Which is the conjugate variable?

[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?

 

[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.

 

[tom.stoer]

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

 

[wdlang]

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.

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.

 

[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.

 

[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:

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

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

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 $\frac{i}{\hbar} \, {\sigma}_{z}$.

 

[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?

 

[wdlang]

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?

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

 

[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) "

 

[tom.stoer]

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

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?

 

[Dickfore]

i think we should concentrate on the original question

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

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

 

[tom.stoer]

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) "

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.

 

[Dickfore]

From the OP (with corrected LaTeX):

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

the general state is of the form

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

where $|g \rangle$ and $|e \rangle$ are the two basis states

 

[wdlang]

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

http://prl.aps.org/abstract/PRL/v105/i4/e040506

 

[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

 

[Dickfore]

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

According to the Heisenberg uncertainty principle, the measurement of a physical variable necessarily perturbs its quantum conjugate variable, imposing a fundamental limit on the precision of the measurement. However, it is in principle possible to decouple the dynamics of the variable which is measured from the perturbed conjugate variable. This is the central idea in the concept of quantum nondemolition (QND) measurement, which was initially developed to perform high precision measurement beyond the standard quantum limit [1].

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.