Quantum Mechanics Solution Manual |
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© Leon van Dommelen |
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2.5.1 Solution eigvals-a
Question:
Show that
, above, is also an eigenfunction of ![${\rm d}^2$](img91.gif)
, but with eigenvalue ![$\vphantom{0}\raisebox{1.5pt}{$-$}$](img9.gif)
. In fact, it is easy to see that the square of any operator has the same eigenfunctions, but with the square eigenvalues.
Answer:
Differentiate the exponential twice, [1, p. 60]:
So ![${\rm d}^2$](img91.gif)
turns
into
; the eigenvalue is therefore
which equals ![$\vphantom{0}\raisebox{1.5pt}{$-$}$](img9.gif)
.