# Transformations in vector space

1. Dec 22, 2009

### Mr confusion

dear all,we know that active transformation refers to action of changing vectors keeping the operators unchanged whereas passive transformation refers to change of operator components keeping vectors unchanged.
what i cannot understand(i am just starting quantum mechanics)is in the former if we introduce change in vectors, then the basis vectors also get changed. so the matrix component of the operators will also change! how can they remain unchanged.??
thanks.

2. Dec 22, 2009

### Truecrimson

Say, we use the standard Cartesian basis. If we rotate a vector (0,2) to a vector (2,0) does the basis change? No. The basis is still (1,0) and (0,1) i.e. (2,0)=2(1,0)+0(0,1)

3. Dec 22, 2009

### Fredrik

Staff Emeritus
Let V be a vector space over $\mathbb F$ (either the complex numbers or the real numbers). For each ordered basis $$E=(\vec e_1,\dots,\vec e_n)[/itex], we define a function $$f_E:V\rightarrow\mathbb F^n$$ by $$f(\vec x)=\begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}$$ where the $x_i$ are defined by $$\vec x=\sum_i x_i\vec e_i$$ The map $E\mapsto f_E$ is clearly bijective, so it identifies functions of this type with ordered bases. The group of invertible linear operators from V into V is called GL(V). There's an obvious left action of GL(V) on V defined by $$(A,\vec x)\mapsto A\vec x$$ and a slightly less obvious right action of GL(V) on the set of ordered bases defined by $$(f_E,A)\mapsto f_E\circ A$$ The former is sometimes called an "active" transformation of V, and the latter is sometimes called a "passive" transformation of V. (Note that a passive transformation of V isn't really a transformation of V at all). Last edited: Dec 22, 2009 4. Dec 23, 2009 ### RedX How are the two transformations equivalent then? Take the expectation value of the operator X on the vector |v> after it has been transformed by A: <v|A-1 X A|v>. According to the passive transformation, |v> doesn't change under the action of A, so the expectation value is <v|X|v>. All the passive transformation seems to do is to change the coordinate labels. 5. Dec 23, 2009 ### Mr confusion thanks to all of u for helping me. truecrimson, i think i am having a wrong concept. i am thinking that active transformation is an act of transforming all vectors,including the basis vectors. can you help me here? frederick, i am really not an expert, i could not make out what are in those black boxes? that is why i cannot gather what you have said. 6. Dec 23, 2009 ### Fredrik Staff Emeritus $$\vec y=A\vec x$$ $$y_i=(Ax)_i=\sum_j A_{ij}x_j$$ $$\vec y=\sum_i y_i \vec e_i=\sum_i\sum_j A_{ij}x_j \vec e_i=\sum_j x_j\left(\sum_i A_{ij}\vec e_i\right)=\sum_j x_j \vec e_j{}'$$ where I have defined $$\vec e_j{}'=\sum_i A_{ij}\vec e_i$$ The thing on the left is $\sum_i x_i\vec e_i$ with the components changed, and the thing on the right is $\sum_i x_i\vec e_i$ with the basis vectors changed. Both changes are induced by A in a natural way. I don't know what you're doing here. If A|v>=|v>, A is the identity operator. What black boxes? Last edited: Dec 23, 2009 7. Dec 23, 2009 ### Truecrimson I suppose we're talking about Schrodinger picture and Heisenberg picture. Imagine you've a quantum state vector. 1. Usually we think of the state vector evolving in time by a unitary transformation $$U$$. $$|x\rangle \to U|x\rangle$$ 2. We're free to choose any basis to represent the vector as a unique linear combination of basis vectors. That is an active transformation or Schrodinger picture. But we can also fix the vector and allow basis vectors to evolve in time which gives an equivalent description. This is a passive transformation or Heisenberg picture. For example, if the vector rotates clockwise in Schrodinger picture, you want the basis to rotate counterclockwise in Heisenberg picture. In doing so, we can't use operators written in the old basis anymore. Since an operator acts on the right, we have to transform an operator $$A\to U^{\dagger} AU$$. So the expectation value stays the same. $$\langle x|U^{\dagger} AU|x\rangle$$. Last edited: Dec 23, 2009 8. Dec 23, 2009 ### Fredrik Staff Emeritus When we talk about the Schrödinger picture vs. the Heisenberg picture, there's no need to mention bases at all. 9. Dec 23, 2009 ### RedX I got it now. An active transformation clearly changes the vector by changing the components but keeping the same basis, i.e., action under GL(V). A passive transformation implicitly changes the vector by changing the basis, as the definition $$(f_E,A)\mapsto f_E\circ A$$ says that f changes and since there is a 1-1 map between basis vectors E and f that means E changes. The two transformations transform the vector in the same way, of course, it's just that one does so obviously and the other implicitly. I had always thought an active transformation was a change on vectors, and a passive transformation was a change on operators. That is wrong as both change vectors. However, in the passive transformation, because the basis are changed, the operators have to be converted to another language via similarity transformation (what truecrimson said). thanks for explaining this and clearing things up! 10. Dec 24, 2009 ### Truecrimson I see. Thanks. 11. Dec 28, 2009 ### Mr confusion i am really very sorry, frederik. it so happened that i was using an older browser version and so all the math part appeared black.i have changed it and now see them clearly. however, please tell me what is meant by "for heisenberg or scroedinger pics we do not need to mention a basis" 12. Dec 28, 2009 ### Fredrik Staff Emeritus Sure. In the Schrödinger picture, state vectors are time dependent and operators are time independent. The time dependence is described by $$|\psi;t\rangle=e^{-iHt}|\psi;0\rangle$$ Let's define $U(t)=e^{-iHt}$ and $|\psi\rangle=|\psi;0\rangle$$, and write the above as $$|\psi;t\rangle=U(t)|\psi\rangle$$ The expectation value of an operator A in this state is $$\langle A\rangle_{|\psi;t\rangle}=\langle\psi;t|A|\psi;t\rangle=\langle\psi|U(t)^\dagger AU(t)|\psi\rangle=\langle A(t)\rangle_{|\psi\rangle}$$ where I have defined $$A(t)=U(t)^\dagger AU(t)$$ As you can see, the expectation value of A in the state [itex]|\psi;t\rangle$ is equal to the expectation value of A(t) in the state $|\psi\rangle$. This is why we have the option to think of the states as time-independent and the operators as time-dependent. This option is called the Heisenberg picture.

