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For a potential of anharmonic oscillator like this: $$V= \frac{1}{2}m \omega^2 x^2+\frac{\lambda}{4}x^4$$

Can eigenfunction and eigenvalue of Hamiltonian be solved non-perturbatively that is analytically? Because I heard the $(0+1)$-dim $\phi^4$ theory, which is this case, can be solved exactly. We can calculate the generating function for this theory analytically. However how to solve the eigenvalue and eigenfunction.

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  • $\begingroup$ Liverts et al. exactly numerically--I hope you have no confusions about analytical expectations; but no pert theory. What are you going to do with eigenvalues and eigenfunctionals? Why give up on covariance? What, exactly, are you up to? $\endgroup$ – Cosmas Zachos Nov 1 '16 at 14:11
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    $\begingroup$ I think a possible source of confusion here is that what you described is not $0$-dim $\phi^4$ QFT. By $0$-dim QFT physicists usually mean a simplified model with no space and no time, where path integrals are just ordinary integrals. What you described is a $1$-dim QFT actually, since it has a single time axis, but I would suggest to just use "quartic oscillator" as it is the best name for the model. $\endgroup$ – Prof. Legolasov Nov 1 '16 at 15:23
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Let's clearify serval important concepts and misunderstandings.

$\bullet$ QM can be viewed as $(0+1)$D QFT no matter non-relativisitc or relativistic. In fact, it is a general point of view that a (d+1) dimensional field $\phi$ is a mapping $(t,x_1,...,x_n)\to\phi(t,x_1,...,x_n)$, among which the trajectory of particle is a mapping $t\to \big(x_1(t),...,x_n(t)\big)$. Therefore $(0+1)$D field is actually $t\to\phi(t)$ with corresponding particle trajectory picture is a mapping $(1+1)$D $t\to x(t)$. Such equivalence works in classical field and mechanics as well: a incompressible fluid field (speed field) under Lagrange formalism is also a mapping $(t,x,y,z)\to (v_1(t,x,y,z),v_2(t,x,y,z),v_3(t,x,y,z))$, while in classical mechanics particle trajectory is $t\to (v_1(t),v_2(t),v_3(t))$, which is effectively a $(0+1)$D fluid.

Taking $(0+1)$D free scalar boson field as example, it is, to the matter of fact, exactly the quantum harmonic oscillator whose hamiltonian is written as $$H_0=\frac{1}{2}\dot{\phi}^2+\frac{1}{2}m^2\phi^2$$ and hence the calssical equation of motion is $(\partial_t^2+m^2)\phi=0$. It worths emphasising that in order to avoid causing confusion, I do not use $x$ symbol since there are people considering the $x(t)$ is coordinate variable which is rather a field strength (please rember the physical picture of oscillator). Under particle number representation, the field is quantized in a familiar form $\phi=\sqrt{\frac{1}{2m}}(a+a^\dagger)$. In interaction picture it becomes $$\phi_I(t)=e^{iH_0t}\phi_S e^{-iH_0t}=\sqrt{\frac{1}{2m}}(a e^{-imt}+a^\dagger e^{imt})$$ ($m$ mass gap is known as $\omega$ in QM description of the harmonic oscillator).

Thereafter the so called two-point correlation function can be written as usual $$G_0(t_1-t_2)=\langle 0|T[\phi_I(t_1)\phi_I(t_2)]|0\rangle$$ where the time-ordering yields $$T[\phi_I(t_1)\phi_I(t_2)]=\theta(t_1-t_2)\phi_I(t_1)\phi_I(t_2)+\theta(t_2-t_1)\phi_I(t_2)\phi_I(t_1).$$ According to thr equation of motion $(\partial_t^2+m^2)G_0(t)=-i\delta(t)$, it is easy to find $$G_0(t_1-t_2)=\frac{1}{2m}e^{-i(m-i0^+)|t_1-t_2|}$$ and in frequency domain $$G_0(t)=\frac{1}{2\pi}\int\widetilde{G}_0(\omega) e^{-i\omega t}d\omega=\frac{1}{2m}e^{-i(m-i0^+)|t|},$$ it is the propagator of the field $$\widetilde{G}_0(\omega)=\int G_0(t) e^{i\omega t}dt=\frac{i}{\omega^2-m^2+i0^+}.$$

