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GR9677 #17
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Quantum Mechanics$\Rightarrow$}Probability

The careless error here would be to just directly square the grids. When one remembers the significance of the meaning of the probability $P=\int |\langle \psi | \psi \rangle|^2 dV$, one finds that one must square the wave function, and not the grids.

The total probability is,

$\int_0^6|\psi|^2 =1+1+4+9+1+0=16
$

The un-normalized probability from $x=2$ to $x=4$ is,

$\int_2^4|\psi|^2 =4+9=13
$

The normalized probability is thus $13/16$, as in choice (E).

Alternate Solutions
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 mhazelm2008-09-08 12:05:19 If you are familiar with probability theory, this amounts to computing the probability of event A, where A is the the particle's being between x =2 and x=4. From probability, this is given as P(A) = # ways A can occur/# total possible outcomes. Now # ways A can occur: if x is between 2 and 3 (an interval of 1) then Psi takes on the value 2 (so Psi^2 = 4). If x is between 3 and 4 (another interval of 1), then Psi^2 = 9. Thus # ways A can occur = 4+9 = 13. Now for the total number of possibilities, we just do the same thing, but for all possibilities of x. Our intervals are all 1, so we get 1(1^2) + 1(1^2) +1(2^2) + 1(3^2) + 1(1^2) = 16. Now just use the fact that P(A) = 13/16 and you're done with answer E being correct. Reply to this comment

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LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$

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