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[tex]i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle = H\left|\psi\right\rangle[/tex]

be consistent with the form that operates on wavefunctions in the position representation:

[tex]i\hbar\frac{\partial}{\partial t}\psi(x) = H\psi(x)[/tex]

If I try to do this by plugging in [tex]\left|\psi\right\rangle =\int dx\,\psi(x)\left|x\right\rangle[/tex] to the abstract form, I end up with a contradiction. Starting with the RHS:

[tex]i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle & = & i\hbar\frac{\partial}{\partial t}\left(\int dx\,\psi(x)\left|x\right\rangle \right)[/tex]

[tex] & = & i\hbar\int dx\,\left(\frac{\partial\psi}{\partial t}\left|x\right\rangle +\psi\frac{\partial\left|x\right\rangle }{\partial t}\right)[/tex]

If we then use the Schrodinger equation again on [tex]\frac{\partial\left|x\right\rangle }{\partial t}[/tex]

[tex]\frac{\partial\left|x\right\rangle }{\partial t} & = & -\frac{i}{\hbar}H\left|x\right\rangle =\frac{i\hbar}{2m}\int dx'\,\left|x'\right\rangle \frac{\partial^{2}}{\partial x'^{2}}\left(\left\langle x'|x\right\rangle \right)[/tex]

[tex] & = & \frac{i\hbar}{2m}\int dx'\,\left|x'\right\rangle \delta''(x-x')[/tex]

[tex] & = & \frac{i\hbar}{2m}\frac{\partial^{2}\left|x\right\rangle }{\partial x^{2}}[/tex]

Where above we use the identity [tex]\int dx\,\phi(x)\delta''(x-a)=\phi''(a)[/tex]. I think we can now use parts a couple of times, together with the fact that [tex]\psi (x)[/tex] and [tex]\psi' (x)[/tex] go to zero at infinity, to say that

[tex]\int dx\,\psi\frac{\partial\left|x\right\rangle }{\partial t} & = & \frac{i\hbar}{2m}\int dx\,\psi\frac{\partial^{2}\left|x\right\rangle }{\partial x^{2}}[/tex]

[tex] & = & \frac{i\hbar}{2m}\int dx\,\frac{\partial^{2}\psi}{\partial x^{2}}\left|x\right\rangle[/tex]

Now, the RHS of the TDSE will go to [tex]-\frac{\hbar^{2}}{2m}\int dx\,\frac{\partial^{2}\psi}{\partial x^{2}}\left|x\right\rangle [/tex].

If we multiply by [tex]\left\langle x\right|[/tex] from the left on both sides,

we end up with

[tex]i\hbar\left(\frac{\partial\psi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}\right) & = & -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}[/tex]

[tex]\Rightarrow\quad\frac{\partial\psi}{\partial t} & = & 0[/tex]

Clearly this is not right, as from the second form of the Schrodinger equation written down above, we get

[tex]\frac{\partial\psi}{\partial t} & = & \frac{i\hbar}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}[/tex]

Can somebody help me find whereabouts I am going wrong here? Thanks.