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We know that total energy of the system is classically continuous, but in quantum mecanics (QM) it is quantised. My question is:

  1. How can we use the conservation of energy equation to derive schrödinger equation in QM?

  2. I mean What is the validity of conservation of energy equation in QM?

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    $\begingroup$ "how can we use the conservation of energy equation to derive schrödinger equation in QM"...why do you think the SE derives from energy conservation? $\endgroup$ – ACuriousMind Jan 16 '16 at 15:05
  • $\begingroup$ I mean SE can also derivable from Conservation of Energy right? How COE is also true in QM? That's my question? $\endgroup$ – Muthu manimaran Jan 16 '16 at 15:10
  • $\begingroup$ @ACuriousMind: huh? See: en.wikipedia.org/wiki/… $\endgroup$ – Gert Jan 16 '16 at 15:12
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    $\begingroup$ @Gert: That's just a weird way of saying that the Hamiltonian is usually the energy operator. You don't derive the Schrödinger equation. It's the quantum mechanical equation of motion which is postulated. You can make various plausibility arguments for it (e.g. classical limit, energy conservation, etc.), but it is not derived from those arguments. $\endgroup$ – ACuriousMind Jan 16 '16 at 15:15
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    $\begingroup$ For that, you first have to say what exactly you mean by conservation of energy in QM. One manifestation would be that its expectation value is conserved as per Ehrenfest's theorem. $\endgroup$ – ACuriousMind Jan 16 '16 at 15:26
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  1. The Schrödinger equation cannot be derived from classical physics. There are various consistency checks and motivations, such as its consistency with conservation of energy, but it is not derived from those considerations. However, that the Schrödinger equation conserves energy is built in when one knows that the Hamiltonian is the energy operator since $$ \partial_t \lvert \psi \rangle = \frac{1}{i\hbar}H\lvert \psi \rangle$$ means that the time evolution operator is $\mathrm{e}^{-\mathrm{i}tH}$, so the energy operator commutes with the evolution operator, so it doesn't matter whether I measure the energy before or after the evolution, so energy is conserved.

  2. The quantum version of conservation laws is a bit subtle, there are various ways in which one can call a quantity "conserved":

    • In the Heisenberg picture, operators are time-dependent and conservation means that the time derivative of the operator is zero. By the Heisenberg equation of motion $$ \frac{\mathrm{d}}{\mathrm{d}t}A = \frac{\mathrm{i}}{\hbar}[H,A]$$ and the correspondence of the quantum commutator with the classical Poisson bracket1 in the classical equation of motion $$ \frac{\mathrm{d}}{\mathrm{d}t}A = \{A,H\}$$ shows that classically conserved quantities will be conserved in the Heisenberg picture in this way.

    • Regardless of picture, one might speak of the expectation values of the operators. By Ehrenfest's theorem $$ \frac{\mathrm{d}}{\mathrm{d}t}\langle A\rangle_\psi = \langle\frac{\mathrm{i}}{\hbar}[H,A]\rangle_\psi$$ so again, if the commutator with the Hamiltonian vanishes, then the expectation value is constant in time.

Altogether, you see that an observable is conserved when it commutes with the Hamiltonian. However, since observables don't have definite values for most states, one must be careful what exactly one means when one says that something is conserved quantum mechanically.

This becomes more subtle in quantum field theories, where the proper statements about conserved quantities are Ward-Takahashi identities.


1The naive canonical quantization prescription is replacing $\{,\}$ by $\frac{1}{\mathrm{i}\hbar}[,]$. This might fail in some cases and require quantum corrections of order $\hbar^2$, see also this Phys.SE post.

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  • $\begingroup$ I see we were working on answers at the same time and covered the same ground (mine a little more historical, yours a little more technical). Never-the-less I like your answer so +1. $\endgroup$ – Lewis Miller Jan 16 '16 at 18:51
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Quantum mechanics is not derived from classical mechanics or energy conservation, but there are "jumping off points" in classical mechanics that may serve to answer your question.

If you study classical mechanics at a sufficiently advanced level you will discover the Hamiltonian formalism. The Hamiltonian for an isolated system with only conservative interactions is the energy and it is conserved. In such systems it is a function of the generalized coordinates (q) and their time derivatives ($H(q, \frac{dq}{dt})$). In the Hamiltonian formalism there are entities called Poisson brackets (I'll leave it to you to look up their definition but they are represented like this [a,b]). Once you understand Poisson brackets, you can show that the equations of motion for your system are: $$\frac{dq_i}{dt}=[q_i,H]$$ where $q_i$ is one of the generalized coordinates.

Now let's jump from classical mechanics to quantum mechanics. The Planck hypothesis was that energies are no longer continuous but discrete (always proportional to the Planck constant h). He was forced to introduce this assumption in order to derive the black body radiation curve. Heisenberg realized that he could preserve most of the classical formalism if he introduced operators for the classical variables and associated their Poisson bracket with the operator commutator like this: $$[a,b]->\frac{2\pi}{ih}(ab-ba).$$ This led to the uncertainty principle and the matrix formulation of quantum mechanics.

