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I am still working on the Feynman path integral, more specifically on the case of a free particle with an infinite potential wall, i.e. the quantum system defined by the Hamiltonian

$$H_1 = \frac{\mathbf{P}^2}{2m} + V(\mathbf{Q})$$

where $V(\mathbf{Q})$ is the potential defined by

$$ V(\mathbf{Q})=\left\{ \begin{array}{cc} \infty, & \mathbf{Q} \leq b \\ 0, & \mathbf{Q}>b. \\ \end{array} \right. $$

It is clear that a general solution for the free particle case is given by $$\psi_E(q) = A_1 e^{i\frac{\sqrt{2mE}}{\hbar}q} + A_2 e^{-i\frac{\sqrt{2mE}}{\hbar}q}.$$ Using the fact that $\psi_E(0)=0$ I get that $A_2=-A_1$ so the general solution is $$A \sin\left(\frac{\sqrt{2mE}}{\hbar}q\right) = A \sin(kq)$$ where $k$ is defined as $$k=\frac{\sqrt{2mE}}{\hbar}.$$ The problem I have is the calculation of the constant $A$. In some references they say this constant should be $$A = \sqrt{\frac{2m}{\pi\hbar^2 k}}$$ and in other references I find $A=2$ or $A=-2i$,...

Could you please help me to understand that point ? Thanks.

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  • $\begingroup$ The Hamiltonian is Euclidean, so the normalising factor is $\sqrt{\frac{2i\pi\hbar}{Nm}}$, . Does that clear the confusion? . $\endgroup$ Jul 20, 2013 at 8:03
  • $\begingroup$ Thanks for the answer, but I don't know what a Euclidean Hamiltonian is, and what $N$ is. The dimension of the space ? $\endgroup$
    – Ludo
    Jul 20, 2013 at 8:15
  • $\begingroup$ No, not the dimension of the space! Maybe I should have been more clear. $\lim_{N\to\infty}\sqrt{\frac{2i\pi\hbar}{Nm}}$ is the infinitesimal reciprocal of the normalising factor of the phase, and a Euclidean Hamiltonian just means that it takes the form $\frac{p^2}{2m}+U$. $\endgroup$ Jul 20, 2013 at 8:17
  • $\begingroup$ Ok, but this constant comes after in the calculation. The constant $A$ coming from the eigenfunctions and your constant coming from the path integral are not the same thing, I guess. $\endgroup$
    – Ludo
    Jul 20, 2013 at 8:19
  • $\begingroup$ Actually, I got this reference from a previous post (reference here) and I am trying to understand it. $\endgroup$
    – Ludo
    Jul 20, 2013 at 8:20

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