# I Two orthonormal bases that span the same space

1. Apr 5, 2017

### mike1000

I just read something that I do not want to misinterpret.

If there are two orthonormal basis that span the same space, which I think implies that each basis can be written in terms of the other basis, then measurements made with respect to each basis will not commute?

Does this mean that position and momentum are two different basis that span the same space? If so, what is that common space?

I think this is a very good introduction to QM.

Last edited: Apr 5, 2017
2. Apr 5, 2017

### Staff: Mentor

Taking the same base twice is compatible with your description, and measurements will certainly commute if you have the same base twice (this is just an example).
The Hilbert space of your particle. Note that position and momentum have eigenstates that are a bit problematic.

3. Apr 5, 2017

### mike1000

Lets restrict my definition to two different bases that span the same space. Would you give the same answer?

4. Apr 5, 2017

### Staff: Mentor

Consider energy and angular momentum in a hydrogen atom, for example. They commute, and they clearly measure different things. You can find a common basis for both (that's guaranteed by the commutation relation, but you don't have to, you have some freedom within the states of the same energy and total angular momentum.

5. Apr 5, 2017

### mike1000

I am not sure I understand what you are saying.

But I am almost sure of what the author of the paper is saying. The author of the paper is saying that if you have two different bases that span the same space and you can express each base in terms of the other base, then the two bases are Fourier Transform pairs. This implies that they will not commute. Please go to pages 22 to 25 in the following paper....

6. Apr 5, 2017

### Staff: Mentor

The author is deliberately choosing operators that do not commute.

Of course you can have that case. But it is not the only option.

7. Apr 5, 2017

### mike1000

I am not so sure. He makes the following statement. He is talking about taking measurements relative to two different axis.
I did make a change to the first post. The two bases have to be orthonormal bases.

Can you tell me what restrictions on the two bases are required to make it the case? For instance, if each basis is orthonormal and you can express each basis in terms of the other basis, would that be enough?

#### Attached Files:

• ###### gkWeoGumxJtioLrCWLqt3gBAHddlhROtylU1UTBGa0QMcgXWJY-KGSKldNZCy80SdWe4BlazQmiyLgdlbbu=w298-h331-no.jpg
File size:
12 KB
Views:
34
Last edited: Apr 5, 2017
8. Apr 5, 2017

### Staff: Mentor

I am. The choice is interesting in the context of what is shown there. It is not the only possible choice.
That doesn't change the conclusion.

Enough for what?
You can always express every vector in a vector space as linear combination of base vectors.

9. Apr 5, 2017

### mike1000

Here is his example from the paper. It shows two orthogonal bases. The $|0\rangle |1\rangle$ basis and the
$|+\rangle |-\rangle$ basis. There is no question that the two bases shown do not commute.

The figure shows the $|+\rangle |-\rangle$ basis oriented at 45° to the $|0\rangle |1\rangle$ basis. If we rotate the $|+\rangle |-\rangle$ basis by some arbritary amount (but less than 45°) will the two bases still not commute? For any angle, except an angle that would make the two bases equal, shouldn't we expect the two not to commute?

What is special about the two bases shown in the figure that, you agree, do not commute?

We can write the $|+\rangle |-\rangle$ basis as
$$|+\rangle = \cos(\theta)|0\rangle + \sin(\theta)|1\rangle |\rangle$$$$|-\rangle = \cos(\theta)|0\rangle - \sin(\theta)|1\rangle |\rangle$$What is special about $\theta=45^\circ$ that makes that particular orientation of the $|+\rangle |-\rangle$ basis non-commutative with the $|0\rangle |1\rangle$ basis?

Last edited: Apr 5, 2017
10. Apr 5, 2017

### Staff: Mentor

Yes. But this is a property of the particular Hilbert space and operators that were chosen. It does not generalize to all possible Hilbert spaces and operators.

11. Apr 5, 2017

### mike1000

I am not so sure that it does not generalize to all Hilbert spaces. I think I can show that graphically.

