# I Change of basis?

1. Mar 30, 2017

### mike1000

I think I do not quite understand the role that change of basis plays is superpositioning of states.

If there is an observable, $A$ which is represented by the operator $\hat{A}$, then the set of observed values for that observable will be the set of eigenvalues defined by the operator $\hat{A}$ and the set of states that the observable can be measured in are the corresponding eigenvectors for the set of eigenvalues of the matrix operator, $\hat{A}$.

Lets pretend we have the matrix operator $\hat{A}$ for some observable. However, before we determine its eigenvalues and eigenvectors we apply a rotation operator, $\hat{R}$ , of some type to it. (I do not know if this makes any sense to do in QM but I know I can do it from a purely linear algebra point of view.)

After I apply the rotation operator to $\hat{A}$ we have the new operator $$\hat{B}=\hat{R}\hat{A}$$Now we determine the eigenvalues and eigenvectors of the $B$ observable.

What is the relationship between the eigenvectors of the $A$ observable and the eigenvectors for the $B$ observable? Will each eigenvector in $A$ be some linear combination(ie superposition) of the eigenvectors in $B$? Are the calculated probabilities of measuring the same event going to be different when measured relative to the two different basis? WIll some states which were observable in the $A$ basis not be observable in the $B$ basis?

Last edited: Mar 30, 2017
2. Mar 30, 2017

### blue_leaf77

You don't apply unitary transformation, such as rotation, that way. When you rotate the system, you rotate the state as well. Therefore the final state after rotation is $|\psi_f \rangle = \hat R |\psi_i\rangle$. Measuring the observable $A$ in the rotated system is represented by $\langle \psi_i | \hat R ^{-1} A \hat R|\psi_i\rangle$, which is equivalent to measuring an observable $\hat R ^{-1} A \hat R$, this is the proper form of the rotated observable. Note that it is also Hermitian whereas $\hat R A$ is not.

3. Mar 30, 2017

### blue_leaf77

"measured relative to the two different basis" is not the standard way, if not wrong, that people normally say in QM. You are measuring an observable with respect to a state and a state need not be a basis state. However, if you mean that the matrix representation which is represented in different bases and subsequently measured relative to the same state, then it would be no. The expectation value of the observable will not change when only the matrix is represented in different basis.
An observable will stay observable forever.

4. Mar 30, 2017

### mike1000

Thank you for this. In this one simple reply you drove home the importance ot Hermitian matrices and unitary transformations and incidentally, the importance of symmetry and its relationship to real eigenvalues.

Last edited: Mar 30, 2017
5. Mar 30, 2017

### mike1000

What I meant by "measured in two different basis" was the following... if we change the basis, vectors which were orthogonal in the first basis may not be orthogonal in the second basis. This suggest to me that the eigenvectors of the $A$ operator would be linear combinations of the the eigenvectors of the $B$ operator which, I now know, is $B=R^{-1}AR$.

Is this correct?

Last edited: Mar 30, 2017
6. Mar 30, 2017

### vanhees71

This is a misconception! An observable can be measured on the system, if it is defined on this system, i.e., if there is a measurement which allows you to measure this observable. You can thus measure the observable, no matter in which state the system is prepared. If it is prepared in a state, described by a statistical operator $\hat{\rho}$, the probability to measure the eigenvalue $a$ of the operator $\hat{A}$ that represents the observable $A$, is given by
$$P(a)=\sum_{\beta} \langle a,\beta | \hat{\rho}|a, \beta \rangle,$$
where $|a,\beta \rangle$ is a complete set of orthornormalized eigenvectors of $\hat{A}$ for the eigenvalue $a$.

All the physical outcomes of QT are of course independent of the basis you've chosen to describe your observables. It's as with usual vectors in 3D Euclidean space: A vector is some directed quantity, e.g., the position vector is the directed connection from the origin of your reference frame to the position of the particle. It doesn't depend on the basis chosen to describe this "arrow" in terms of three real numbers.

7. Mar 30, 2017

### mike1000

I hope I understand correctly what you are saying.

A state vector $|\psi\rangle$ defines a point in Hilbert Space. The state vector has components given by the projection of that point onto some orthonormal basis. For operator $A$ the orthonormal basis are the eigenvectors of $\hat{A}$. The length squared of the state vector (which is the distance of the point from the origin) is given by the inner product of the vector with itself, $$L^2=\langle\psi|\psi\rangle$$ When we change the coordinate system, which I am calling a change of basis, by some unitary transformation (ie a transformation which preserves the inner product) then the same point will be described by a new state vector in the new basis as $|\phi\rangle$. However the length of the vector will be invariant under the transformation, which implies that $$L^2 = \langle\psi|\psi\rangle=\langle\phi|\phi\rangle$$Which means that I should calculate the same probabilities for the same observable regardless of the basis in which the state vectors are described.

I know I am probably not completely correct, but I think I am getting close.

Last edited: Mar 30, 2017
8. Mar 30, 2017

### mikeyork

Your $\psi$ and $\phi$ are the same state and, if properly normalised, $L=1$. But the state descriptions differ between bases and you express this by projection into either basis:

$|\psi> = \sum_i |\phi^A_i> <\phi^A_i|\psi> = \sum_j |\phi^B_j> <\phi^B_j|\psi>$

where $\phi^A_i$ are the eigenstates in A and similarly for B.

