klepkowy7c
2022-07-19
Answered

Why can't $i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}$ be considered the Hamiltonian operator?

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Marshall Mcpherson

Answered 2022-07-20
Author has **11** answers

If one a priori wrongly declares that the Hamiltonian operator $\hat{H}$ is the time derivative $i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}$, then the Schrödinger equation

$\begin{array}{}\text{(1)}& \hat{H}\mathrm{\Psi}\text{}=\text{}i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}\end{array}$

would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction $\mathrm{\Psi}(\mathbf{r},t)$

On the contrary, the Hamiltonian operator $\hat{H}$ is typically a function of the operators $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$, and the Schrödinger equation

$\begin{array}{}\text{(2)}& \hat{H}\mathrm{\Psi}\text{}=\text{}i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}\end{array}$

One may then ask why is it then okay to assign the momentum operator as a gradient

$\begin{array}{}\text{(3)}& {\hat{p}}_{k}\text{}=\text{}\frac{\hslash}{i}\frac{\mathrm{\partial}}{\mathrm{\partial}{r}^{k}}\text{}?\end{array}$

(This is known as the Schrödinger representation.) The answer is because of the canonical commutation relations

$\begin{array}{}\text{(4)}& [{\hat{r}}^{j},{\hat{p}}_{k}]\text{}=\text{}i\hslash \text{}{\delta}_{k}^{j}\text{}\hat{\mathbf{1}}.\end{array}$

On the other hand, the corresponding commutation relation for time t is

$\begin{array}{}\text{(5)}& [\hat{H},t]\text{}=\text{}0,\end{array}$

because time t is a parameter not an operator in quantum mechanics. Note that in contrast

$\begin{array}{}\text{(6)}& [i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t},\text{}t]\text{}=\text{}i\hslash ,\end{array}$

which also shows that one should not identify $\hat{H}$ and $i\hslash {\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}}$

$\begin{array}{}\text{(1)}& \hat{H}\mathrm{\Psi}\text{}=\text{}i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}\end{array}$

would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction $\mathrm{\Psi}(\mathbf{r},t)$

On the contrary, the Hamiltonian operator $\hat{H}$ is typically a function of the operators $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$, and the Schrödinger equation

$\begin{array}{}\text{(2)}& \hat{H}\mathrm{\Psi}\text{}=\text{}i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}\end{array}$

One may then ask why is it then okay to assign the momentum operator as a gradient

$\begin{array}{}\text{(3)}& {\hat{p}}_{k}\text{}=\text{}\frac{\hslash}{i}\frac{\mathrm{\partial}}{\mathrm{\partial}{r}^{k}}\text{}?\end{array}$

(This is known as the Schrödinger representation.) The answer is because of the canonical commutation relations

$\begin{array}{}\text{(4)}& [{\hat{r}}^{j},{\hat{p}}_{k}]\text{}=\text{}i\hslash \text{}{\delta}_{k}^{j}\text{}\hat{\mathbf{1}}.\end{array}$

On the other hand, the corresponding commutation relation for time t is

$\begin{array}{}\text{(5)}& [\hat{H},t]\text{}=\text{}0,\end{array}$

because time t is a parameter not an operator in quantum mechanics. Note that in contrast

$\begin{array}{}\text{(6)}& [i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t},\text{}t]\text{}=\text{}i\hslash ,\end{array}$

which also shows that one should not identify $\hat{H}$ and $i\hslash {\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}}$

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