When we measure a quantum entity using two different bases, it is equivalent to measuring two different properties of the quantum entity. The axis of each basis are the eigenvectors for the corresponding property. In order to know the exact value of both properties, simultaneously, the eigenvectors of the two basis have to coincide. This is to say that when the measurement of one property puts you in a collapsed state for that property (on one of its eigenvectors), that vector must coincide with an eigenvector for the other property. If it does not, then the state vector for the other property is a linear combination of it basis states and not a collapsed state for that property. This means that both properties cannot be in a collapsed state at the same time, which implies that you cannot know the exact values of the two properties simultaneously.

12. Apr 5, 2017

### Staff: Mentor

Your graphic applies to the particular case of the spin of a spin-1/2 particle. That is not the same as applying to all possible Hilbert spaces and operators. See below.

More precisely, one particular vector must be an eigenvector of both operators. But your reasoning assumes that once we know that particular vector, we also know a unique vector that is orthogonal to it, so that finding one particular vector is equivalent to finding a basis for the entire Hilbert space. But it should be evident that this reasoning is only valid for a Hilbert space with two dimensions (i.e., for which any operator has at most two distinct eigenvectors). For any Hilbert space with more than two dimensions, there will be an infinite number of possible sets of orthonormal vectors that are orthogonal to a given vector, so picking out a particular vector does not pick out a unique basis.

13. Apr 6, 2017

### edguy99

Nothing. As I understand it, they will always be non-commutative unless the basis vectors of the two orthonormal measurement systems are orthonormal to each other (90 degrees sideways or verticle) and in your example, they are not.

14. Apr 6, 2017

### vanhees71

I don't know, what this confusion is all about. If you have two orthonormal bases $|u_k \rangle$ and $|v_k \rangle$, then you can express each Hilbert-space vector in terms of each of the bases using completeness relations, and the components wrt. to these bases transform by an infinite matrix. To see this, let's define
$$\psi_k=\langle u_k|\psi \rangle, \quad \psi_k'=\langle v_k|\psi \rangle.$$
Now, using the completeness of the bases
$$|\psi \rangle=\sum_{k} |u_k \rangle \langle u_k |\psi \rangle=\sum_{k} |u_k \rangle \psi_k = \sum_{k} \sum_j |v_j \rangle \langle v_j|u_k \rangle \psi_k.$$
On the other hand you find in the same way
$$|\psi \rangle = \sum_j |v_j \rangle \psi_j'.$$
Since the decomposition of the vectors wrt. to the bases is unique this implies
$$\psi_j'=\sum_k U_{jk} \psi_k \quad \text{with} \quad U_{jk} = \langle v_j|u_k \rangle.$$
It is easy to show that $U_{jk}$ is indeed a unitary "matrix" by using the orthonormality of the bases, i.e.,
$$\sum_k U_{kj}^* U_{kl}=\delta_{jl}.$$
Things change a bit when you use generalized bases, i.e., generalized eigenstates of self-adjoint operators that have a continuous spectrum (or a continuous part in their spectrum). Let's take position and momentum as an example. These eigenstates are normalized to a $\delta$ distribution and instead of sums you have to use integrals:
$$\langle x|x' \rangle=\delta(x-x'), \quad \langle p|p' \rangle=\delta(p-p'),$$
and from the commutation relations ("Heisenberg algebra"), using natural units such that $\hbar=1$,
$$[\hat{x},\hat{p}]=\mathrm{i},$$
you get
$$\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p).$$
Now the formalism is analogous to the case of a true orthonormal basis. The transformation from the momentum to the position representation is simply found by inserting an identity operator:
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p) \tilde{\psi}(p),$$
where
$$\tilde{\psi}(p)=\langle p|\psi \rangle.$$
The inverse transformation follows in the same way, leading to
$$\tilde{\psi}(p)=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} x p) \psi(x).$$
Of course, again these Fourier transformations are unitary mappings from $\mathrm{L}^2(\mathbb{R},\mathbb{C}) \rightarrow \mathrm{L}^2(\mathbb{R},\mathbb{C})$, $\psi \mapsto \tilde{\psi}$.

15. Apr 6, 2017

### Truecrimson

This is false. An arbitrary unitary rotation of a basis gives another basis. But not all unitary rotations are Fourier transformation.

Two Hermitian operators can commute even if they are different in a nontrivial way because of degeneracy (repeated eigenvalues). In 2 dimensions, the only degenerate operators are multiples of the identity operator. You have to go to higher dimensions to have nontrivial degenerated Hermitian operators.