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So, I'm reading through Polyakov's paper from 1974, "Particle Spectrum in Quantum Field Theory." I'm trying to work through all of the steps and properly understand everything. For context, this is one of the papers along with 't Hooft's that the name 't Hooft-Polyakov monopoles came from. The paper starts out with the interacting scalar Hamiltonian $$H = \frac{1}{2} \int_{-\infty}^{\infty} dx \left[ \pi^2 + \left( \frac{\partial\phi}{\partial x}\right)^2 -\mu^2 \phi^2 + \frac{\lambda}{2}\phi^4\right]. $$ There's an extremal of the potential energy solved by $$\phi_c''+\mu^2\phi_c -\lambda\phi_c^3=0.$$ with boundary conditions $$\phi_c^2(\pm \infty)=\mu^2/\lambda.$$ This equation then has solution $$\phi_c = \frac{\mu}{\sqrt{\lambda}} \tanh{\left( \frac{\mu x}{\sqrt{2}} \right)}.$$ Then what we want to do is calculate the vibrational modes surrounding this solution by making the substitution $\phi \rightarrow \phi_c + \phi$. Now this is the part where I've gotten lost. He claims that by

Diagonalizing the obtained quadratic Hamiltonian, we obtain the mass spectrum $$ M_n = \frac{2\sqrt{2}}{3} \frac{\mu^3}{\lambda} + \frac{\sqrt{3} + 2}{2\sqrt{2}}\mu + \frac{\mu}{\sqrt{2}}\sqrt{n(4-n)} $$ for $n=0,1,2.$

In the paper you're supposed to ignore the $\phi^3$ and $\phi^4$ terms that come out of the substitution. This is easy enough to do. I get that, $$ H = \frac{1}{2} \int_{-\infty}^{\infty} dx \left[ \pi^2 + \left( \frac{\partial\phi}{\partial x}\right)^2 + 2 \left( \frac{\partial\phi_c}{\partial x}\right) \left( \frac{\partial\phi}{\partial x}\right)+ \left( \frac{\partial\phi_c}{\partial x}\right)^2 +\phi^2 \left( 3\lambda \phi_c^2 -\mu^2\right) + \phi \left( 2\lambda\phi_c^3 -\mu^2 \phi_c \right) +\left(\frac{\lambda}{2}\phi_c^4 -\mu^2 \phi_c^2 \right) \right]. $$ But I'm lost on how to do this diagonalization. All of the stackexchange posts on here that I've looked at for diagonalizing a Hamiltonian start out with it in terms of creation/annihilation operators. So can someone help diagonalizing the modified Hamiltonian to find the mass spectrum?

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  • $\begingroup$ The article is here. If you look above (1.26) in this you'll see the first term, then in (1.35) and (1.36) you'll see the first term and half of the second term, so this may be an indication of how to approach things (terms in there seems to go wrong after this...). $\endgroup$ – bolbteppa Oct 25 '20 at 0:33
  • $\begingroup$ Unfortunately it seems that your link is broken. Do I need to use a VPN or something? $\endgroup$ – PaulPhy Oct 25 '20 at 4:35
  • $\begingroup$ However, the paper by Rajantie's student looks excellent. Let me take a look through tomorrow and see if that clarifies it for me $\endgroup$ – PaulPhy Oct 25 '20 at 6:05
  • $\begingroup$ The start of his paper "Isomeric States of Quantum Fields" (pdf here: www.jetp.ac.ru/cgi-bin/dn/e_041_06_0988.pdf ) seems to explain in more detail what he means in his Particle Spectrum paper (www.jetpletters.ac.ru/ps/1789/article_27297.shtml), e.g. what he means by mass spectrum, (the https at the start of the first link I sent above is why it doesn't work, removing it makes it work). If you end up working out equation (4) for $M_n$ explicitly please post it here. $\endgroup$ – bolbteppa Dec 5 '20 at 19:01
  • $\begingroup$ So, is phi_c to be treated as "a constant" with regards to the action principle thereafter? $\endgroup$ – daydreamer Dec 5 '20 at 19:39
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can someone point me in the right direction at least to diagonalize the modified Hamiltonian?

