Question

When a particle is dropped into a 1-D infinite potential well of finite width L, by solving the Schrodinger's equation, \(\LARGE {-h^2\over 2m} {d^2 \Psi
\over dx^2}+U\Psi=E\Psi\), the result of the energy E becomes \(\LARGE E=\dfrac {h^2k^2}{2m}=(\dfrac {h^2}{8mL^2})n^2\).

This shows that energy is quantized and depends only on the integer value n. However, if energy is quantized, then assuming that the Energy is at it's lowest state of \(E_1\) where n=1, then the energy is fixed. But if E is fixed, and the potential energy inside the well is 0, then E=K where K is fixed. (cont.)

Answer 1

But if K is fixed, and we know that \(\Large K=\dfrac {p^2}{2m}\), then the momentum is also fixed.

However, this violates the Heissenberg Uncertainty Principle.

Does this argument in fact hold and violate the Heissenberg Uncertainty Principle. If not, where is the hole in the argument? Explain.

Answer 2

@Vincent-Lyon.Fr

Answer 3

@Saeeddiscover

Answer 4

No, you have missed a very important fact. even though momentum is fixed, but you still have uncertainty in the position of the object. We really don't know where the particle is located.

Answer 5

But the width is finite. While I agree we do not know it's location, \(\Delta x\Delta p \) is not greater than h/2. But

Answer 6

Ignore the "but"

Answer 7

be back

Answer 8

No, you can't name it a violation of the principle. the product of the two quantities MUST always be greater than or equal to hbar/2. so I cannot see any violation....

Answer 9

Right, it MUST be greater, but yet the solution is not so wouldn't that be a paradox?