Spectral theory in quantum mechanics

1. Jan 11, 2009

guhan

Can someone please tell me what the integrand in the below equation mean?
$$1 = \langle \psi | \psi \rangle = \int_{-\infty}^{\infty} d \langle \psi |E_\lambda \psi \rangle$$
where,
$$E_\lambda$$ is an increasing (and absolutely continuous) function of projection operators such that $$\int_{-\infty}^{\infty} dE_\lambda = I$$

( I read the integrand as a differential (or measure) of a complex constant, which should have been zero!? So I am certainly wrong in interpreting it)

2. Jan 11, 2009

jensa

I would say that $$E_\lambda$$ corresponds to an element of a continuous set of projection operators i.e. $$E_\lambda=|\lambda\rangle\langle\lambda|$$ with the completeness relation
$$\int dE_\lambda\equiv \int d\lambda |\lambda\rangle\langle\lambda|=1$$.

In this sense the equation you wrote simply means

$$1=\langle\psi|\psi\rangle=\langle\psi|(\int dE_\lambda )|\psi\rangle =\int d\underbrace{\lambda\langle\psi|\lambda\rangle\langle \lambda|\psi\rangle}_{\langle \psi |E_\lambda\psi\rangle}$$

Hope this helps

Last edited: Jan 11, 2009
3. Jan 11, 2009

guhan

Thanks!.. I get the drift.

As in, in the case of $$\lambda$$ belonging to the spectrum of position operators,
$$\langle \psi | E_\lambda \psi \rangle = \int_{-\infty}^{\lambda} |\psi |^2 dx$$
which is absolutely continuous and its differential is well defined ($$= | \psi |^2$$)

But, is writing $$E_\lambda=|\lambda\rangle\langle\lambda|$$ valid even for $$\lambda \in$$ spectrum, which can be continuous?
I was of the opinion that it is valid only when $$\lambda$$ is an eigenvalue (and hence discrete).

And when you wrote $$E_\lambda=|\lambda\rangle\langle\lambda|$$, I assume you meant the following:
$$E_\lambda=\sum_{\lambda_i \le \lambda} |\lambda_i\rangle\langle\lambda_i|$$

4. Jan 12, 2009

jensa

Actually I think the answer I gave was not really what you were looking for. I just took a shot in the dark. To answer what the integrand means I would have to know in what context it appears. And I have a feeling even then I probably am not the right person to answer.

I think that, in a "rigged" Hilbert space, continuous eigenvalues are allowed and one can safely write $$E_\lambda=|\lambda\rangle\langle \lambda$$ and even $$1=\int d\lambda |\lambda\rangle\langle\lambda|$$ but I don't know how to prove this or where proof could be found.

Sorry for not being able to help you. I hope somebody comes along who can.