# Classical Limits of Quantum Mechanics

$$\hbar\to0$$.

 Classical Quantum $$\frac{d A}{dt}=[A,H]_{classical}$$ $$\frac{d A_H}{dt}=\frac{i}{\hbar}[H,A_H]$$

Where $$[A,H]_{classical}$$ are poisson brackets, $$\frac{\partial H}{\partial q}\frac{\partial A}{\partial p}-\frac{\partial A}{\partial q}\frac{\partial H}{\partial p}$$.

Dirac’s rule: $$[]_{classical}\Leftrightarrow[]_{QM}/(i\hbar)$$. Some things in QM cannot be expressed classically: i.e. spin.

 CM QM No Uncertainty Uncertainty Zero Energy of Oscillator Finite Ground State of Oscillator Continuous Spectra of Oscillator Discrete Spectra of Oscillator Hamilton’s Principle Function QM Wavefunction Phase

$$\Psi(\vec{x},t) = \sqrt{\mathcal{P}(\vec{x},t)}\exp\left(\frac{i}{\hbar}S(\vec{x},t)\right)$$, $$\mathcal{P}(\vec{x},t)=|\Psi(\vec{x},t)|^2$$.

$$\vec{j}(\vec{x},t)=\frac{\hbar}{m}\Im(\Psi^*\vec{\nabla}\Psi) = \frac{\hbar}{m}\Im(\sqrt{\mathcal{P}}\exp(-iS/\hbar)(\exp(iS/\hbar)(\vec{\nabla}\sqrt{\mathcal{P}}+i/\hbar \sqrt{\mathcal{P}}\exp(iS/\hbar)\vec{\nabla}S))) = \frac{\hbar}{m}\Im(\sqrt{\mathcal{P}}\vec{\nabla}\sqrt{\mathcal{P}} + \frac{i}{\hbar}\mathcal{P}\vec{\nabla}S) = \frac{\mathcal{P}}{m}\vec{\nabla}S$$.

### Example

$$\Psi \sim \exp i\frac{\vec{p}\cdot\vec{x}-Et}{\hbar}$$, $$\vec{j}=\frac{\hbar}{m}\frac{\vec{p}}{\hbar}=\frac{\vec{p}}{m} = \frac{\mathcal{P}}{m}\vec{\nabla}S= \frac{1}{m}\vec{\nabla}S$$. So, $$\vec{\nabla}S=\vec{P}$$.

## Inserting into Schrodinger’s Equation - Hamilton-Jacobi

$$\Psi(\vec{x},t) = \sqrt{\mathcal{P}(\vec{x},t)}\exp\left(\frac{i}{\hbar}S(\vec{x},t)\right)$$, $$\mathcal{P}(\vec{x},t)=|\Psi(\vec{x},t)|^2$$.

$$i\hbar\frac{\partial}{\partial t}\Psi=\left(-\frac{\hbar^2}{2m}\vec{\nabla}^2+V\right)\Psi = i\hbar\left[\frac{\partial}{\partial t}\sqrt{\mathcal{P}}+\frac{i}{\hbar}\sqrt{\mathcal{P}}\frac{\partial S}{\partial t}\right]\exp\frac{i}{\hbar}S = -\frac{\hbar^2}{2m}\vec{\nabla}^2(\sqrt{\mathcal{P}}\exp\frac{i}{\hbar}S)+V\sqrt{\mathcal{P}}\exp\frac{i}{\hbar}S$$.

$$i\hbar\left[\frac{\partial}{\partial t}\sqrt{\mathcal{P}}+\frac{i}{\hbar}\sqrt{\mathcal{P}}\frac{\partial S}{\partial t}\right] =-\frac{\hbar^2}{2m}\left[\vec{\nabla}^2(\sqrt{\mathcal{P}}) + \left(\frac{i}{\hbar}\sqrt{\mathcal{P}}\vec{\nabla}^2S+\frac{2i}{\hbar}\vec{\nabla}(\sqrt{\mathcal{P}}(\vec{\nabla}S) - \frac{1}{\hbar^2}\sqrt{\mathcal{P}}|\vec{\nabla}S|^2)\right)\right]$$.

In the classical limit, we see $$0=\sqrt{\mathcal{P}}\frac{\partial S}{\partial t} + \frac{\sqrt{\mathcal{P}}}{2m}|\vec{\nabla}S|^2 + V\sqrt{\mathcal{P}}$$. So, $$\frac{\partial S}{\partial t} + \frac{1}{2m}|\vec{\nabla}S|^2 + V=0$$ which is the Hamiltonian-Jacobi equation with $$S$$ the Hamilton’s principle function (action?).

### Special Case

If $$H$$ is time independent then $$S(\vec{x},t)=W(\vec{x})-Et$$. So, $$\Psi(\vec{x},t)=\sqrt{\mathcal{P}}\exp\frac{i}{\hbar}(W(\vec{x})-Et)$$, which gives a time independent and time dependent state. In HJ, $$\frac{1}{2m}|\vec{\nabla}W(\vec{x})|^2 + V(\vec{x}) = E$$. Thus, $$|\vec{\nabla}W(\vec{x})|^2 = 2m(E-V(\vec{x}))$$.

#### One Dimensional

$$\frac{d}{dx}W = \pm\sqrt{2m(E-V(x))}$$, $$E>V(x)$$. So, $$W(x)=\pm\int^x dx'\sqrt{2m(E-V(x'))}$$.

### Helper Info

$$\vec{\nabla}^2(\sqrt{\mathcal{P}}\exp\frac{i}{\hbar}S) = \vec{\nabla}^2(\sqrt{\mathcal{P}})\exp\frac{i}{\hbar}S + \left(\frac{i}{\hbar}\sqrt{\mathcal{P}}\vec{\nabla}^2S+\frac{2i}{\hbar}\vec{\nabla}(\sqrt{\mathcal{P}}(\vec{\nabla}S) - \frac{1}{\hbar^2}\sqrt{\mathcal{P}}|\vec{\nabla}S|^2)\right)\exp\frac{i}{\hbar}S$$

#### Examples

• Quantum Harmonic Oscillator

At the classical limit, we see that the time evolved uncertainty of position follows a localized mass exactly.

Created: 2024-05-30 Thu 21:18

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