Spectral theory in quantum mechanics

[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)

 

[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

 

[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|

 

[jensa]

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)

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.

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).

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.