Re: Sakurai Advanced QM

Todd wrote:
<phd15@xxxxxxxxxxxxxxxx> wrote in message
news:e28bp6$iaj$1@xxxxxxxxxxxxxxxxxxxxxxxxx

A question for the great and good among you,
can anybody show how to derive eq. 4.133 of Sakurai's Advanced QM ?
Thanks, Zak



It appears that understanding 4.133 doesn't require understanding
much else
on that page. Thank goodness, or I wouldn't be able to offer any help.

4.133 is simply making use of a well-known property of the Dirac delta
function. If a function f(x) has one zero at x = x0 (i.e., f(x0) =
0), then

Delta[f(x)] = Delta[x-x0]/|f '(x0)|

yes this makes sense i've worked through that and convinced myself that
it's consistent with 4.133

On the left hand side of 4.133 you have a delta function written as a
function of the energies of the pion and proton. The proton energy is a
function of the magnitude of the proton momentum according to the basic
relation

E = Sqrt[p^2+m^2].

Likewise for the pion energy. But note that the magnitude of the pion
momentum is equal to the magnitude of the proton momentum due to momentum
conservation (see third line on page 203.) Hence, both the pion and
proton
energies are functions of the proton momentum.

So, you can use the above property of the delta function to rewrite the
delta function expression on the left side of 4.133 as

Delta[p-p0]/|derivative of (M_hyperon - E_proton - E_pion with respect to
p)|

Here, p0 is the magnitude of the proton momentum that makes E_proton +
E_pion = M_hyperon.

so this exactly expresses conservation of 4-momentum ?

Using E = Sqrt[p^2+m^2], the absolute value of the derivative in the
denominator is easily shown to equal

p/E_pion + p/E_proton.

Because of the Delta[p-p0] factor, these expressions should be
evaluated at
p = p0. But Sakurai doesn't bother to state this explicitly.

Hope this helps,

Todd

a couple of other questions eq 4.128 which reads

dw = (2 pi)^4 (1/2 E_Lambda) [(d^3 p_proton)/((2 pi)^3 2 E_proton]
X [(d^3 p_pion)/((2 pi)^3 2 E_pion] 4 m_proton m_pion
X sum{curly{M}_fi|^2} delta^4{p_Lambda - p_pion - p_proton}

(here X denotes multiplication)

is it true that d^3 p_proton is a volume element in momentum space such
that ?

d^3 p_proton = |bold{p}_proton|^2 d|bold{p}_proton| d Omega

which due to the frame of reference taken (rest frame of the Lambda
hyperon) ie bold{p} = bold{p}_proton = -bold{p}_pion , bold{p}_lamda = 0
=> E_lambda = m_Lambda

d^3 p_proton = |bold{p}|^2 d|bold{p}| d Omega

when dw is integrated over d^3 p_pion am i right in beliving that this
kills off the spatial parts of delta^4{p_Lambda - p_pion - p_proton}
leaving only the Energy parts of the four momentum ?

So that ?

p_Lambda -> m_Lambda
p_proton -> E_proton
p_pion -> E_pion

Incidentally is it correct to associate a photons momentum with the
spatial parts of the k-vector of that photon, where the components of
the four k-vector are given by

k = (k_x,k_y,k_z,i omega/c)

by the way, many thanks todd, your post was a great help. Although i
must admit to feeling a little stupid. I should read up on my basic
relativistic kinematics !
Zak

.

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