Question

A fellow student proposes that a possible wave function for a free particle with mass \(m\) (one for which the potential-energy function \(U(x)\) is zero ) is

$$ \psi(x)=\left\{\begin{array}{ll} e^{-k x}, & x \geq 0 \\ e^{+\kappa x}, & x<0 \end{array}\right. $$

where \(\kappa\) is a positive constant. (a) Graph this proposed wave function.

(b) Determine the energy of the particle if the proposed wave function satisfies the Schrödinger equation for \(x<\)0.

(c) Show that the proposed wave function also satisfies the Schrödinger equation for \(x \geq 0\) with the same energy as in part (b).

(d) Explain why the proposed wave function is nonetheless not an acceptable solution of the Schrödinger equation for a free particle.

Answer 1

For the free particle of mass of \(m\), the wave function is given as

\(\psi(x)=\left\{\begin{array}{l}e^{+k x}, x \leq 0 \\ e^{-k x}, x \geq 0\end{array}\right.\)

Where \(k\) is a positive constant

(a) The graph of the function stated in step 1:

(b) The Schrödinger wave equation in one dimension is

\(-\frac{\hbar^{2}}{2 m}\left(\frac{d^{2} \psi}{d x^{2}}\right)+V(x) \psi(x)=E \psi(x)\)

For \(\psi(x)=e^{k x}, x<0\)

\(\frac{d \psi}{d x}=-k e^{k x},\) As \(x<0\)

\(\frac{d^{2} \psi}{d x^{2}}=k^{2} e^{k x},\) As \(x<0\)

\(=k^{2} \psi(x)\)

For the free particle \(U(x)=0\)

Putting these values in the Schrödinger wave equation,

$$ \begin{array}{l} -\frac{\hbar^{2}}{2 m}\left(k^{2}\right) \psi(x)+0 \cdot \psi(x)=E \psi(x) \\ V=0 \end{array} $$

That is, \(-\frac{\hbar^{2}}{2 m} k^{2} \psi(x)=E \psi(x)\)

If \(E=-\frac{\hbar^{2} k^{2}}{2 m},\) then the Schrödinger equation is satisfied.

(c) Also, the second part of the function

\(\psi(x)=e^{-k x}\) If \(x>0\)

\(\frac{d \psi}{d x}(x)=(-k) e^{-k x}\)

\(\frac{d^{2} \psi}{d x^{2}}=(-k)(-k) e^{-k x}\)

\(=k^{2} e^{-k x}\)

\(=k^{2} \psi(x)\)

That is,

\(-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}=\frac{-h^{2}}{2 m} k^{2} \psi(x)\)

In order that \(\psi(x)=e^{-k x}, x \geq 0\) satisfies the Schrödinger wave equation,

\(E=\frac{-\hbar^{2} k^{2}}{2 m}\)

(d) We see from part (b) and

(c) that the energy of the free particle is negative.

\(\psi(x)\) satisfies Schrödinger's equation if \(E=\frac{-\hbar^{2} k^{2}}{2 m}\) and \(\frac{d \psi}{d x}\) is

discontinuous at \(x=0\)