Exponential projection operator in Dirac formalism

[Dixanadu]

Homework Statement

Hey guys.

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

[itex]\hat{P}_{+}=|+><+|[/itex] and [itex]\hat{P}_{-}=|-><-|[/itex].

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

Compute the following:

[itex]<+|\hat{O}|+>[/itex]

[itex]<-|\hat{O}|->[/itex]

[itex]<-|\hat{O}|+>[/itex]

Homework Equations

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

Series representation of e^x:

[itex]e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k!}[/itex]

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:

[itex]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)[/itex]

Similarly:

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

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

[itex]\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)[/itex]

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

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

[itex]<+|\hat{O}|+>=e^{i\alpha}[/itex]. I haven't done the rest yet, but is this one right guys...?

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

 

[Goddar]

[itex]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)[/itex]

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