- #1
fscman
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I am currently reading through Griffiths Quantum Mechanics textbook, and on page 14, Griffiths proves that
[itex]\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi(x,t)|^2 \, dx = \left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex]
and he claims that the right hand side evaluates to zero since [itex]\Psi[/itex] must be normalizable and hence [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex].
My question is whether [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] must also equal zero. I ask this because:
Suppose we take the function [itex]\Psi (x,t)=\frac{1}{x} \sin(x^9)[/itex]. Then, we can calculate [itex] \int_{-\infty}^{\infty} \Psi^2 \, dx = 1.14...[/itex] so [itex]\Psi[/itex] IS normalizable. We can find that [itex]\frac{\partial \Psi}{\partial x}=\frac{1}{x^2}\left(9x^9 \cos(x^9)-\sin(x^9)\right)=9x^7 \cos(x^9)-\frac{\sin(x^9)}{x^2}[/itex],
so if we take [itex] \Psi^* \frac{\partial \Psi}{\partial x}=9 x^6 \sin(x^9) \cos (x^9) - \frac{\sin^2(x^9)}{x^3}[/itex], we find that this expression does not approach 0 as x approaches infinity (which renders Griffiths' argument using the fact [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex] to explain why the original integral false; there are functions that tend to zero as x approaches infinity but the derivative can grow arbitrarily large). Of course, in this case, [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] happens to vanish since [itex]\Psi^*=\Psi[/itex], but if we choose some general complex function, then I would guess that [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] might not vanish if we do not set [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] equal to zero. However, I have never seen this condition in quantum mechanics textbooks. Is there something wrong with my analysis, or is [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}=0[/itex] an implicit assumption physicists make? Thanks.
[itex]\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi(x,t)|^2 \, dx = \left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex]
and he claims that the right hand side evaluates to zero since [itex]\Psi[/itex] must be normalizable and hence [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex].
My question is whether [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] must also equal zero. I ask this because:
Suppose we take the function [itex]\Psi (x,t)=\frac{1}{x} \sin(x^9)[/itex]. Then, we can calculate [itex] \int_{-\infty}^{\infty} \Psi^2 \, dx = 1.14...[/itex] so [itex]\Psi[/itex] IS normalizable. We can find that [itex]\frac{\partial \Psi}{\partial x}=\frac{1}{x^2}\left(9x^9 \cos(x^9)-\sin(x^9)\right)=9x^7 \cos(x^9)-\frac{\sin(x^9)}{x^2}[/itex],
so if we take [itex] \Psi^* \frac{\partial \Psi}{\partial x}=9 x^6 \sin(x^9) \cos (x^9) - \frac{\sin^2(x^9)}{x^3}[/itex], we find that this expression does not approach 0 as x approaches infinity (which renders Griffiths' argument using the fact [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex] to explain why the original integral false; there are functions that tend to zero as x approaches infinity but the derivative can grow arbitrarily large). Of course, in this case, [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] happens to vanish since [itex]\Psi^*=\Psi[/itex], but if we choose some general complex function, then I would guess that [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] might not vanish if we do not set [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] equal to zero. However, I have never seen this condition in quantum mechanics textbooks. Is there something wrong with my analysis, or is [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}=0[/itex] an implicit assumption physicists make? Thanks.
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