Recent content by Alexitron

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    Square root of negative complex exponential

    The square root of a complex number has two solutions. The attachment will help you
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    Graduate Average value in superposition question

    Thanks for your time. You helped me a lot to figure it out.
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    Graduate Average value in superposition question

    Well, i know how the values of <x> and <E> are calculated and that energy is quantized but the position for example is not , but to be honest i don't understand why the conjugate ψ* has different index. The only thing i can think is that we lose the factors A12 and A21 when calculating <A> (what...
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    Graduate Average value in superposition question

    Thanks a lot for your reply. it bothers me for some time So for example,(ψ=c1ψ1 + c2ψ2) when calculating the average position (or any other Item A) of a particle, if we had the same summation index we wouldn't have the terms X_12 and X_21. But when we calculate the energy we don't have...
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    Graduate Average value in superposition question

    Hi there. A quick question: When calculating the average energy (or any value) of a particle which is in superposition, why the sums have different summation index (n,m)? Thanks!
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    Graduate What is the relationship between symmetry energy and asymmetry energy?

    Thanks.
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    Graduate What is the relationship between symmetry energy and asymmetry energy?

    Hi Can someone please explain me what is the symmetry energy? I know asymmetry energy in the weizsacker semiempirical mass formula which is the gap between a nucleus and it's isobar with Z=N.
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    Solve this with lower incomplete gamma function

    Homework Statement \int e^{-\frac{2Zr}{a}}*r^{-1}dr Boundaries:[0,R] Homework Equations --- The Attempt at a Solution I tried to solve this with lower incomplete gamma function and got \gamma(0,\frac{2ZR}{a}) which is infinite i think. Z=81,a:Bohr radius,R=r0*A^(1/3)...
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    So, (nx+L)*[L^(n-1)] = [L^(n-1)]*(nx+L)Therefore, [L^(n-1)]*(nx+L) is symmetric.

    Solved! Abstract the n-1 column from the n column, then the n-2 from the n-1, then the n-3 from the n-2 etc.You get: |x+L -L 0 0 . . .0 0 0 | | x L -L 0 . . .0 0 0| | x 0 L -L . . .0 0 0| | x 0 0 L . . .0 0 0 | | . . . . . . . . . . . | | . . . . . ...
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    So, (nx+L)*[L^(n-1)] = [L^(n-1)]*(nx+L)Therefore, [L^(n-1)]*(nx+L) is symmetric.

    I guess real numbers.
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    So, (nx+L)*[L^(n-1)] = [L^(n-1)]*(nx+L)Therefore, [L^(n-1)]*(nx+L) is symmetric.

    Can someone help me proove this one please? If M is an nxn matrix _____|x+L x x . . . x | _____| x x+L x . . . x | |M|= | x x x+L. . . x | = [L^(n-1)]*(nx+L) _____| : : : : : : : : | _____| x x x . . . x+L|
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    Graduate Why Are Nuclei with Excess Neutrons Unstable?

    Hi there First of all excuse my English. My question is: Why a nucleus with more neutrons than these in the valley of nuclear stability are not stable and are beta(-)? I know that a nucleus with high mass number have more neutrons than protons because the more the protons the...