18/06/17 16:59:56.91 QjvB9Pxf.net
>>172
ψ(x) = ∫[ {1/(p^2 + a^2)} * exp(ipx) ] dp
=2�
18/06/17 16:59:56.91 QjvB9Pxf.net
>>172
ψ(x) = ∫[ {1/(p^2 + a^2)} * exp(ipx) ] dp
=2�
18/06/17 16:59:56.91 QjvB9Pxf.net
>>172
ψ(x) = ∫[ {1/(p^2 + a^2)} * exp(ipx) ] dp
=2