Active and Passive Transformations Confusion

[ThereIam]

Hi all,

I'm sort of struggling to understand the difference between active and passive transformations. I'm working with two books, and both say that in one case the vectors are transformed (active) and in the other the operators are transformed (passive). That's fine and dandy, but in terms of being able to actually identify a transformation as one or the other, I am at a loss.

For example, in Boas's Mathematical Methods in the Physical Sciences, she uses two 2x2 rotation matrices as examples. The first has a -sinΘ in the upper right and the sinΘ in the lower left, and is the vector rotation, and the second is the same but with sine terms swapped, and is claimed to be "axes" rotated.

Now, if you just switch the direction of rotation, one effectively becomes the other... i.e., change the sign of Θ and suddenly you have the other matrix. So what make one passive and the other active, really?

In my other book, Shankar's Principles of Quantum Mechanics, the treatment is a little different. He says,

Suppose we subject all the vectors |V> in a space to a unitary transformation (analogous to a rotation, I understand)

|V> → U|V>

Under this transformation the matrix elements of any operator Ω are modified as follows:

<V'|Ω|V> → <UV'|Ω|UV> = <V'|U⁺ΩU|V>

And then he goes on to talk about how it would be the same story if we just went straight to subjecting all operators to the change U⁺ΩU which is fine. What I don't understand is something pretty elementary - why is the matrix element of the operator Ω = <V'|Ω|V>. Where the hell did the <V'| come from... why not just <V|?

Thanks for reading my novel. Any help would be extremely appreciated.

Edit: So I think I figured out what the <V'| is... just any other arbitrary row vector in the space... but why isn't it specified as a basis vector? Or is this implied? Like if <V'| was a row basis vector, and |V> was another column basis vector, then I could see how it would be selecting an element of the matrix... but if they're both just some arbitrary vectors, couldn't they have some scaling effect on the matrix element?

 

[micromass]

I'm sort of struggling to understand the difference between active and passive transformations. I'm working with two books, and both say that in one case the vectors are transformed (active)

All I know about this that it is a distinction that physicists like to make. I have never read about it in math books. So I'm moving this to the physics section.

 

[ThereIam]

Well, that explains why I couldn't find anything about it in Anton's Elementary Linear Algebra book, haha.

 

[Fredrik]

This terminology confuses the hell out of me too, but I think I understand the following at least. Let R be a rotation operator on $\mathbb R^3$.

The active transformation by R of a vector x is the map $x\mapsto x'=Rx$. Note that we have
$$x_i=(Rx)_i =R_{ij}x_j.$$ The passive transformation by R of the vector x is the map $x\mapsto x'$ induced by the transformation $e_i\mapsto e_i'=Re_i$ and the equalities $x=x_ie_i=x'_ie'_i$. We have
$$x_je_j=x'_i e'_i =x'_i Re_i=x'_i (Re_i)_j e_j =x'_i R_{ji} e_j =x'_i (R^{-1})_{ij} e_j.$$ People never seem to understand the third equality above, so I'll explain it: I'm just rewriting $Re_i$ as a linear combination of basis vectors. The fourth equality requires you to understand the relationship between linear operators and matrices. (If you don't, see the https://www.physicsforums.com/showthread.php?t=694922 about it).

The above implies that $x_j=x'_i (R^{-1})_{ij}$. Multiply this by $R_{jk}$. We get
$$x_j R_{jk}=x'_i (R^{-1})_{ij}R_{jk} =x'_i (R^{-1}R)_{ik} =x'_i \delta_{ik}=x'_k.$$ So
$$x'_i=x_j R_{ji}=(R^{-1})_{ij}x_j =(R^{-1}x)_i,$$ i.e. $x'=R^{-1}x$.

So a passive transformation by $R$ of x is an active transformation by $R^{-1}$ of x.

If we view x not as the vector, but as the vector's matrix of components with respect to the ordered basis $(e_1,e_2,e_3)$, then we can describe the above as follows: An active transformation by R acts on each vector's matrix of components. A passive transformation by R acts on the basis vectors.

It's also useful to think about this in physical terms. Suppose that R is a clockwise rotation by 30°. Then you can think of the active transformation by R as a clockwise rotation by 30° of the object on which you intend to do a measurement, and a passive transformation by R as a counterclockwise rotation by 30° of the measuring device (or the entire laboratory, or the entire universe...except for the object on which you intend to do a measurement of course).

 

[Fredrik]

In my other book, Shankar's Principles of Quantum Mechanics, the treatment is a little different.

You didn't explain what he says about active vs. passive transformations. My first thought is that whatever he says, it has something to do with the fact that
$$\langle\alpha|X|\alpha\rangle =\langle\alpha|U^\dagger U X U^\dagger U|\alpha\rangle = \langle\alpha'|X'|\alpha'\rangle$$ with $|\alpha'\rangle=U|\alpha\rangle$ and $X'=UXU^\dagger$.

Note that $|\alpha\rangle$ represents the object on which we intend to do measurements, and $X$ represents the measuring device.

 

[Joker93]

You didn't explain what he says about active vs. passive transformations. My first thought is that whatever he says, it has something to do with the fact that
$$\langle\alpha|X|\alpha\rangle =\langle\alpha|U^\dagger U X U^\dagger U|\alpha\rangle = \langle\alpha'|X'|\alpha'\rangle$$ with $|\alpha'\rangle=U|\alpha\rangle$ and $X'=UXU^\dagger$.

