Agreed, the energy levels of the Casimir force are expectation values of f(r)E, so in Eq (18) del squared acts on (f(r) psi / m(r) power half ) and the Leibniz Theorem applies. So to be completely clear, the bracket in Eq. (19) should be placed between psi and d tau. This integral reduces to the usual energy integral when f(r) / m(r) half is unity. Then <E> = – h bar squared / (2m) integral psi* del squared psi dtau as usual.

Note 430(2)

ok, the expectation value <f(r) E> only refers to the kinetic energy. According to eq.(63) of UFT 428, the kinetic energy is determined by the first two terms. I guess that the factor 1/m(r)^(1/2) has to be replaced by f(r)/m(r)^(1/2). This gives more than one term. In eq.(18) of the note you have simply evaluated the expectation value of

<nabla^2 (f(r)/m(r)^(1/2)) >.

I do not see the connection to the kinetic energy terms which would require to apply nabla^2 also to the right wave function in the integral.

Horst

Am 02.02.2019 um 15:36 schrieb Myron Evans:

Note 430(2)

OK thanks again! Eq. (13) can be written as dm / dr1= f(r) dm / dr. On the left hand side m is a function of r1, and on the right hand side m is a function of r, but it is the same m on both sides. This point was discussed in the preparation for UFT417. If m = m(r1) and m = m(r) then m(r1) = m(r). In general m = m(r1(r)). In Eq. (17), E is the total relativistic energy, so the expectation value does not contain the potential energy term.

Note 430(2)

The computations are verified up to eq.(13). In (14) you have used

eq.(7), but with replacing m(r1) by m(r). In (7) the factor

partial m(r1)/partial r1

was replaced by r dependence but the factor 1/2m(r1) it was not. A

similar problem occurs in (8) where m(r1) was replacec by m(r) on the

RHS. Is it coorect to do this? We arrive at the problem of endless

recursion here again.

Another point: Shouldn’t eq. (18) contain all 3 terms of the total

energy as given in eq. (6) of note 430(1) ?

Horst

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