So generally speaking, the n-point correlation function is $$\langle \phi(t_1)...\phi(t_{N-1})\phi(t_N)\rangle.$$

In path integral form, the generating functional (or sometime phrased as partition function) is $$Z=\int\mathcal{D}\phi(t)\exp\Big(i\int dt\frac{1}{2}[\dot{\phi}^2-m^2\phi^2]\Big),$$ two-point correlation function can therefore be generated as $$G(t_n,t_m)=\langle\phi(t_n)\phi(t_m)\rangle=\frac{\int\mathcal{D}\phi(t)\phi(t_n)\phi(t_m)e^{iS}}{\int\mathcal{D}\phi(t)e^{iS}}.$$ Through Wick theorem, n-point correlation can bedecopoed into mutiplication structure of two-point correlation $\langle \phi^2n\rangle\to\prod_n\langle\phi^2\rangle$: $$\langle\phi^2\rangle=\frac{\int d\phi\,\phi^2\,e^{-\frac{1}{2}\phi G^{-1}\phi}}{\int d\phi\,e^{-\frac{1}{2}\phi G^{-1}\phi}}=G.$$

Fermion field can also be treated in same way which leads to relativstic Dirac QM.

reference Quantum field theory in $(0 + 1)$ dimensions

$\bullet$ The question which you have described on the quartic oscillator is not about $0$-dimensional scalar field, instead, as talked above, is a $(0+1)$-dimensional field. Such anharmonic oscillator is normaly handled with perturbation theory (The anharmonic oscillator in quantum mechanics). On the exact solution discussion you could refer to Quasi-Exactly Solvable Models in Quantum Mechanics and On the Exact Solution of the Schrödinger Equation with a Quartic Anharmonicity .

What treated as a $0$-dimensional analog of the $\phi^4$ theory, is something like $H=\phi^2+g\phi^4$ in which the field $\phi$ is located merely in $0$-dimensional space with neither spatial nor time configuration. And the genrating functional $Z(g)=\int_{-\infty}^{+\infty}d\phi\,e^{-\phi^2-g\phi^4}$ can be calculated explictly $Z(g)=\frac{e^{1/8g}}{2\sqrt{g}}K_{1/4}(1/8g)$ where $K_\alpha$ is the second modied Bessel function ($0$-dimensional QFT).

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  • $\begingroup$ Yes, I knew how to calculate the generating function analytically. Does it imply that I can solve the spectrum analytically? Or at least ground state's energy. $\endgroup$ – 346699 Nov 2 '16 at 4:20
  • $\begingroup$ No you don't solve any eigenstates or eigenvalues out from such $0$-dimensional analog model like oscillator in QM does, beacuse it is not a QM problem. The reason I have already told you above. $\endgroup$ – Tom Gao Nov 2 '16 at 6:38
  • $\begingroup$ On another hand, if you consider the anhramonic quartic oscillator, then I think normally you spare no work on calculating the particion function with functional integral, and it is mathemtical equivalent to solve it in Schödinger's equation. I also gave you materials on it already. $\endgroup$ – Tom Gao Nov 2 '16 at 6:51
  • $\begingroup$ Hi! I would like to rate your answer, but unfortunately it is hard to read :( Could you please put your latex in double dollars instead of single dollars in order to forbid the markup system to layout your formulas as inline? $\endgroup$ – Prof. Legolasov Nov 2 '16 at 13:47
  • $\begingroup$ @Solenodon Paradoxus Sure, I now have altered it. $\endgroup$ – Tom Gao Nov 2 '16 at 19:18

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