Note that $|\alpha\rangle$ represents the object on which we intend to do measurements, and $X$ represents the measuring device.

This has to do with active or passive transformation?

 

[stevendaryl]

Active versus passive transformation for vectors has to do with changing vectors versus changing coordinates. If you are moving East, and you turn counterclockwise through 90 degrees, then you'll be heading North. That's a different velocity vector than the one you started with. But now look at your journey on a map. Hold the map one way, and you're traveling horizontally across the map. Turn the map clockwise through 90 degrees, and now you're traveling vertically across the map. Your velocity hasn't changed, but its representation in terms of horizontal and vertical has changed.

Mathematically, an active change of a vector means an operation that takes a vector and returns a different vector. A passive change leaves the vector alone, but changes the basis that you use to describe that vector.

So if v is a vector, and R is a rotation, a linear transformation on vectors, then R(v) is a new vector, call it \tilde{v},different from v. In terms of components, let v^i be the components of v, so
v = v^i e_i, where e_i is the i^{th} basis vector, and where by convention, repeated indices are summed over. Let \tilde{v}^i, be the components of \tilde{v}. Since R is a linear transformation, we can write:

\tilde{v}^i = R^i_j v^j (again, j is summed over)

So,

\tilde{v} = \tilde{v}^i e_i = R^i_j v^j e_i

Now, we can define a new basis, \tilde{e}_j via: \tilde{e}_j = R^i_j e_i. So we can also write:

\tilde{v} = v^j \tilde{e}_j

So there are two ways to represent the rotated vector \tilde{v} in terms of components:

1. Keep the basis e_i and rotate the components, \tilde{v}^i = R^i_j v^j
2. Keep the components v^i and rotate the basis, \tilde{e}_i = R^j_i e_j

These are both active transformations--they are changing the vector v to a different vector, \tilde{v}. Now, let's use \tilde{v} in the form \tilde{v}^i e_i and rotate back using the inverse transformation R^{-1}. Again, we have the option of writing R^{-1}(\tilde{v}) by keeping the components the same, \tilde{v}^i and rotating the basis, or the other way around. Let's keep the components the same. So we have:

v = R^{-1}(\tilde{v}) = \tilde{v}^i (R^{-1})^j_i e_j

So we have:

\tilde{v} = \tilde{v}^i e_i
v = \tilde{v}^i (R^{-1})^j_i e_j

So using the tranformed coordinates \tilde{v}^i might mean two different things:

1. We might be talking about the vector \tilde{v} in the original basis, e_i. That's an active transformation.
2. We might be talking about the vector v in the counter-rotated basis, (R^{-1})^i_j e_i. That's a passive transformation.

 

[Superresistivity]

This terminology confuses the hell out of me too, but I think I understand the following at least. Let R be a rotation operator on $\mathbb R^3$.

The active transformation by R of a vector x is the map $x\mapsto x'=Rx$. Note that we have
$$x_i=(Rx)_i =R_{ij}x_j.$$ The passive transformation by R of the vector x is the map $x\mapsto x'$ induced by the transformation $e_i\mapsto e_i'=Re_i$ and the equalities $x=x_ie_i=x'_ie'_i$. We have
$$x_je_j=x'_i e'_i =x'_i Re_i=x'_i (Re_i)_j e_j =x'_i R_{ji} e_j =x'_i (R^{-1})_{ij} e_j.$$ People never seem to understand the third equality above, so I'll explain it: I'm just rewriting $Re_i$ as a linear combination of basis vectors. The fourth equality requires you to understand the relationship between linear operators and matrices. (If you don't, see the https://www.physicsforums.com/showthread.php?t=694922 about it).

The above implies that $x_j=x'_i (R^{-1})_{ij}$. Multiply this by $R_{jk}$. We get
$$x_j R_{jk}=x'_i (R^{-1})_{ij}R_{jk} =x'_i (R^{-1}R)_{ik} =x'_i \delta_{ik}=x'_k.$$ So
$$x'_i=x_j R_{ji}=(R^{-1})_{ij}x_j =(R^{-1}x)_i,$$ i.e. $x'=R^{-1}x$.

So a passive transformation by $R$ of x is an active transformation by $R^{-1}$ of x.

If we view x not as the vector, but as the vector's matrix of components with respect to the ordered basis $(e_1,e_2,e_3)$, then we can describe the above as follows: An active transformation by R acts on each vector's matrix of components. A passive transformation by R acts on the basis vectors.

It's also useful to think about this in physical terms. Suppose that R is a clockwise rotation by 30°. Then you can think of the active transformation by R as a clockwise rotation by 30° of the object on which you intend to do a measurement, and a passive transformation by R as a counterclockwise rotation by 30° of the measuring device (or the entire laboratory, or the entire universe...except for the object on which you intend to do a measurement of course).

But an Active transformation actually changes a vector (it's direction in the case of rotation) which is a physical thing while a passive transformation is just changing our description of it. So, shouldn't the former be different from the latter? And therefore this "So a passive transformation by $R$ of x is an active transformation by $R^{-1}$ of x" is a little misleading?

 

[Demystifier]

in Boas's Mathematical Methods in the Physical Sciences, she uses

I haven't noticed before that Boas is a female.