Checking $U^{\dagger} U=I$ for Lorentz boost

Question

The boost is given by $|p\rangle \rightarrow |\Lambda p\rangle$. I can understand that this is unitary because the inner product :

$$\int dp \frac{1}{2\omega _p} f(p) g^*(p) ....(1)$$

is invariant. I'd also like to check unitarity using $U^{\dagger} U=I$.

$U$ can be represented as :

$$f(p')=U[f(p)]=\int dp \delta (\Lambda p - p') f(p) $$

Is this correct? This is designed to ensure that $f(p')=f(\Lambda p)$

Then, I took the above representation of $U$ to compute $U^{\dagger} U$

$$=\int dp' \delta (\Lambda p_1 - p')\delta (\Lambda p_2 -p')$$

$$= \delta (\Lambda p_1 - \Lambda p_2)$$

This is not equal to Identity, which is $\delta (p_1 -p_2)$ .

How to fix this analysis?

According to the comment below, this is equal to identity. One problem is that i never had to use the definition of an inner product in $(1)$ to prove this. i only had to use $U[f(p)]=f(\Lambda p)$ This proof implies that the following inner product should be Lorentz invariant:

$$\int dp f(p) g^* (p)$$

$$=\int dp f(p) I g^* (p)$$

$$=\int dp f(p) UU^{\dagger} g^* (p)$$

$$=\int dp f(\Lambda p) g (\Lambda p)$$

which would be an incorrect conlusion

How to avoid this conclusion while showing $U^{\dagger} U=I$ for boosts?

Answer 1

Let's write down all the right formulas. The inner product is $$ (f,g) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} f^*(p) g(p) , \qquad \omega_p = \sqrt{ | \vec{p} |^2 + m^2 } . $$ The action of a unitary operator is $$ \tag{1} U_\Lambda f(p) = f(\Lambda p) = \int d^3 p' \delta^3( \Lambda p-p') f(p') . $$ Using the definition of $U^\dagger$ \begin{align} (U_\Lambda^\dagger f , g) &= ( f , U_\Lambda g) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} f^*(p) U_\Lambda g (p) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} f^*(p) g (\Lambda p) \end{align} We now change the integration variable to $p \to \Lambda^{-1} p$. Using the fact that the integration measure is Lorentz invariant, we find $$ (U_\Lambda^\dagger f , g) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} f^*(\Lambda^{-1} p) g (p) $$ Since this is true for any $g(p)$, we must have $$ U_\Lambda^\dagger f(p) = f(\Lambda^{-1} p). $$ We can now check that the operator is unitary $$ U_\Lambda^\dagger U_\Lambda f(p) = U_\Lambda^\dagger f(\Lambda p) = f( \Lambda^{-1} \Lambda p ) = f(p) $$ so that $U_\Lambda^\dagger U_\Lambda = I$.

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Alternatively, we can follow your way of deriving unitarity. The idea is that we must find the momentum representation of the operator $U_\Lambda(p,p')$. For this, we must be careful about the normalization of the inner product. Rewriting (1), we have $$ U_\Lambda f(p) = \int \frac{d^3 p'}{(2\pi)^3} \frac{1}{2\omega_{p'}} [ (2\omega_{p'} )(2\pi)^3 \delta^3( \Lambda p-p') ] f(p') . $$ Therefore, $$ U_\Lambda(p,p') = (2\omega_{p'} )(2\pi)^3 \delta^3( \Lambda p-p') . $$ Similarly, $$ U^\dagger_\Lambda(p,p') = (2\omega_{p'}) ( 2\pi)^3 \delta^3( \Lambda^{-1} p - p') = (2\omega_p) ( 2\pi)^3 \delta^3( p - \Lambda p') $$ where in the last equality, we used the fact that the Dirac delta function is Lorentz invariant, i.e. $$ (2\omega_{\Lambda p}) \delta^3 ( \Lambda p - \Lambda p' ) = (2\omega_p) \delta^3( p - p') $$ We can now check unitarity, \begin{align} U_\Lambda^\dagger U_\Lambda (p,p') &= \int \frac{d^3p''}{(2\pi)^3} \frac{1}{2\omega_{p''}} U_\Lambda^\dagger(p,p'') U_\Lambda (p'',p') \\ &= \int \frac{d^3p''}{(2\pi)^3} \frac{1}{2\omega_{p''}} [ (2\omega_{p''} )(2\pi)^3 \delta^3( \Lambda p-p'') ] [ (2\omega_{p''}) ( 2\pi)^3 \delta^3( p'' - \Lambda p') ] \\ &= (2\omega_{\Lambda p}) ( 2\pi)^3 \delta^3( \Lambda p - \Lambda p') \\ &= (2\omega_p) ( 2\pi)^3 \delta^3( p - p') \\ &= I(p,p') . \end{align} The last object is the identity operator in the momentum representation.

Answer 2

The analysis in the latter half of the post is completely incorrect. It arises by using a mix of integral and matrix notation.

Let's just stick to one notation at a time. The inner product is given by :

$$\int d^3p \frac{1}{2\omega _p} f(p) g^*(p)$$

In Einstein notation, this may be re-written as:

$$f_p M^{pp'} g^* _{p'}$$

where $M^{pp'}=\frac{1}{2\omega _p}\delta (p-p')$. Repeated indices are integrated over.

If $U^a_b$ is a transformation, then the transformed inner product is:

$$f_a U^a _b M^{bc} U^{*d}_c g^* _d$$

We can see that the inner product is invariant iff :

$$U^a_b M^{bc} U^{*d}_c= M^{ad}$$

The above is the condition of unitarity. We can see that it explicitly involves the inner product measure encoded in $M$. So when a transformation matrix satisfies this condition, the inner product with the specific measure given by $M$ is invariant.

Here, $*$ represents a complex conjugate rather than the adjoint.

Answer 3

Regarding the last part of your expression you have $$\intop dpf\left(\Lambda p\right)g\left(\Lambda p\right)=\intop\det\left(\Lambda^{-1}\right)dp'f\left(p'\right)g\left(p'\right)=\intop dp'f\left(p'\right)g\left(p'\right)$$ Which is just a coordinate transformation with nothing special about. I think the confusion rises from the fact that the state is not described by a 4-vector but a 3-momentum vector when you canonically quantize it. You have $\left|\overrightarrow{p}\right\rangle =a_{\overrightarrow{p}}^{\dagger}\left|0\right\rangle$, we define $\left\langle 0|0\right\rangle =1$ and we have $$\left\langle 0\right|a_{\overrightarrow{q}}a_{\overrightarrow{p}}^{\dagger}\left|0\right\rangle =\left\langle 0\right|\left[a_{\overrightarrow{q}},a_{\overrightarrow{p}}^{\dagger}\right]\left|0\right\rangle =\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\overrightarrow{p}-\overrightarrow{q}\right)$$ This delta function is not Lorentz invariant but $E_{\overrightarrow{p}}\delta^{\left(3\right)}\left(\overrightarrow{p}-\overrightarrow{q}\right)$ is! You can check it by taking a Lorentz transformation in some direction (breaking the delta function into $\delta\left(p_{\parallel}-q_{\parallel}\right)\delta^{\left(2\right)}\left(p_{\perp}-q_{\perp}\right)$) and it will result in $$E'_{\overrightarrow{p}}\delta^{\left(3\right)}\left(\overrightarrow{p}'-\overrightarrow{q}'\right)=E_{\overrightarrow{p}}\delta^{\left(3\right)}\left(\overrightarrow{p}-\overrightarrow{q}\right)$$ That's why you're going to see it next to the delta function a lot during quantization and the results will be Lorentz invariant.