Exponential projection operator in Dirac formalism

[Dixanadu]

Homework Statement

Hey guys.

So here's the situation:
Consider the Hilbert space H_{\frac{1}{2}}, which is spanned by the orthonormal kets |j,m_{j}> with j=\frac{1}{2}, m_{j}=(\frac{1}{2},-\frac{1}{2}). Let |+> = |\frac{1}{2}, \frac{1}{2}> and |->=|\frac{1}{2},-\frac{1}{2}>. Define the following two projection operators:

\hat{P}_{+}=|+><+| and \hat{P}_{-}=|-><-|.

Now consider the operator \hat{O}=e^{i\alpha \hat{P}_{+}+i\beta \hat{P}_{-}}.

Compute the following:

<+|\hat{O}|+>

<-|\hat{O}|->

<-|\hat{O}|+>

Homework Equations

orthonormality stuff: <+|+> = <-|-> = 1, <-|+> = <+|-> = 0.

Series representation of e^x:

e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k!}

The Attempt at a Solution

So here's what I've done. Of course you gota represent the operator O in a nicer way, and this is what I need to know if I've done right:

e^{i\alpha \hat{P}_{+}}=I+(i\alpha) \hat{P}_{+} + \frac{(i\alpha)^{2}}{2}\hat{P}_{+}+...=I+\hat{P}_{+}(i\alpha + \frac{(i\alpha)^2}{2}+...)=I+\hat{P}_{+}(e^{i\alpha}-1)

Similarly:

e^{i\beta \hat{P}_{-}}=I+\hat{P}_{-}(e^{i\beta}-1)

where I is the identity matrix. Now, O is a product of these two:

\hat{O}=[I+\hat{P}_{+}(e^{i\alpha}-1)][I+\hat{P}_{-}(e^{i\beta}-1)]=I+\hat{P}_{+}(e^{i\alpha}-1)+\hat{P}_{-}(e^{i\beta}-1)

because the cross-terms in the product vanish as \hat{P}_{+}\hat{P}_{-}=0

So that's my expression for O. Using that, I find that

&lt;+|\hat{O}|+&gt;=e^{i\alpha}. I haven't done the rest yet, but is this one right guys...?

please tell me if I've made any math errors!

 

[Goddar]

e^{i\alpha \hat{P}_{+}}=I+(i\alpha) \hat{P}_{+} + \frac{(i\alpha)^{2}}{2}\hat{P}_{+}+...=I+\hat{P}_{+}(i\alpha + \frac{(i\alpha)^2}{2}+...)=I+\hat{P}_{+}(e^{i\alpha}-1)

Hi.
That's right because (\hat{P}_{\pm})^{n} = \hat{P}_{\pm}, otherwise you would have to determine all exponents of the operators.
You're good so far!