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Yemon Choi
Yemon Choi
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Self-adjointness Selfadjointness of hamiltonian with $11/x$x potential

Let us consider the Hilbert space $L^2([0,\infty))$ and the operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$, smooth (smooth functions with compact support away from $0$).

Is the operator $H$H essentially self-adjoint? What is the domain of its self-adjoint extension?

Self-adjointness of hamiltonian with $1/x$ potential

Let us consider the Hilbert space $L^2([0,\infty))$ and the operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain $C^{\infty}_0((0,\infty))$, smooth functions with compact support away from $0$.

Is the operator $H$ essentially self-adjoint? What is the domain of its self-adjoint extension?

Selfadjointness of hamiltonian with 1/x potential

Let us consider the Hilbert space $L^2([0,\infty))$ and operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$ (smooth functions with compact support away from $0$).

Is the operator H essentially self-adjoint? What is the domain of its self-adjoint extension?

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edit approved May 9, 2016 at 8:45
Amir Sagiv
edit approved May 9, 2016 at 8:45
Amir Sagiv
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Selfadjointness Self-adjointness of hamiltonian with 1$1/xx$ potential

Let us consider the Hilbert space $L^2([0,\infty))$ and the operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$ (smooth, smooth functions with compact support away from $0$).

Is the operator H$H$ essentially self-adjoint? What is the domain of its self-adjoint extension?

Selfadjointness of hamiltonian with 1/x potential

Let us consider the Hilbert space $L^2([0,\infty))$ and operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$ (smooth functions with compact support away from $0$).

Is the operator H essentially self-adjoint? What is the domain of its self-adjoint extension?

Self-adjointness of hamiltonian with $1/x$ potential

Let us consider the Hilbert space $L^2([0,\infty))$ and the operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain $C^{\infty}_0((0,\infty))$, smooth functions with compact support away from $0$.

Is the operator $H$ essentially self-adjoint? What is the domain of its self-adjoint extension?

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user72829
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Selfadjointness of hamiltonian with 1/x potential

Let us consider the Hilbert space $L^2([0,\infty))$ and operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$ (smooth functions with compact support away from $0$).

Is the operator H essentially self-adjoint? What is the domain of its self-adjoint extension?