This answer explains how to diagonalize the Hamiltonian and how to derive the $n$-dependent term in $M_n$. That's the interesting part, because the $n$-independent part is just an overall constant term in the Hamiltonian, which has no observable consequences in a non-gravitational theory.

Eliminating the linear terms

Sometimes a little extra generality can help. Start with $$ H = \frac{1}{2}\int dx\ H(x) \tag{1} $$ where $$ \newcommand{\pl}{\partial} H(x) = \pi^2+(\phi')^2+V(\phi) \tag{2} $$ with $\phi'\equiv\pl\phi/\pl x$. Suppose $\phi_c$ solves $$ 2\phi_c'' - V'(\phi_c)=0 \tag{3} $$ with $V'\equiv \pl V/\pl\phi$. Replace $\phi\to\phi_c+\phi$ in (2) to get $$ H(x) = \pi^2+(\phi')^2+2\phi'\phi_c'+V'(\phi_c)\phi +\frac{V''(\phi_c)}{2}\phi^2 + \cdots \tag{4} $$ where the omitted terms are either independent of $\phi$ or are of order $\phi^3$ and higher. After subsituting this back into (1), we can integrate-by-parts to convert the term $2\phi'\phi_c'$ to $-2\phi''\phi_c$. Then equation (3) says that the terms linear in $\phi$ add up to zero, which leaves $$ H = \frac{1}{2}\int dx\ \left(\pi^2+(\phi')^2 +\frac{V''(\phi_c)}{2}\phi^2 + \cdots\right). \tag{5} $$ As before, the terms represented by "$\cdots$" are either independent of $\phi$ or are of order $\phi^3$ and higher.

The $x$-dependent "mass" term

The Hamiltonian (5) looks like a harmonic oscillator but with an $x$-dependent coefficient $$ \frac{V''(\phi_c)}{2} \tag{6} $$ for the $\phi^2$ term (the "mass" term). The function $V(\phi)$ in the question is $$ V(\phi)=-\mu^2\phi^2+\frac{\lambda}{2}\phi^4, \tag{7} $$ which gives $$ \frac{V''(\phi_c)}{2}=-\mu^2+3\lambda\phi_c^2. \tag{8} $$ The question also provides this expression for $\phi_c$: $$ \phi_c=\frac{\mu}{\sqrt{\lambda}} \tanh\left(\frac{\mu x}{\sqrt{2}}\right). \tag{9} $$ Use (9) in (8) to get $$ \frac{V''(\phi_c)}{2} =\mu^2\left(3\tanh^2\left(\frac{\mu x}{\sqrt{2}}\right)-1\right). \tag{10} $$ The goal is to diagonalize the Hamiltonian (5) with $V''(\phi_c)$ given by (10).

Diagonalization

Here's a summary of what we've derived so far: after omitting higher-than-quadratic terms, the Hamiltonian is $$ H = \frac{1}{2}\int dx\ \left(\pi^2+(\phi')^2 +f(x)\phi^2\right) \tag{5b} $$ with $$ f(x) =\mu^2\left(3\tanh^2\left(\frac{\mu x}{\sqrt{2}}\right)-1\right). \tag{10b} $$ Integrate-by-parts to put (5b) in the form $$ H = \frac{1}{2}\int dx\ \Big(\pi^2+\big(-\phi'' +f(x)\phi\big)\phi\Big). \tag{11} $$ This shows that we can finish diagonalizing the Hamiltonian by solving the eigenvalue equation $$ -\phi''+f(x)\phi=E\phi, \tag{12} $$ because solutions of (12) with different eigenvalues $E$ are mutually orthogonal.