In particular, if $\psi = \phi^A_k$ is the $k^{th}$ eigenstate in basis A, then it is, in general, a superposition in B:

$|\phi^A_k> = \sum_j |\phi^B_j> <\phi^B_j|\phi^A_k>$

9. Mar 30, 2017

### mike1000

You are confusing me.

When you say that "your $\psi$ and $\phi$ are the same state" what do you mean? I am pretty sure they have different elements. So why do you say they are the same state?

Please try to keep it simple.

10. Mar 30, 2017

### mikeyork

I have kept it as simple as possible and you are confusing yourself with notions like "I am pretty sure they have different elements" without apparently knowing what your "elements" are (I don't either). It is the observables that have different eigenvalues.

11. Mar 30, 2017

### Staff: Mentor

I live in a town whose street grid is laid out with avenues running from southwest to northeast, with the cross streets running from southeast to northwest. If I am walking uptown with speed $v$, I can write my velocity vector as $(v,0)$ in the uptown/crosstown basis or as $(v/\sqrt{2},v/\sqrt{2})$ in the compass basis. Different elements, but same vector with the same physical significance... and the same general idea works for vectors in Hilbert space, which is what the states are.

12. Mar 30, 2017

### mike1000

No, I am not confusing myself. The components of the state vectors are the projection of the point on each of the basis vectors. If you change the basis vectors the projection of the same point on the new set of base vectors will be different, ie two state vectors with different elements. That is how I view it currently.

13. Mar 30, 2017

### mike1000

The way you describe it I agree with. Its the same point described in two different bases, meaning two different state vectors.

I did get something from mikeyorks description. The point in Hilbert Space is the state.

14. Mar 30, 2017

### mikeyork

Please make up your mind about whether you are talking about "points" or states, "elements" or eigenstates. Once you get that clear you will understand that when you say "the same point" you mean the same state because QM is about states and state vectors, not "points" or "elements". It is only the descriptions of the state in terms of the eigenstates that differ and I fully explained this in post #8.

15. Mar 30, 2017

### mike1000

Well, I could say the same to you because a state is a point in Hilbert Space! When I say element I mean element of a vector. As far as this statement goes,
I think that was the entire point of my post #7

16. Mar 30, 2017

### mikeyork

No. A state is represented (not "is") by a vector in Hilbert space.
The vectors $|\psi>$ and $|\phi>$ in your #7 are the same vector. Since each vector represents a unique state, that means $\psi$ and $\phi$ are the same state. That is one of your areas of confusion, since you insisted they are different states.

I have tried to help you, but it looks to me like you insist on remaining confused between states and their description/expression in a particular basis. Good luck with that.

17. Mar 30, 2017

### mike1000

The vector defines a point in Hilbert Space. It is that point that is invariant under a unitary transformation. If what you said is true then when I change the basis I must change the state, but you told me before that when I change the basis I do not change the state. I am referring to post #8 where you said my $\psi$ and $\phi$ are the same state.

When you say the same vector, that implies to me that corresponding elements of the vectors are equal. That is not true. What is true is that the inner products of each vector with itself are equal. (ie the length of the two vectors is the same)

Last edited: Mar 30, 2017
18. Mar 30, 2017

### mikeyork

That is just another example of your confusion. Can someone else help this guy? I'm giving up.

19. Mar 30, 2017

### mike1000

So you are agreeing that the corresponding elements of the two vectors are not equal?

I don't think you were trying to help me.

20. Mar 30, 2017

### mikeyork

The projections of the state vector onto a basis vector are not equal for different bases -- as is evident from my post #8. If you insist on using your own private language, rather than the accepted language of QM, you will remain confused about this and no doubt carry on claiming that a state vector can represent two different states.
If I look at an object in fron of me and choose my z-axis as the direction that points to it, then I will have defined its projection onto the z-axis. If I turn my head to the left, but keep my z-axis pointing straight ahead of me then the projection of the object onto the z-axis will have changed, but the object is in exactly the same position (has the same position vector, is the same state) as before but is expressed in a different frame (basis).

That is why I'm going to stop trying.

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21. Mar 30, 2017

### mike1000

I do not think I ever claimed that a state vector can represent two different states.

I am not insisting on using my own private language. I will get the language right I have no doubt about that.

22. Mar 30, 2017

### mike1000

I know the object is in the exact same position. I have been saying all along that the "point" is invariant under the transformation. Are the eigenvectors of the operator in the two different coordinate systems the same?

23. Mar 30, 2017

### mikeyork

In post #9, you wrote
And again you are evading the fact that the same point describes the same vector and the same state. Turning my head doesn't change the state it just changes the frame of reference.
No but the state is the same. Just because you change the representation (basis, co-ordinate frame, or whatever) it does not mean you have changed the state.

24. Mar 30, 2017

### Staff: Mentor

Yes. (The eigenvectors are vectors, so that answer could reasonably have been "yes, of course").
Of course the components are different in different coordinate systems, but the vectors are the same. Another example from my diagonal street grid: The eigenvectors of $\hat{N}$, the "north" operator, are $(1,0)$ and $(-1,0)$ in the compass basis and $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ in the avenue/street basis - one points due north and one points due south, no matter the choice of basis.

25. Mar 30, 2017

### mike1000

Of course I know that the head and tail of the vector are the same. I have said all along that the point does not move. And, also the origin does not move. The vector from the origin to the point remains the same under the transformation.

Some how, I was attaching significance to the values of the individual components of the state vectors. Isn't the value of the individual components interpreted as the probability amplitude? (I know it is the projection of the observable operator onto the corresponding basis vector) but it is interpreted as a probability amplitude in that direction.