Why can't i ℏ ∂ ∂ t be considered the Hamiltonian operator?

If one a priori wrongly declares that the Hamiltonian operator H ^ is the time derivative i ℏ ∂ ∂ t , then the Schrödinger equation (1) H ^ Ψ = i ℏ ∂ Ψ ∂ t would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction Ψ ( r , t )On the contrary, the Hamiltonian operator H ^ is typically a function of the operators r ^ and p ^ , and the Schrödinger equation (2) H ^ Ψ = i ℏ ∂ Ψ ∂ t One may then ask why is it then okay to assign the momentum operator as a gradient (3) p ^ k = ℏ i ∂ ∂ r k ? (This is known as the Schrödinger representation.) The answer is because of the canonical commutation relations (4) [ r ^ j , p ^ k ] = i ℏ δ k j 1 ^ . On the other hand, the corresponding commutation relation for time t is (5) [ H ^ , t ] = 0 , because time t is a parameter not an operator in quantum mechanics. Note that in contrast (6) [ i ℏ ∂ ∂ t , t ] = i ℏ , which also shows that one should not identify H ^ and i ℏ ∂ ∂ t

Key to "Why can't i ℏ ∂ ∂ t be..."

klepkowy7c

Answered question

2022-07-19

Why can't

i ℏ \frac{\partial}{\partial t}

be considered the Hamiltonian operator?

Answer & Explanation

Marshall Mcpherson

Beginner2022-07-20Added 11 answers

If one a priori wrongly declares that the Hamiltonian operator

\hat{H}

is the time derivative

i ℏ \frac{\partial}{\partial t}

, then the Schrödinger equation

\begin{matrix} (1) & \hat{H} Ψ = i ℏ \frac{\partial Ψ}{\partial t} \end{matrix}

would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction

Ψ (r, t)

On the contrary, the Hamiltonian operator

\hat{H}

is typically a function of the operators

\hat{r}

and

\hat{p}

, and the Schrödinger equation

\begin{matrix} (2) & \hat{H} Ψ = i ℏ \frac{\partial Ψ}{\partial t} \end{matrix}

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

\begin{matrix} (3) & {\hat{p}}_{k} = \frac{ℏ}{i} \frac{\partial}{\partial r^{k}} ? \end{matrix}

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

\begin{matrix} (4) & [{\hat{r}}^{j}, {\hat{p}}_{k}] = i ℏ δ_{k}^{j} \hat{1} . \end{matrix}

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

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

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

\begin{matrix} (6) & [i ℏ \frac{\partial}{\partial t}, t] = i ℏ, \end{matrix}

which also shows that one should not identify

\hat{H}

and

i ℏ \frac{\partial}{\partial t}

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