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Suppose I have a wave function $\Psi$ (which is not an eigenfunction) and a time independent Hamiltonian $\hat{\mathcal{H}}$. Now, If I take the classical limit by taking $\hbar \to 0$ what will happen to the expectation value $\langle\Psi |\hat{\mathcal{H}}|\Psi\rangle$? Will it remain the same (as $\hbar = 1.0$) or will it be different as $\hbar\to 0$? According to correspondence principle this should be equal to the classical energy in the classical limit.

What do you think about this? Your answers will be highly appreciated.

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For a particle in ground state of a box the expectation value of hamiltonian is $\frac{\pi^2\hbar^2}{2mL^2}$ which tends to zero when $\hbar \rightarrow 0$. – richard Oct 31 '13 at 17:34
Related: and links therein. – Qmechanic Nov 1 '13 at 14:54
up vote 1 down vote accepted

The above posters seem to have missed the fact that $\Psi$ is not an eigenfunction, but an arbitrary wavefunction. The types of wavefunctions we normally see when we calculate things are usually expressed in terms of eigenfunctions of things like energy or momentum operators, and have little to do, if anything, with classical behaviour (e.g. look at the probability density of the energy eigenstates for the quantum harmonic oscillator and try to imagine it as describing a mass connected to a spring).

What you might want to do is construct coherent states which are states where position and momentum are treated democratically (uncertainty is shared equally between position and momentum).

Then, the quantum number that labels your state might be thought of as the level of excitation of the state. For the harmonic oscillator, this is roughly the magnitude of the amount of energy in the state in that $E = \langle n \rangle \hbar= |\alpha^2| \hbar$. If you naively take $\hbar \to 0$ then everything vanishes. But if you keep, say, the energy finite, while taking $\hbar \to 0$, then you can recover meaningful, classical answers (that don't depend on $\alpha$ or $\hbar$).

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First of all I would like to thank all of you for sharing your valuable ideas. I have asked this question, in connection with the coherent state. If you have a pure classical phase-space variables (p,q) then you can find a coherent state with $\alpha=\frac{1}{2}(p+i q)$. Classical energy is given by the classical hamiltonian $H(p,q)$ and the quantum energy can be computed from the $\langle\alpha|H|\alpha \rangle$. These classical and quantum energy values will be different due to zero-point energy. In this case as $\hbar \to 0$ it seems that quantum energy decreases. – Sijo Joseph Nov 8 '13 at 1:31

For the case of a particle in a potential, $\hat{\mathcal H} = \frac{\hat{p}^2}{2m}+V({\mathbf x})$, let an arbitrary wavefunction be written in the form $$\Psi({\mathbf{x}},t) = \sqrt{\rho({\mathbf x},t)}\exp\left(i\frac{S({\mathbf x},t)}{\hbar}\right)\text{,}$$ where $\rho \geq 0$. Then it becomes a simple calculus exercise to derive: $$\Psi^*\hat{\mathcal{H}}\Psi = \rho\left[\frac{1}{2m}\left|\nabla S({\mathbf x},t)\right|^2 + V(\mathbf{x})\right] + \mathcal{O}(\hbar)\text{,}$$ where I'm omitting terms that have at least one power of $\hbar$. Since $\langle\Psi|\hat{\mathcal H}|\Psi\rangle$ is the spatial integral of this quantity, integrating this instead is what we want for an $\hbar\to 0$ limit of energy.

[Edit] As @Ruslan says, the wavefunction would have to oscillate faster to have a kinetic term. In the above, keeping $S$ independent of $\hbar$ means increasing the phase at the same proportion that $\hbar$ is lowered.

Additionally, substituting this form for $\Psi$ into the Schrödinger equation gives, after similarly dropping $\mathcal{O}(\hbar)$ terms, $$\underbrace{\frac{1}{2m}\left|\nabla S({\mathbf x},t)\right|^2 + V(\mathbf{x})}_{{\mathcal H}_\text{classical}} + \frac{\partial S({\mathbf x},t)}{\partial t} = 0\text{,}$$ which is the classical Hamilton-Jacobi equation with $S$ taking the role of the Hamilton's principal function.

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That's the right answer. UpVote $0$k. – Felix Marin Nov 2 '13 at 9:48
I like this, thank you. I partially agree with you. Normally quantum energy of a coherent state will be higher than the energy of the corresponding classical counter part due to zero point energy. As $\hbar\to0$ the zero point energy vanishes hence quantum energy should depend on the $\hbar$ value. That gives me a contradiction. – Sijo Joseph Nov 8 '13 at 1:51

Normal time-independent hamiltonian looks like $\hat H=\hat T+\hat V$, where $\hat T=-\frac{\hbar^2}{2m}\nabla^2$ is kinetic energy operator and $\hat V=V(\hat x)$ is potential energy operator. As seen from these expressions, only kinetic energy operator changes with $\hbar$.

Now we can see that

  1. Quantum mechanical expectation value of particle total energy is sum of expectation values for kinetic and potential energies: $$\langle\Psi\left|\hat H\right|\Psi\rangle=\langle\Psi\left|\hat T\right|\Psi\rangle+\langle\Psi\left|\hat V\right|\Psi\rangle$$

  2. Taking $\hbar\to0$, we get $\hat T\to \hat 0\equiv0$. Now expectation value for particle total energy becomes equal to expectation value of its potential energy: $$\langle\Psi\left|\hat H_{\hbar=0}\right|\Psi\rangle=\langle\Psi\left|\hat V\right|\Psi\rangle$$

From this follows immediate answer: no, the expectation value will not remain the same. And interesting result is that for any smooth wavefunction expectation value of kinetic energy is zero when $\hbar$ is zero.

This implies that for classical limit the wavefunction must oscillate infinitely fast (i.e. have zero wavelength) to remain at the same total energy. As you make $\hbar$ smaller, the state with given total energy gets larger quantum number - i.e. becomes more excited.

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Yes, this can be answered using a classical perspective. We all know the electromagnetic or optical equation: $$ E =\nu h = \omega \hbar \longrightarrow 0 = \omega 0 $$ As Richard has indicated the answer to this can be produced from a visit to wiki, "the Hamiltonian is commonly expressed as the sum of operators corresponding to the kinetic and potential energies" $$ \mathcal{ \hat H } =\hat T + \hat V = {{\hat p^2} \over {2 m}}+V = V - { \hbar^2 \bigtriangledown^2 \over 2m} $$ For this case: $$\mathcal{ \hat H } \rightarrow \hat V = V=0$$ "V" is just the potential the system is placed at, and for our universe we can assume V=0. $$\Psi=\Psi( \vec{r} ) \space \space \space \space and \space thus: \space \space \space \space \mathcal{ \hat H } \mid \Psi \rangle = i \hbar {{\partial \Psi } \over {\partial \vec{r}}} \rangle$$ $$ \langle \Psi \mid \mathcal{ \hat H } \mid \Psi \rangle =\int \Psi^* \mathcal{ \hat H } ( \Psi ) d \vec{r} = \int \Psi^* i \hbar ( \Psi ' ) d \vec{r} $$ So it does not matter what Psi is or what the derivative of Psi over some dimension is or what dimensions Psi is existent in, or what the complex conjugate of Psi is or what limits we integrate over. The solution is a multiple of h.

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