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r1 - 23 Jul 2007 - 19:45:32 - ThanhTruongYou are here: TWiki >  Main Web > LectureNotes > UndergraduatePChem > Chapter5 > FreeParticle1D

Free Particle in One-Dimension

The above two particle in a box systems show that energy quantization arises from the boundary conditions, or from the particle being confined in a finite region.  We also pointed out in the particle in 1D box that as the length of the box L increases to infinity, the energy spacing decreases to zero, i.e. the energy spectrum of the system becomes continuous.   In other words, free particle has the classical mechanics picture!   Let solve the free particle in 1D quantum mechanically to verify this.

The potential energy V(x) is zero for all x. The Schrödinger equation is

\[- \frac{ \hbar ^2}{ 2m} \frac{ d^2 \psi(x)}{ dx^2} = E \psi(x) \; \; \; \; (- \infty < x < \infty) \]

This differential equation has two general solutions

\[\psi_1(x) = A_1 e^{ikx} \; \; \; \; and \; \; \; \; \psi_2(x) = A_2 e^{-ikx}\]

where

\[k = \frac {\sqrt {2mE}}{\hbar}\]

Thus,

\[E = \frac{ \hbar ^2 k^2}{ 2m} \]

There is no restriction on the value of k and thus the energy is not quantized (consistent with the classical picture).

Since V = 0, the total energy of the system, E, is the kinetic energy and thus must be positive. What happen if E is negative?

\[ \lim_{ x \rightarrow \infty } \psi_2(x) = \lim_{ x \rightarrow \infty } A_2 e^{ (\sqrt {2m \left | E\right |}/ \hbar)x} = \infty \]

Thus if E is negative the wavefunction becomes unbound and thus is not a valid wavefunction (Postulate 1).

To provide a physical interpretation for $\psi_1(x)$ and $\psi_2(x)$, let operate them with the momentum operator $\hat P$.

\[\hat P \psi_1 = -i \hbar \frac{ d\psi_1}{ dx} = -i \hbar \frac{ d}{ dx} \left ( A_1 e^{ikx}\right ) = (\hbar k) \psi_1 \]

\[\hat P \psi_2 = -i \hbar \frac{ d\psi_2}{ dx} = -i \hbar \frac{ d}{ dx} \left ( A_2 e^{-ikx}\right ) = (-\hbar k )\psi_2 \]

Since $\psi_1$ and $\psi_2$ are also the eigenfunctions of the momentum operator $\hat P$, the energy and momentum of the particle can be measured simultaneously and precisely at arbitrary precision (Postulate 4). Thus, we can interprete that $\psi_1$ describes the particle moving to the right with the fixed momentum $\hbar k$ and $\psi_2$ describes the particle moving to the left with the fixed momentum $-\hbar k$.

Finally, the probability density for finding the particle at x

\[\psi_1^* (x) \psi_1 (x) = \left (A_1^* e^{-ikx} \right ) \left (A_1 e^{ikx} \right )= A_1^* A_1 = \left |A_1 \right |^2 = constant \]

\[\psi_2^* (x) \psi_2 (x) = \left (A_2^* e^{ikx} \right ) \left (A_2 e^{-ikx} \right )= A_2^* A_2 = \left |A_2 \right |^2 = constant \]

Since the probability density is constant in both cases, the particle is equally likely to be found anywhere along the x axis. In other words, there is an infinite uncertainty on the location of the particle. This is consistent with the Uncertainty Principle since the momentum of the particle is known exactly.

-- ThanhTruong - 23 Jul 2007

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