Quantum Harmonic Oscillator

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The following discussion of the quantum harmonic oscillator relies on the article Mathematical Formulation Of Quantum Mechanics . ONE-DIMENSIONAL HARMONIC OSCILLATOR Hamiltonian and energy Eigenstate s In the one-dimensional harmonic oscillator problem, a particle of mass ''m'' is subject to a potential ''V''(''x'') = (1/2)''m''ω2 ''x''2. The Hamiltonian of the particle is: : where ''x'' is the Position operator, and ''p'' is the Momentum operator ($p\; =\; -i\; \backslash hbar\; \{\backslash partial\; \backslash over\; \backslash partial\; x\}$). The first term represents the kinetic energy of the particle, and the second term represents the potential energy in which it resides. In order to find the Energy levels and the corresponding energy eigenstates, we must solve the time-independent Schrödinger Equation ,

:$n\; =\; 0,\; 1,\; 2,\; \backslash ldots$

The first six solutions (''n'' = 0 to 5) are shown on the right. The functions $H\_n$ are the Hermite Polynomials :

:

They should not be confused with the Hamiltonian, which is also denoted by ''H''. The corresponding energy levels are

:$E\_n\; =\; \backslash hbar\; \backslash omega\; \backslash left(n\; +\; \{1\backslash over\; 2\}\; ight)$.

This energy spectrum is noteworthy for two reasons. Firstly, the energies are "quantized", and may only take the discrete values of $\backslash hbar\backslash omega$ times 1/2, 3/2, 5/2, and so forth. This is a feature of many quantum mechanical systems. In the following section on ladder operators, we will engage in a more detailed examination of this phenomenon. Secondly, the lowest achievable energy is not zero, but $\backslash hbar\backslash omega/2$, which is called the "ground state energy" or Zero-point Energy . It is not obvious that this is significant, because normally the zero of energy is not a physically meaningful quantity, only differences in energies. Nevertheless, the ground state energy has many implications, particularly in Quantum Gravity .

Note that the ground state probability density is concentrated at the origin. This means the particle spends most of its time at the bottom of the potential well, as we would expect for a state with little energy. As the energy increases, the probability density becomes concentrated at the "classical turning points", where the state's energy coincides with the potential energy. This is consistent with the classical harmonic oscillator, in which the particle spends most of its time (and is therefore most likely to be found) at the turning points, where it is the slowest. The Correspondence Principle is thus satisfied.


Ladder operator method

The power series solution, though straightforward, is rather tedious. The "ladder operator" method, due to Paul Dirac , allows us to extract the energy eigenvalues without directly solving the differential equation. Furthermore, it is readily generalizable to more complicated problems, notably in Quantum Field Theory . Following this approach, we define the operators ''a'' and its Adjoint ''a''^†

:$\backslash begin\{matrix\}$
a &=& \sqrt{m\omega \over 2\hbar} \left(x + {i \over m \omega} p ight) \
a^{\dagger} &=& \sqrt{m \omega \over 2\hbar} \left( x - {i \over m \omega} p ight)
\end{matrix}

The operator ''a'' is not Hermitian since it and its adjoint ''a''^† are not equal.

In deriving the form of ''a''^†, we have used the fact that the operators x and p, which represent observables, ''are'' Hermitian. These observable operators can be expressed as a linear combination of the ladder operators as

:$\backslash begin\{matrix\}$
x &=& \sqrt{\hbar \over 2m\omega} \left( a^{\dagger} + a ight) \

:<math>\left(a \left\psi E Ight Angle, A \left\psi E Ight Angle Ight) \left\langle\psi_E ight a^\dagger a \left \psi_E ight angle \ge 0</math>   :<math>\left\langle\psi E Ight {H \over \hbar \omega} - {1 \over 2} \left\psi E Ight Angle \left({E \over \hbar \omega} - {1 \over 2} ight) \ge 0</math>,   So That <math>E \ge \hbar \omega / 2</math> Note That When (<math>a \left \psi E Ight Angle</math>) Is The Zero Ket (ie A Ket With Length Zero), The Inequality Is Saturated, So That <math>E \hbar \omega / 2</math> It is straightforward to check that there exists a state satisfying this condition it is the ground (''n'' = 0) state given in the preceding section   & & (\left[H,a ight] + a H) \left\psi_E ight angle \   & & (- \hbar\omega a + a E) \left\psi_E ight angle \   & & (E - \hbar\omega) (a\left\psi_E ight angle)   :<math>H (a^\dagger \left \psi E Ight Angle) (E + \hbar\omega) (a^\dagger \left \psi_E ight angle)</math>   Given Any Energy Eigenstate, We Can Act On It With The Lowering Operator, ''a'', To Produce Another Eigenstate With <math>\hbar \omega</math>, Less Energy By Repeated Application Of The Lowering Operator, It Seems That We Can Produce Energy Eigenstates Down To ''E'' &minus&infin However, this would contradict our earlier requirement that <math>E \ge \hbar \omega / 2</math> Therefore, there must be a ground-state energy eigenstate, which we label <math>\left 0 ight angle</math> (not to be confused with the zero ket), such that   :<math>a \left 0 Ight Angle 0 \hbox{(zero ket)}</math>   :<math>H \left0 Ight Angle (\hbar\omega/2) \left0 ight angle</math>   :<math> H \leftn Ight Angle \hbar\omega (n + 1/2) \leftn ight angle </math>

  \prod_{i=1}^N\langle x_i\psi_{n_i} angle
  :<math> \Delta E^{(2)} \lambda^2 \left\langle \psi_E ight x^3 {1 \over E - H_0} x^3 \left \psi_E ight angle </math>