# Do Coherent States Imply 0 Energy Uncertainty?

• uxioq99
In summary: With this you can rewrite the energy equation$$\langle E \rangle = \hbar \omega (|\lambda|^2 + 1/2)$$as$$\langle \Psi_\lambda | a^\dagger a^\dagger a | \Psi_\lambda \rangle= (a^\dagger a | \Psi_\lambda \rangle)^\dagger (a^\dagger a | \Psi_\lambda \rangle)$$Which is just the energy equation in terms of the energy eigenstates.
uxioq99
Homework Statement
Consider the state ##\psi_\lambda = N e^{\lambda a^\dagger} \phi_0## where ##\phi_0## is the ground state of the harmonic well and ##a^\dagger##. What is the energy uncertainty ##\Delta E## of ##\psi_\lambda##?
Relevant Equations
##\psi_\lambda = N e^{\lambda a^\dagger} \phi_0##
##\Delta E = \sqrt{\langle E^2 \rangle - \langle E \rangle^2}##
By considering the power series for ##e^x##, I assert that ##N=e^{-\lambda^2/2}## and that ##a\Psi_\lambda = \lambda \Psi_\lambda##. Because the Hamiltonian may be written ##\hbar \omega(a^\dagger a + 1/2)##, ##\langle E \rangle = \hbar \omega(\langle a \Psi_\lambda, a \Psi_\lambda \rangle + 1/2)## by the definition of the adjoint. Then, ##\langle E \rangle = \hbar \omega (|\lambda|^2 + 1/2)##. Likewise, ##E^2 = \hbar^2 \omega^2((a^\dagger a)^2 + a^\dagger a + 1/4)##.

##
\begin{align}
\langle \Psi_\lambda | a^\dagger a a^\dagger a | \Psi_\lambda \rangle
&= (a^\dagger a | \Psi_\lambda \rangle)^\dagger (a^\dagger a | \Psi_\lambda \rangle) \\
&= |\lambda|^2 \langle a^\dagger \Psi_\lambda, a^\dagger \Psi_\lambda \rangle \\
&= |\lambda|^2 \langle a\Psi_\lambda, a\Psi_\lambda \rangle \\
&=|\lambda|^4 \langle \Psi_\lambda, \Psi_\lambda \rangle \\
\end{align}
##

Therefore, ##\langle E^2 \rangle = (|\lambda|^4 + |\lambda|^2 + 1/4)=\langle E \rangle^2##. Is it really possible that a coherent state has ##0## energy uncertainty? How would that not contradict the energy time uncertainty principle?

Last edited:
uxioq99 said:
Is it really possible that a coherent state has ##0## energy uncertainty? How would that not contradict the energy time uncertainty principle?

Can you state, precisely, the energy time uncertainty principle? As it applies to the harmonic oscillator?

topsquark
Coherent states are not eigenstates of the Hamiltonian and thus the energy does not take a determined value. I'd not bring in the energy-time uncertainty relation, which is subtle and has nothing to do with the simpler question about the energy uncertainty of a coherent state.

The most simple representation of the coherent state is in terms of the energy eigenstates ("Fock states"):
$$|\Phi(\alpha) \rangle=c_0 \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle, \quad c_0 = \exp(-|\alpha^2|/2), \quad \alpha \in \mathbb{C}.$$
From this you can get the probaility to find ##n## "phonons" by
$$P(n)=|\langle n|\Phi(\alpha)|^2,$$
and from this you can evaluate ##\langle n \rangle## and ##\Delta n^2## easily. Then you only need
$$\hat{H}=\hbar \omega (\hat{n}+1/2 \rangle)$$
to get ##\langle E \rangle## and ##\Delta E^2=\langle (E-\langle E \rangle)^2 \rangle.##

topsquark
uxioq99 said:
##
\begin{align}
\langle \Psi_\lambda | a^\dagger a a^\dagger a | \Psi_\lambda \rangle
&= (a^\dagger a | \Psi_\lambda \rangle)^\dagger (a^\dagger a | \Psi_\lambda \rangle) \nonumber\\
&= |\lambda|^2 \langle a^\dagger \Psi_\lambda, a^\dagger \Psi_\lambda \rangle \nonumber \\
&= |\lambda|^2 \langle a\Psi_\lambda, a\Psi_\lambda \rangle \nonumber\\
&=|\lambda|^4 \langle \Psi_\lambda, \Psi_\lambda \rangle \nonumber \\
\end{align}
##
Note ##\langle a^\dagger \Psi_\lambda, a^\dagger \Psi_\lambda \rangle \neq \langle a\Psi_\lambda, a\Psi_\lambda \rangle##

topsquark, uxioq99 and MatinSAR
@TSny Thank you, I forgot that they didn't commute. My brain was still operating in "elementary mode".

TSny
And also note
$$\hat{a}^{\dagger} |\Psi_{\lambda} \rangle \neq \lambda^* |\Psi_{\lambda} \rangle.$$
Note the INequality sign!

The trick in evaluating expectation values or matrix elements of operators between coherent states is to bring everything in normal ordering and then act with the creation operators on the bra and the annihilation operators on the ket. For this you only need
$$\hat{a} |\Psi_{\lambda} \rangle=\lambda |\Psi_{\lambda} \rangle.$$

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