As I said, no need to involve bases. The difference between the Schrödinger picture and the Heisenberg picture is a matter of where you put the parentheses:

$$\Big(\langle\psi|U(t)^\dagger\Big) A\Big(U(t)|\psi\rangle\Big)=\langle\psi|\Big(U(t)^\dagger AU(t)\Big)|\psi\rangle$$

13. Dec 28, 2009

### Mr confusion

thanks. this is really a bliss to be among so many quantum experts while taking this course. this is removing the fears i had earlier that i will not understand this course at all.

14. Jan 9, 2010

### Mr confusion

fredrik, my course is in quantum mech is in full run. looking at this again, i have something to ask you. hope you will help me again.
as far as i have understood, in scroedinger picture state vectors undergo a rotation (as the operator ie. the propagator is unitary) in hilbert space.
but in heisenberg picture, it is the operators change with time while state vector remains fixed.you have beautifully showed how the expectation value of observable remains the same i both pictures.
but i was wondering whether time dependent scroedinger equation will remain valid in hiesenberg picture? because the state vectors are now fixed and their time derivatives will be zero!! then how will i progress with my statistical calculations of observables in such a picture?
i hope my question is understandable. i have built up my english quite recently.

15. Jan 9, 2010

### Mr confusion

truecrimson, will you please confirm me about this- i think i need to rotate the basis clockwise too, because then only the the state vector will appear static.if i move basis counterclock, then it seems to me that the state vector will approach me with double the speed. or am i wrong in visualisasion?

Last edited: Jan 9, 2010
16. Jan 9, 2010

### Phrak

This is the first time I've heard of "active" and "passive" transformations, though Wikipedia has done a quick job of letting me know that these are calculationally equivalent changes where a state vector evolves over time, or equivalently a basis evolves over time.

I think the issue has been clouded by too much quantum mechanical machinery.

In the most comprehensible case one should simply consider vector rotations in Euclidian two-space. Rotating a vector to the left is quantitatively equivalent to rotating the coordinate system the right. These are synonymous with the Schrödinger and Heisenberg approach respectively, no?

17. Jan 9, 2010

### Mr confusion

it seems to me you are correct . but i am getting confused!!
if i rotate the basis to right, ..... then yes.....it is equivalent to vector rotated to left.
but my book says in heisenberg representation state vectors remain frozen. now if i rotate a basis to right at ang speed w while the vector was rotating originally to the left at w, then i will see the vector rotate with angular velocity 2w?

however i am sure the discussion is not clouded at all. my book has stressed very much on these transformations before describing scroedinger picture. and after all , it is a quantum mechanics text.

Last edited: Jan 9, 2010
18. Jan 9, 2010

### Truecrimson

After Fredrik replied that we don't need to mention about basis at all in Schrodinger vs Heisenberg picture, I found a passage in Peres that confirms that both Schrodinger and Heisenberg picture are active transformations. My second reply is partly wrong. Sorry about that.

So your first question is about Schrodinger (change in states) vs Heisenberg picture (change in operators), not active vs passive transformation.

Edit: What is the book you're using? At first, I assumed (wrongly) that the two pictures and the two transformations are synonymous because I searched through Sakurai and found only one place where he mentions passive transformation in the context of usual rotations in vector spaces. And he discusses Schrodinger vs Heisenberg pictures before that, if I recall correctly.

Last edited: Jan 9, 2010
19. Jan 9, 2010

### Mr confusion

truecrimson, you have truly killed me down!!!
all along my course , i am sharing the idea that in heisenberg picture, state vectors remain frozen and the operators change. how can it be an active transformation? how?
http://electron6.phys.utk.edu/qm1/modules/m5/pictures.htm [Broken]

please noteur text is a material given by our professor.he accumulates and compiles it himself. but he has referred us W. GREINER AND F.SCHWABL, i think he mainly prepares it from these two books.

Last edited by a moderator: May 4, 2017
20. Jan 9, 2010

### Truecrimson

Seeing that the explanation on the webpage is more or less the same as mine, I think it's a valid reasoning. The confusion probably just arises from various usages of the words "active" and "passive" transformations by different authors. Sometimes a passive transformation means the inverse of the correspending active transformation so it gives $$UAU^{\dagger}$$ instead of $$U^{\dagger}AU$$.

IMO, there's nothing to worry about.