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Chapter 35: Problem 31

Suppose \(\psi_{1}\) and \(\psi_{2}\) are solutions of the Schrodinger equation for the same energy \(E .\) Show that the linear combination \(a \psi_{1}+b \psi_{2}\) is also a solution, where \(a\) and \(b\) are arbitrary constants.

Short Answer

Expert verified
Yes, the linear combination \(\psi = a \psi_{1} + b \psi_{2}\) where \(a\) and \(b\) are arbitrary constants is also a solution of the Schrödinger equation for the same energy \(E\).

Step by step solution

01

Write down the Schrodinger equation

The time-independent Schrödinger equation is given by:\[H \psi = E \psi\]where \(H\) is the Hamiltonian operator, \(\psi\) is the wave function, and \(E\) is the energy of the system.
02

Substitute the solutions into the equation

\(\psi_{1}\) and \(\psi_{2}\) are solutions for the same energy \(E,\) so:\[H \psi_{1} = E \psi_{1}\]\[H \psi_{2} = E \psi_{2}\]where \(H\) is the Hamiltonian.
03

Form the Linear Combination

Now, the problem states that a linear combination of these solutions is also a solution. We write down this combination as:\[\psi = a \psi_{1} + b \psi_{2}\]where \(a\) and \(b\) are arbitrary constants.
04

Apply the Hamiltonian to the Linear Combination

Now, we apply the Hamiltonian operator to this linear combination:\[H \psi = H (a \psi_{1}+ b \psi_{2})\]Using the linearity properties of the Hamiltonian operator, we can write this as:\[H \psi = a H \psi_{1} + b H \psi_{2}\]substitute the earlier derived Schrödinger equations for \( \psi_{1} \) and \( \psi_{2} \) into this expression to get:\[H \psi = a E \psi_{1} + b E \psi_{2}\]This can be rearranged into:\[H \psi = E (a \psi_{1} + b \psi_{2})\]which we recognize as Schrödinger's equation but with the wave function being the linear combination \(\psi = a \psi_{1} + b \psi_{2}\). This confirms that the linear combination is indeed a solution to the Schrödinger equation with energy \(E\).

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