# Angular momentum operator in terms of ladder operators

I wanted to show that the angular momentum of the particle state with zero momentum $| \vec{0} \rangle$ is $0$, that is to say the intrinsic spin of a scalar field is $0$ using a mode expansion. There is a clever argument which is essentially that

$e^{i \vec{\theta} \cdot \hat{\vec{J}}} \, | \vec{p} \rangle = | R(\vec{\theta}) \vec{p} \rangle$

and thus we have that for any $\vec{\theta}$

$e^{i \vec{\theta} \cdot \hat{\vec{J}}} \, | \vec{0} \rangle = |\vec{0} \rangle + i \vec{\theta} \cdot \hat{\vec{J}} |\vec{0} \, \rangle + \cdots = |\vec{0} \rangle$

and hence it is clear that

$\hat{\vec{J}} |\vec{0} \, \rangle = 0$

which shows the desired result. I would like to show the same thing by using

$\hat{\vec{J}} = -\int d^3x \, \vec{x} \times ( \hat{\pi} \, \nabla \hat{\phi})$

and use the mode expansions for $\hat{\phi}$ and $\hat{\pi}$:

$\hat{\phi} = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\vec{p}}}} (\hat{a}_{\vec{p}} + \hat{a}^{\dagger}_{-\vec{p}})e^{i \hat{p} \cdot \hat{x}}$

$\hat{\pi} = \int \frac{d^3p}{(2 \pi)^3} \sqrt{\frac{E_{\vec{p}}}{2}} (\hat{a}_{\vec{p}} - \hat{a}^{\dagger}_{-\vec{p}})e^{i \hat{p} \cdot \hat{x}}$

Doing this (and normal ordering) I get the following

$\hat{\vec{J}} = - i \int \frac{d^3p}{(2 \pi)^3} \hat{a}^{\dagger}_{\vec{p}} ( \vec{p} \times \nabla_{\vec{p}}) \hat{a}_{\vec{p}}$

I am unsure of how to proceed with derivatives of the ladder operators and how to apply them to states in the Fock space.

Could someone argue as to how one can apply this mode expansion to the state $| \vec{0} \rangle$ and show that it annihilates it, as it must.

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A reference to where this is treated in detail would also be much appreciated. –  Kyle May 7 '12 at 3:02

First a note about normalization. When doing mode expansions, for your own sanity, you should use relativistic normalization:

$$\alpha_p = \sqrt{2E_p} a_p$$

and the same for the Hermitian conjugate. This makes the mode expansion a manifestly invariant expansion in terms of the invariant measure on the hyperbola $d^3p\over 2E_p$ and makes all the manipulations transparent enough to do in the head. This is not important here, the relativistic ladder operators have unusual state normalization, but it is indispensible for linking up the ladder operators to the covariant Feynman methods you learn later.

The main problem you are having is how to deal with the derivatives of operators which produce singular states. The states produced by $a_p$ are delta-functions in momentum space, they are not normalized states. So acting with a p-derivative produces a derivative of delta-function. This derivative of a delta function serves to rotate p, just like in the elementary quantum mechanics case.

The expression you got is the operator that rotates p by an infinitesimal amount. The way to see this is to use L as a Hamiltonian, and move the operators an infinitesimal amount by the Heisenberg equations of motion (generate the unitary transformation using L_z, for example)

$${da_p\over ds} = i [L_z ,a_p] = p_x \partial_{p_y} a_p - p_y \partial_{p_x} a_p$$

Where you use the commutation relations of a and a-dagger. This can be done intuitively if you know the commutation relation of the number operator $a^\dagger_p a_p$ with $a_p$ is just $a_p$ once again. The resulting equation for the canonical transformation effect on the operators a is solved by rotating the p-vector of the a operator around the z axis

$$a_p(s) = a_{p(s)}$$

Where the rotating momentum vector is rotating around the z axis

$$p_x(s) = p_x cos(s) - p_y sin(s)$$ $$p_y(s) = p_x \sin(s) + p_y cos(s)$$

If you differentiate the operator $a_{p(s)}$, you find the rate of change of $p_x$ times the derivative of the $a$ operator with respect to $p_x$, plus the rate of change of $p_y$ times the derivative of the $a$ operator with respect to $p_y$, which, when you write down the rates of change of the p's reproduces the right hand side of the canonical transformation.

The same holds for the $a^\dagger$, so the $L_z$ operator does the same thing as the nonrelativistic example, it rotates the operators around the z axis. The rotation of $a^\dagger_0$ does nothing because the zero vector is rotationally invariant, and you get that the $L_z$ operator annihilates the state $a^\dagger_0 |0>$.

This derivation is kind-of hokey, you probably wanted a direct application of the operator to the state. This is tricky because the singular delta-function derivatives are multiplying the vanishing p right at this point. Nevertheless, if you do it, you get the rate of change of the x and y components of the 0 vector under a z rotation, times the derivative of the a operator in the x and y directions, which gives zero because the first factor is zero.

You should be aware that it is possible to make nonzero angular momentum states with arbitrarily slowly varying waveforms, but not ones they have to vanish at the origin. This is why the 0 momentum state is special--- it doesn't vanish at the origin.

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