# Which is the conjugate variable?

1. Aug 26, 2010

### wdlang

take a 1/2 spin, that is, a qubit

the general state is of the form

psi= \cos(\theta /2) |g>+ e^{i\phi} \sin(\theta/2) |e>

where |g> and |e> are the two basis states

it is stated in a PRL paper that \phi is the conjugate variable to \sin^2(\theta/2)

why?

by the way, for a 1/2 spin, what is the conjugate variable to the operator \sigma_z ? Is this question meaningful?

2. Aug 27, 2010

### tom.stoer

The conjugate momentum is originally defined as the variable you get by differentiation of the Lagrangian w.r.t. the velocity.

Starting with the Hamiltonian instead of the Lagrangian the conjugate variable is defined as the variable for which the Poisson bracket is equal to 1.

In quantum mechanics the Poisson bracket is replaced by the commutator and the 1 is replaced by i. So starting in QM means to construct a self-adjoint operator which has the correct commutator. Be careful: it is not always possible to complete this construction; one famous obstacle is the phase operator.

For the spin 1/2 problem you have to construct a 2*2 matrix X which satisfies [σ³,X] = i.

3. Aug 27, 2010

### tom.stoer

btw.: I think one can easily prove that no variable X conjugate to σ³ exists.

4. Aug 27, 2010

### wdlang

i cannot agree with you

according to you, the conjugate variable to some variable depends on the Lagrangian

i think the conjugate variable to some variable should be an intrinsic one, not dependent on any system or L or H.

5. Aug 27, 2010

### tom.stoer

Of course the original derivation of conjugate variables depends on the Lagrangian.

If you have x and p (as symbols), how do you know that their commutator is i? How can you check that the p = -id/dx is the correct choice? The starting point is always L and/or H.

Of course you can define operators, observe that by some magic reason they have the correct commutation relation and say that they are conjugate variables. But I doubt that this will clarify the situation. Originally the concept is rooted in the symplectic structure of classical mechanics in Hamiltonian formulation.

6. Aug 27, 2010

### Dickfore

Let's find the (anti)commutator of an arbitrary $2 \times 2$ matrix $\hat{X}$ with the matrix corresponding to the operator $\hbar \, \hat{s}_{z} = \frac{\hbar}{2} \, \sigma_{3}$. We get:

$$\left[\frac{\hbar}{2} \, \hat{\sigma}_{3}, \hat{X} \right] = \left[\begin{array}{cc} \frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \end{array}\right] \cdot \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] - \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \cdot \left[\begin{array}{cc} \frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \end{array}\right] = \hbar \, \left[\begin{array}{cc} 0 & b \\ -c & 0 \end{array}\right]$$

But, this matrix has no diagonal elements, so it is never proportional to the unit matirx.

However, if you take the anticommutator, then you will get:

$$\left\{\frac{\hbar}{2} \, \hat{\sigma}_{3}, \hat{X} \right\} = \left[\begin{array}{cc} \frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \end{array}\right] \cdot \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] + \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \cdot \left[\begin{array}{cc} \frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \end{array}\right] = \hbar \, \left[\begin{array}{cc} a & 0 \\ 0 & -d \end{array}\right]$$

Now, choosing $a = -d = i$, we see that:

$$\left\{\hbar \, \hat{\sigma}_{3}, i \, \hat{\sigma}_{3}\right\} = i \, \hbar \, \hat{1}$$

So, if we accept that for the particles with spin-1/2, the corresponding canonical relations between the operators are anticommutations, then, we might say that the conjugate variable to $\sigma_{z}$ is $\frac{i}{\hbar} \, {\sigma}_{z}$.

7. Aug 27, 2010

### tom.stoer

I agree with the first step.

My proof would have been slightly different: any 2*2 matrix can be written in terms of the Pauli matrices and the identity matrix as a basis. Commutators of Pauli matrices with Pauli matrices generate new Pauli matrices, commutators of Pauli matrices with the identity generate zero. Therefore a conjugate variable (based on commutators) does not exist.

I have questions regarding the second step = regarding the construction using anti-commutators:
- is there any physical insight from the fact that Pauli matrices are "self-conjugate"?
- doesn't it matter that one has to introduce an anti-hermitian operator?

8. Aug 27, 2010

### wdlang

but there is no refer to L or H in his proof

9. Aug 27, 2010

### wdlang

i think we should concentrate on the original question

"it is stated in a PRL paper that \phi is the conjugate variable to \sin^2(\theta/2) "

10. Aug 27, 2010

### tom.stoer

You don't need L or H to show that two operators satisfy a specific communtation relation, but in order to understand their dynamics and their meaning either L or H are required. Which physical information can a 2*2 matrix have if no context is specified?

11. Aug 27, 2010

### Dickfore

Can you give us a reference to the particular PRL paper?

12. Aug 27, 2010

### tom.stoer

We cannot answer this question as long as we don't know which variables \phi and \theta are, how the Lagrangian or something like that looks like, how they are related via Legendre transformation or something like that. They are just symbols w/o any meaning.

13. Aug 27, 2010

### Dickfore

From the OP (with corrected LaTeX):

14. Aug 27, 2010

### wdlang

15. Aug 27, 2010

### tom.stoer

I am no expert on the Bloch-sphere representation of qbit states, but to me it seems that the two angles are just coordinates; I have no idea what this "perturbed conjugate variable" means. To me it seems that is nothing to with a canonically conjugate variable as there are no commutators specified.

http://www.vcpc.univie.ac.at/~ian/hotlist/qc/talks/bloch-sphere.pdf

16. Aug 28, 2010

### Dickfore

According to the the paper you had cited, the introductory sentence is:

and reference [1] is:

[1] V. Braginsky and F. Khalili, Quantum Measurement (Cambridge University Press, Cambridge, England, 1992).

So, it does have to do with commutators. If you keep reading through the fourth and the fifth paragraphs, you will see what their physical realization of qubits is and what is the Hamiltonian governing their dynamics. Therefore, it seems people were right when they asked you for a Lagrangian (or Hamiltonian).

As for the note I made about the anticommutator, I was confused with the canonical anticommutation relations between the field operators for half-integer spin fields. However, there should still be a commutator between operators corresponding to real observables.