined $x'=z(p)$ and $\psi'=f\circ z$)

This gives us $\psi'(x')=\psi(x)$. Now if $x'=\Lambda x$, we also have $x=\Lambda^{-1}x'$ and therefore $\psi'(x')=\psi(\Lambda^{-1}x')$. Since the point p was arbitrary, this holds for all x' in $\mathbb R^3$, so it no longer matters what we call the variable. We might as well call it x. Therefore, we have

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

for all x. So the wavefunction changes according to

$$\psi\rightarrow\psi'=\psi\circ\Lambda^{-1}$$

(Replace "space" with "spacetime" and $\mathbb R^3$ with $\mathbb R^4$ if time is involved).

Last edited: Nov 12, 2009
9. Nov 12, 2009

### neelakash

Look at what I have said just before the statement you quoted:

When I said,

,what I really meant is the functional form of $$\psi$$ remains unchanged...dependence on the variable is precisely the same as earlier...Like peak is shifted to a constant distance.

Last edited: Nov 12, 2009
10. Jan 31, 2010

### neelakash

I wish to mention that the precise answer to my question can be found at Ballentine's book,p-176.And there is no hard and fast rule that says

How you will writw about it,depends on you...

Last edited: Jan 31, 2010