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In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |ik\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of the) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |ik\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |k\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |ik\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

In standard quantum mechanics Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |k\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |k\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.