- #1
Tony Hau
- 101
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- TL;DR Summary
- I don't know where this question suits; does it belong to calculus or quantum mechanics? But anyway, I would like to ask a question about deriving the expected value of velocity of wave function.
So I am confused about a proof in which the formula for expected value of velocity, ##\frac{d\langle x \rangle}{dt} ##, is derived.
Firstly, because the expected value of the position of wave function is $$\langle x \rangle =\int_{-\infty}^{+\infty} x|\Psi(x,t)|^2 dx$$Therefore, $$\frac{d\langle x \rangle}{dt} = \int x\frac {\partial } {\partial t} |\Psi|^2 dx = \frac{i \hbar}{2m}\int x\frac {\partial } {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(1)$$
The author then writes:
This expression can be simplified using integration-by-parts, which gives: $$\frac{d\langle x \rangle}{dt} =-\frac{i\hbar}{2m}\int (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(2)$$
(By the way, he has also written this line of notes: I used the fact that ##\frac{\partial x} {\partial x} =1##, and threw away the boundary term, on the grounds that ##\Psi## goes to zero at ##(\pm)## infinity.)
At the bottom of the page, he makes a note as follows:
For $$ \int_a^b f\frac{dg}{dx}dx = -\int_a^b\frac{df}{dx}g dx +fg |_a^b$$
Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one - it'll cost you a minus sign, and you'll pick up a boundary condition.
To be honest with, I have no idea of what it means by "you can peel a derivative off one factor in a product, and slap it onto the other one". Plus, I don't know how the author transits from eq(1) to eq(2). According to my previous calculus II knowledge, whenever we see an integrand composed of a product of two functions, we should try to evaluate the antiderivative of one of the functions and expand it into the integration by parts. However, it seems that there is a more sophisticated way of understanding it in physics.
Firstly, because the expected value of the position of wave function is $$\langle x \rangle =\int_{-\infty}^{+\infty} x|\Psi(x,t)|^2 dx$$Therefore, $$\frac{d\langle x \rangle}{dt} = \int x\frac {\partial } {\partial t} |\Psi|^2 dx = \frac{i \hbar}{2m}\int x\frac {\partial } {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(1)$$
The author then writes:
This expression can be simplified using integration-by-parts, which gives: $$\frac{d\langle x \rangle}{dt} =-\frac{i\hbar}{2m}\int (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(2)$$
(By the way, he has also written this line of notes: I used the fact that ##\frac{\partial x} {\partial x} =1##, and threw away the boundary term, on the grounds that ##\Psi## goes to zero at ##(\pm)## infinity.)
At the bottom of the page, he makes a note as follows:
For $$ \int_a^b f\frac{dg}{dx}dx = -\int_a^b\frac{df}{dx}g dx +fg |_a^b$$
Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one - it'll cost you a minus sign, and you'll pick up a boundary condition.
To be honest with, I have no idea of what it means by "you can peel a derivative off one factor in a product, and slap it onto the other one". Plus, I don't know how the author transits from eq(1) to eq(2). According to my previous calculus II knowledge, whenever we see an integrand composed of a product of two functions, we should try to evaluate the antiderivative of one of the functions and expand it into the integration by parts. However, it seems that there is a more sophisticated way of understanding it in physics.
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