Schrodinger realized that he could form a differential equation for something called a wave function by substituting Heisenberg's operators into the classical Hamiltonian function. It took a while for the probability interpretation of the wave function to jell and to prove that the Schrodinger equation approach was completely equivalent to Heisenberg's matrix mechanics, but that in a nutshell is how quantum mechanics was born (pun intended). Lest I be down voted for slighting Bohr and the Old Quantum Theory (OQT), I should add that I consider OQT the gestation (not the birth) of quantum theory.

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Let's make things a little more fun.

  1. How can we use the conservation of energy equation to derive Schrödinger equation in QM?

Let's say we know the system's Hilbert space $\mathcal H$ and we know how to define a Hamiltonian $H:\mathcal H \rightarrow \mathcal H$ whose average value $\langle \psi | H | \psi\rangle$ provides the average energy in state $|\psi\rangle \in \mathcal H$.

We want an equation of evolution for $|\psi\rangle$ that satisfies the following simple conditions:

  • It conserves energy: $$ \frac{d}{dt}\langle \psi | H | \psi\rangle = \langle \dot\psi | H| \psi \rangle + \langle \psi | H | \dot\psi \rangle = 0 $$

  • It conserves probability: $$ \frac{d}{dt}\langle \psi | \psi\rangle = \langle \dot\psi | \psi \rangle + \langle \psi | \dot\psi \rangle = 0 $$

  • It does the above in a non-trivial way, such that $| \dot \psi \rangle \neq 0$, or equivalently, $\langle \dot \psi | \dot \psi \rangle > 0$. We may require in fact that $| \dot \psi \rangle$ be such that $\langle \dot \psi | \dot \psi \rangle > 0$ is maximal under the above constraints (the minimum is boring) and so demand the following variational equation: $$ \delta_{\dot\psi^*, \dot\psi} \left[ \langle \dot \psi | \dot \psi \rangle - \lambda \left( \langle \dot\psi | H| \psi \rangle + \langle \psi | H | \dot\psi \rangle \right) - \mu \left( \langle \dot\psi | \psi \rangle + \langle \psi | \dot\psi \rangle \right) \right] = 0 $$ where $\lambda$, $\mu$ are for the moment real variational parameters. Let us notice, however, that transformations of the form $|\dot\psi\rangle \rightarrow |\dot\psi\rangle - i\alpha H |\psi\rangle - i\beta |\psi\rangle$, for arbitrary real $\alpha$, $\beta$, leave both conservation constraints invariant, while changing the variational equation to $$ \delta_{\dot\psi, \dot\psi^*} \left[ \langle \dot \psi | \dot \psi \rangle - \lambda \langle \dot\psi | H| \psi \rangle - \lambda^* \langle \psi | H | \dot\psi \rangle - \mu \langle \dot\psi | \psi \rangle - \mu^* \langle \psi | \dot\psi \rangle \right] = 0 $$ where we set $\lambda + i \alpha \rightarrow \lambda$, $\mu + i \beta \rightarrow \mu$ and ignored terms not containing $\dot\psi$, $\dot\psi^*$ since they do not contribute to variations. Requiring that this variational equation also be left invariant by such transformations tells us that the correct form must be the above, with $\lambda$, $\mu$ complex parameters. Taking now the variations on $\langle \dot\psi |$, $|\dot\psi\rangle$ gives respectively $$ |\dot\psi\rangle = \lambda H |\psi\rangle + \mu |\psi\rangle \\ \langle \dot\psi | = \lambda^* \langle \psi | H + \mu^* \langle \psi | $$ The two conservation constraints then yield $$ Re\lambda \;\langle \psi | H | \psi\rangle + Re\mu \;\langle \psi | \psi\rangle = 0 \\ Re\lambda \;\langle \psi | H^2 | \psi\rangle + Re\mu \;\langle \psi | H | \psi\rangle = 0 $$ and therefore $$ Re\lambda = Re\mu = 0 $$ unless $\langle \psi | (\Delta H)^2 | \psi\rangle = 0$. This leaves us $$ |\dot\psi\rangle = i\; Im\lambda \;H |\psi\rangle + i\;Im\mu \;|\psi\rangle $$ or after absorbing $Im\mu$ as a phase factor ( $|\psi\rangle \rightarrow e^{iIm\mu t}|\psi\rangle$, assuming $Im\mu$ time independent; otherwise use the time integral in the exponent), $$ |\dot\psi\rangle = i \;Im\lambda \;H |\psi\rangle $$ At last, nature tells us to identify $i\; Im\lambda = \frac{1}{i\hbar}$, and behold the Schroedinger equation: $$ i\hbar|\dot\psi\rangle = H |\psi\rangle $$

    1. What is the validity of conservation of energy equation in QM?

Very valid in the sense that the average is conserved, as above. The answer by ACuriousMind covers the details in the larger context.

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