$$\frac{4k_{0}p_{1}+p_{0}\left ( k_{0}^{2}-2k_{0}k_{1}+k_{1}^{2} \right )}{\left ( k_{0}+k_{1} \right )^{2}p_{0}}$$

I need to show that this is equal to $1$ but for my life I can't figure how to arrive at one despite multiple attempts at completing the square. This is an assignment in Quantum mechanics so obviously the focus is not completing the square. But I'd like to see how to arrive at one and the trick(s) involved.

Thanks in advance.

  • 3
    $\begingroup$ It is only true if the $4k_0 p_1$ was $4k_0 k_1 p_0$ in the numerator $\endgroup$ – lEm Apr 11 '16 at 9:15
  • $\begingroup$ @Bubububu Correct. Is there no other way to show that they cancel out after manipulation? $\endgroup$ – Mathematicing Apr 11 '16 at 9:16
  • 1
    $\begingroup$ Maybe it was a typo? Because it depends on the value of $k_0 ,k_1, p_0, p_1$ specifically. It is obvious that they would not simply cancel out because there is a $p_1$ in the numerator but not in the denominator. $\endgroup$ – lEm Apr 11 '16 at 9:21
  • $\begingroup$ P0 and P1 are not the same variable. You need to find out the relationship between these two variables. They cannot cancel out the way the question is. Could it be a typo? $\endgroup$ – Josh Apr 11 '16 at 9:22
  • $\begingroup$ What if $p_{0}=k_{0}$ $\space$ and $p_{1}=k_{1}$? Would anything change? $\endgroup$ – Mathematicing Apr 11 '16 at 9:29

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