Some intuition about the spectrum

The function $f(x)$ approaches a positive constant $2\mu^2$ as $x\to\pm\infty$ but dips down to $-\mu^2$ at $x=0$. Since (12) has the familiar form of a time-independent Schrödinger equation for a particle in the potential well $f(x)$, we expect that equation (12) will have a finite number of solutions with discrete eigenvalues. These eigenvalues are determined below.

Calculation of the spectrum

To save some writing, define $y\equiv \mu x / \sqrt{2}$ so that equation (12) becomes $$ \newcommand{\pl}{\partial} (-\pl_y^2+6\tanh^2 y-2)\phi=2(E/\mu^2)\phi. \tag{13} $$ To save even more writing, define $t\equiv \tanh y$. The identity $$ -\pl_y^2+6t^2-2 = (-\pl_y+2t)(\pl_y+2t) \tag{14} $$ immediately gives us one eigenfunction, namely the function $$ \phi_2(y) \propto \exp\left(-\int_0^y dz\ \tanh(z)\right), \tag{15} $$ which is annihilated by $\pl_y+2t$ and which approaches $0$ for $y\to\pm\infty$. The corresponding eigenvalue of (14) is $0$. Use this in (13) to get $E=0$. We can generate more eigenvalues using this identity: $$ (-\pl_y+nt)(\pl_y+nt)+n = \big(\pl_y+(n+1)t\big)\big(-\pl_y+(n+1)t\big)-(n+1), \tag{16} $$ which holds for every integer $n$. To use this, let $\phi_n$ be the function annihilated by $\pl_y+nt$. Using (16), we can verify that $$ (-\pl_y+2t)\phi_1 \tag{17} $$ is an eigenfunction of (14) with eigenvalue $3$. Use this in (13) to get $E=(3/2)\mu^2$. Again using (16), we can verify that $$ (-\pl_y+2t)(-\pl_y+t)\phi_0 \tag{18} $$ is an eigenfunction of (14) with eigenvalue $4$. Use this in (13) to get $E=2\mu^2$. This pattern cannot be continued, because the only function annihilated by $\pl_y+0t=\pl_y$ is a constant function, which is not normalizable unless it is zero. Thus we have found exactly three discrete values for $E$, namely $$ E = 0,\ \frac{3}{2}\mu^2,\ 2\mu^2, \tag{19} $$ which can be summarized as $$ E = \frac{n(4-n)}{2}\mu^2 \tag{20} $$ with $n=0,1,2$. For each of these values of $E$, equation (11) gives a corresponding harmonic-oscillator mode with energy $\sqrt{E}$. This reproduces the $n$-dependent term in $M_n$ shown in the question, which is the interesting part.

References

This general approach to diagonalizing the Hamiltonian is consistent with section 2 of ref 1. The approach I used to solve equation (12) with the potential (10b) was inspired by sections 2 and 3 in ref 2. The same equation is also treated in problem 39 in ref 3, starting on page 94. For more information about solvable special cases of the one-dimensional Schrödinger equation, see refs 4 and 5.

  1. Polyakov (1976), "Isomeric states of quantum fields," Sov. Phys. JETP, 41:988-995 (http://jetp.ac.ru/cgi-bin/dn/e_041_06_0988.pdf)

  2. Bougie et al (2012), "Supersymmetric Quantum Mechanics and Solvable Models," Symmetry 4:452-473 (https://www.mdpi.com/2073-8994/4/3/452/htm)

  3. Flügge (1999), Practical Quantum Mechanics, Springer (https://books.google.com/books?id=VpggN9qIFUcC&pg=PA94)

  4. Alvarez-Castillo (2007), "Exactly Solvable Potentials and Romanovski Polynomials in Quantum Mechanics" (https://arxiv.org/abs/0808.1642)

  5. Dereziński and Wrochna (2010), "Exactly solvable Schrödinger operators" (https://arxiv.org/abs/1009.0541)

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