- #1
paweld
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What is the probability of finding particle of Klein-Gordon field in some
region of space V (in an arbitrarily chosen inertial reference frame
at time t=0). According to me it is a following integral:
<br /> P(V) = \int_V d^3 x \overline{\psi^{NW}(x,0) } \psi^{NW}(x,0)~~~~<br />
where \psi^{NW}(x,t) is a Newton Wigner operator, i.e.:
\psi^{NW}(x,t) = \int \frac{d^3 p}{(2 E_p)^{3/2}}} <br /> \exp(i \vec{p} \vec{x} - E_p t) \hat{\psi}(\vec{p})
(\hat{\psi}(\vec{p}) is a Klein-Gordon operator in momentum representation)
On the other hand ususal Klein-Gordon field operator is connected to
its momentum counterpart by equation:
\psi(x,t) = \int \frac{d^3 p}{(2 E_p)}} <br /> \exp(i \vec{p} \vec{x} - E_p t) \hat{\psi}(\vec{p})
Notice that since the scalar product for \hat{\psi}(\vec{p}) is simply:
<br /> ( \hat{\psi_1}|\hat{\psi_2} ) = \int \frac{d^3 p}{(2 E_p)}} <br /> \overline{\hat{\psi_1}(p)}\hat{\psi_2}(p)<br /> ~~~~(1)<br />
then the scalar product for \psi^{NW}(x,t):
<br /> ( \psi^{NW}_1| \psi^{NW}_2) = \int d^3 x \overline{\psi_1(x,0)} \psi^{NW}_2 (x,0)<br /> ~~~~(2)<br />
in comparison to the expression for scalar product for \psi(x,t) which involves
some time derivatives:
<br /> ( \psi_1| \psi_2) = i \int d^3 x \overline{\psi_1(x,0)} (\partial_0 \psi_1)(x,0)<br /> - \overline{(\partial_0 \psi_1)(x,0)} \psi_2(x,0) ~~~~(3)<br />
This explains why we cannot assign probabilistic interpretation
to \psi(x,t) but we can do it for \psi^{NW}(x,t).
I wonder what happens if we consider negative energy states. Equation
(3) can be easiliy generalised for such solution of K-G eq. but it gives negative values.
What is the precise definition of NW operator for such states (accoridng to eq. (2)
norm of any solution is positive). Why some people say that probability of
finding a particles of negative energy (antiparticles) is negative? For me it doesn't
make any sense.
What are your opinion about it?
region of space V (in an arbitrarily chosen inertial reference frame
at time t=0). According to me it is a following integral:
<br /> P(V) = \int_V d^3 x \overline{\psi^{NW}(x,0) } \psi^{NW}(x,0)~~~~<br />
where \psi^{NW}(x,t) is a Newton Wigner operator, i.e.:
\psi^{NW}(x,t) = \int \frac{d^3 p}{(2 E_p)^{3/2}}} <br /> \exp(i \vec{p} \vec{x} - E_p t) \hat{\psi}(\vec{p})
(\hat{\psi}(\vec{p}) is a Klein-Gordon operator in momentum representation)
On the other hand ususal Klein-Gordon field operator is connected to
its momentum counterpart by equation:
\psi(x,t) = \int \frac{d^3 p}{(2 E_p)}} <br /> \exp(i \vec{p} \vec{x} - E_p t) \hat{\psi}(\vec{p})
Notice that since the scalar product for \hat{\psi}(\vec{p}) is simply:
<br /> ( \hat{\psi_1}|\hat{\psi_2} ) = \int \frac{d^3 p}{(2 E_p)}} <br /> \overline{\hat{\psi_1}(p)}\hat{\psi_2}(p)<br /> ~~~~(1)<br />
then the scalar product for \psi^{NW}(x,t):
<br /> ( \psi^{NW}_1| \psi^{NW}_2) = \int d^3 x \overline{\psi_1(x,0)} \psi^{NW}_2 (x,0)<br /> ~~~~(2)<br />
in comparison to the expression for scalar product for \psi(x,t) which involves
some time derivatives:
<br /> ( \psi_1| \psi_2) = i \int d^3 x \overline{\psi_1(x,0)} (\partial_0 \psi_1)(x,0)<br /> - \overline{(\partial_0 \psi_1)(x,0)} \psi_2(x,0) ~~~~(3)<br />
This explains why we cannot assign probabilistic interpretation
to \psi(x,t) but we can do it for \psi^{NW}(x,t).
I wonder what happens if we consider negative energy states. Equation
(3) can be easiliy generalised for such solution of K-G eq. but it gives negative values.
What is the precise definition of NW operator for such states (accoridng to eq. (2)
norm of any solution is positive). Why some people say that probability of
finding a particles of negative energy (antiparticles) is negative? For me it doesn't
make any sense.
What are your opinion about it?