# Is $F=ma$ an assumption in the original Schrodinger equation?

I believe the answer is yes, because in $\hat{H}\psi=E\psi$ he initially used, $\hat{H}=\frac{\hat{p}^2}{2m}+\hat{V}$ was clearly derived from classical mechanics by Hamilton.

Could you please clarify this for me?

In Sakurai's Modern Quantum Mechanics, he "derived" the Schr eq. I did not completely understand the processes, though.

• I think the answer is going to be very subjective. The only assumption in Schrodingers equation is the equation itself! – user12029 Jun 17 '17 at 2:56

It is dangerous to think about the Schrodinger equation and other quantum mechanical ideas from simple Newtonian physics. It is more natural to consider classical Hamiltonian mechanics. For example, in Hamiltonian mechanics, we have the Poisson bracket, which one might think as the classical analog of the quantum mechanical commutator:

$$\{A,\,B\} = \sum_i \left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial B}{\partial q_i}\frac{\partial A}{\partial p_i} \right)$$

where $A$ and $B$ are some physical quantities. From the above, we have the interesting property that the Poisson bracket of a time-independent quantity $A$ with the Hamiltonian is the negative of the total time derivative:

\begin{align} \{H,\,A\}=-\frac{dA}{dt} \end{align}

Now, let's go to the quantum mechanical world, and change our Poisson brackets to true-blue quantum commutators:

$$\{A,\,B\}\rightarrow \frac{1}{i\hbar}[A,\,B]$$

Plugging this into our equation for the derivative above, taking $A$ and $H$ to be average values with respect to some wave function $|\psi\rangle$, and taking $A$ to unity, we obtain the Schrodinger's equation:

$$i\hbar \frac{d}{dt}|\psi\rangle = H|\psi\rangle$$

As ZeroTheHero said, Schrodinger's equation is not limited to simple Hamiltonians of the form $p^2/2m+V$--it is much more general than that, and to understand it as the limit of some classical theory, we have to dive into Hamiltonian mechanics as opposed to Newtonian.

I leave it up to a historians to say whether or not the scientists used $\mathbf{F} = m\mathbf{a}$ literally or not. One of the bases of quantum mechanics, though, is the correspondence principle. This is the name of a broader principle in physics that any new theory has to match older theories in the regime where the older theories have been demonstrated to be valid. In that sense, Einstein "used" Newtonian mechanics to make special relativity, and then special relativity in the production of general relativity. Back to quantum mechanics, the most often used version of the correspondence principle is that the expectation values of quantum operators have to obey the classical equations of motion. In this case, the more accurate version is the pair of differential equations: \begin{align} \mathbf{F} &= \frac{\operatorname{d} \mathbf{p}}{\operatorname{d} t},\ \mathrm{and} \\ \mathbf{p} & = m \frac{\operatorname{d} \mathbf{x}}{\operatorname{d} t}. \end{align} What is $\mathbf{F}$, though? Well, we get from classical Hamiltonian mechanics that $\mathbf{F}_i = -\frac{\partial H}{\partial x_i}$ and $\frac{\mathbf{p}_i}{m} = \frac{\partial H}{\partial p_i}$ changing the equations to: \begin{align} -\frac{\partial H}{\partial x_i} &= \frac{\operatorname{d} p_i}{\operatorname{d} t},\ \mathrm{and}\\ \frac{\partial H}{\partial p_i} &= \frac{\operatorname{d} x_i}{\operatorname{d} t}. \end{align}

For quantum mechanics these equations of motion are the form that the quantum mechanical expectation values, \begin{align} -\left\langle \frac{\partial H}{\partial x_i} \right\rangle &= \frac{\operatorname{d} \langle p_i \rangle}{\operatorname{d} t},\ \mathrm{and}\\ \left\langle \frac{\partial H}{\partial p_i}\right\rangle &= \frac{\operatorname{d} \langle x_i \rangle}{\operatorname{d} t}, \end{align} obey according to the Ehrenfest theorem, which is one of the most frequently quoted versions of the correspondence principle.

The Schrodinger equation is not in any way limited to Hamiltonians of the form $$H=\frac{p^2}{2m} + V(q)$$ and so need not have any connection with $F=ma$.

• I meant the "original", or the initial Schr. Eq. – High GPA Jun 17 '17 at 2:55

I think a flat no is not a proper answer. Schroedinger was definitely guided by classical mechanics, more precisely by Hamilton-Jacobi formalism, in his formulation of the wave mechanics. On the other hand, analytical mechanics can be traced back to Newton's second law by means of d'Alembert Principle. Hence in this sense Schroedinger equation is indirectly related to second Newton's law

Hamilton himself put much effort in understanding and developing an analogy between classical mechanics and geometric optics. He noticed that in Hamilton-Jacobi theory, the momentum of the particle is given by $\vec\nabla S$, where the Hamilton's principal function $S$ is the action viewed as a function of the coordinates. By looking at level surfaces $S=\mathrm{const.}$ we see that particle's trajectory is orthogonal to the level surfaces. This is similar to light rays which travel perpendicularly to level surfaces corresponding to constant phase (wave fronts).

Schroedinger in 1926 conjectured that the action $S$ was indeed a phase of some wave process. Hence this wave should look like $$\psi=\psi_0\exp{\frac{iS}{\hbar}}=\psi_0\exp{\frac{i}{\hbar}\left[W(x)-Et\right]},$$ where $W(x)$ is the Hamilton's characteristic function. The constant $\hbar=h/2\pi$ is chosen so that this wave has frequency $\nu=E/h$, the Planck relation, which was known by that time. Plugging this wave into a wave equation one gets finally the Schroedinger equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=i\hbar\frac{\partial\psi}{\partial t}.$$ Classical mechanics can be understood as a limit case of Quantum Mechanics by plugging $\psi=\psi_0e^\frac{iS}{\hbar}$ into Schroedinger equation and taking the limit $\hbar\rightarrow 0$. The result is the Hamilton-Jacobi equation.

Hamilton-Jacobi is a formulation of analytical or variational mechanics and therefore has its roots in d'Alembert principle. This principle states that the virtual work of the effective force - applied force minus rate of change of momentum - is zero. To arrive at this principle one explicitly assumes Newton's second law $\vec F=\dot p$.

It is interesting to note that Hamilton was close to formulating wave mechanics a century before Schroedinger . He did not though, probably by lack of any experimental evidence.

No. There isn't really a well-defined concept of "force" in quantum mechanics. Assuming that the Hamiltonian for a free particle takes the form $H = p^2/(2m)$ is not the same thing as assuming that $F = ma$, because Hamilton's equations do not apply in quantum mechanics (except in certain limits).