- #1

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In my book it says

[tex] \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx [/tex]

is equivalent to

[tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t}|\Psi(x,t)|^2 dx [/tex]

I understand how It becomes a partial derivative, since I'm differentiating psi with respect to x and t

But what I don't understand Is how you can just move [tex] \frac{d}{dt} [/tex]

into an integral, why can you do that? I don't understand the rules behind it,

for example

(i'm just making this up)

[tex] \frac{d}{dt} \int_0^2 2xt dx [/tex]

I'm first going to integrate the inside function then differentiate it with respect to t

before I try the other way of moving dt inside the integral

[tex] \int_0^2 2xt dx = \left \frac{2x^2t}{2} \right|_0^2 [/tex]

the equation becomes

[tex] \left x^2t\right|_0^2 [/tex]

putting in 2, I get 2^2 *t

= 4t

differentiating that,

[tex] \frac{d}{dt} 4t = 4 [/tex]

now i'll try the other way by moving in the partial derivative

[tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx [/tex]

I've only done one math paper before, and 5 physics papers,

so I don't know how to solve partial derivatives in the conventional sense,

but what I have picked up is, that a physicists first line of attack to any partial derivative, is separation of variables,

so uh, i'm not too sure how i do it in this context, since I can't separate x,t the way I can separate [tex] \Psi , \textrm{ which I can rewrite as } \psi(x)\phi(x) [/tex]