A Dimension of \lambda constant in \delta potential

LagrangeEuler
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Time independent Schroedinger equation in ##\delta## potential ##V(x)=-\lambda \delta(x)##, where ##\lambda >0## is given by
-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}\psi(x)-\lambda \delta(x)\psi(x)=E\psi(x).
How to find dimension of ##\lambda##? Dimension of ##V(x)## is
[V(x)]=ML^2T^{-2}.
Because it is one dimensional problem dimension of ##\psi(x)## is
[\psi(x)]=L^{-\frac{1}{2}}.
Is then also
[\delta(x)]=L^{-\frac{1}{2}}?
 
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The dimension of ##\psi## is irrelevant. From ##\int dx\,\delta(x)=1## it follows that the dimension of ##\delta## is ##L^{-1}##. From that, the dimension of ##\lambda## should be easy.
 
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Demystifier said:
The dimension of ##\psi## is irrelevant. From ##\int dx\,\delta(x)=1## it follows that the dimension of ##\delta## is ##L^{-1}##. From that, the dimension of ##\lambda## should be easy.
Thank you very much.
 
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