Nondegenerate Eignefunctions as Linear Combinations

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SUMMARY

The discussion confirms that a linear combination of nondegenerate eigenfunctions cannot yield another eigenfunction. Specifically, if two eigenfunctions correspond to different eigenvalues, their linear combination does not satisfy the eigenfunction condition. The equation aE_1 ψ_1 + bE_2 ψ_2 = E(aψ_1 + bψ_2) can only hold true if E_1 equals E_2, which contradicts the premise of nondegeneracy. Thus, nondegenerate eigenfunctions maintain linear independence, preventing such combinations from being eigenfunctions.

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zachzach
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It is easily shown that two eigenfunctions with the same eigenvalues can be combined in a linear combination so that the linear combination is itself an eigenfunction. But what if the two eigenvalues are not the same? Can you still find a linear combination of the two functions that is an eigenfunction?

<br /> <br /> aE_1 \psi_1+ b E_2 \psi_2 = E(\psi_1 + \psi_2)<br /> <br />
 
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The equation you have written down you can set aE_1=bE_2 and it will satisfy the equation with E=aE1.

But what you're really looking for is:

<br /> <br /> <br /> aE_1 \psi_1+ b E_2 \psi_2 = E(a\psi_1 + b\psi_2)<br /> <br /> <br />

which can only be satisfied if E_1=E_2=E
 
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Yes, I meant to write the equation that you did. So the answer is no then. You can never have a linear combination of non degenerate eigenfunctions that is itself an eigenfunction.
 
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zachzach said:
Yes, I meant to write the equation that you did. So the answer is no then. You can never have a linear combination of non degenerate eigenfunctions that is itself an eigenfunction.

The idea is that the set of nondegenerate eigenfunctions is linearly independent. One eigenfunction cannot be the sum of two others, or else the set would not be linearly independent.
 

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