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Let $H$ be a diagonalizable matrix (not necessarily Hermitian). Then, it induces a biorthogonal left and right vectors, such that $$ H\left|\lambda\right\rangle=\lambda\left|\lambda\right\rangle,\quad\left\langle\mu\right|H=\mu\left\langle\mu\right|\\ \left\langle\mu\big|\lambda\right\rangle=\delta_{\mu\lambda},\quad \sum_\lambda\left|\lambda\right\rangle\!\!\left\langle\lambda\right|=\mathbb{I}. $$ Extending to the infinite case for time-independent Schrodinger equation with a complex potential, $V$ $$ E\left|E\right\rangle=\left[-\nabla^2+V(x)\right]\!\!\left|E\right\rangle\equiv \hat{H}\left|E\right\rangle . $$ How do i define the left eigenvectors? what sort of differential equation do they satisfy (if any)?

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What you're really doing here is computing (right) eigenvectors for the Hermitian conjugate $H^\dagger$ of $H$, for if $\left| f \right\rangle$ is an arbitrary ket vector, then, by definition of $H^\dagger$, $\left\langle f \right| H$ is the bra vector corresponding to the ket vector $H^\dagger \left| f \right\rangle$. So, the key is to compute $H^\dagger$, given your Hamiltonian $H$, for then your left eigenvalue equation will amount to $$ H^\dagger \left| \mu \right\rangle = \mu \left|\mu\right\rangle. $$

Now, given the form of your Hamiltonian, you have to work with the concrete Hilbert space $L^2(\mathbb{R}^3)$, endowed with the inner product $$ \left\langle f \!\mid\! g \right\rangle = \iiint_{\mathbb{R}^3} \overline{f(x)}g(x)\,\mathrm{d}^3 x. $$ Then, for any ket vectors $\left| f \right\rangle$ and $\left| g \right\rangle$ (taken, for theoretical reasons, to be smooth functions with compact support), $$ \left\langle f \!\mid\! H \!\mid\!g \right\rangle = \iiint_{\mathbb{R}^3} \overline{f(x)}\left(-\Delta g(x) + V(x)g(x)\right)\,\mathrm{d}^3 x = \iiint_{\mathbb{R}^3} \overline{h(x)}g(x)\,\mathrm{d}^3 x = \left\langle h \!\mid\! g\right\rangle, $$ where $\left|h\right\rangle = H^\dagger \left|f\right\rangle$. Can you use integration by parts to compute $\left|h\right\rangle$ from $\left|f\right\rangle$, and hence compute $H^\dagger$?

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  • $\begingroup$ Thanks. Indeed, i solve for $H^\dagger$ and then take the complex conjugate. $\endgroup$ – PhysicistRRE May 29 '14 at 18:39

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