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Let $E_0$ be the ground state energy of $\phi^4$ Hamiltonian, namely \begin{equation} H:=\int d^{3} x\left[\frac{1}{2} \pi^{2}+\frac{1}{2}(\nabla \phi)^{2}+\frac{1}{2} m^{2} \phi^{2} + \lambda \phi^4\right], \end{equation} Is it possible to calculate $\lambda_0$ such that $|\lambda_0-E_0|<\epsilon$ ?

Also, assume the ground state of the hamiltonian is $|\lambda_0\rangle$. Is it possible to calculate a close form solution for ground state denoted by $|\phi_0\rangle$ of this hamiltonian such that $\langle \lambda_0|\phi_0\rangle \leq \epsilon'$ ? I would be thankful if you introduce some good resources that talk about these approximation. Particularly, I want to know what are the most accurate results on these two problems?

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  • $\begingroup$ Depends on epsilon. $\endgroup$ May 19, 2021 at 23:00
  • $\begingroup$ Can you please introduce some resources if any that discuss this? $\endgroup$
    – Heisenberg
    May 19, 2021 at 23:30
  • $\begingroup$ How small the $\epsilon$ can be? is perturbation theory the only options or there are other options out there? $\endgroup$
    – Heisenberg
    May 19, 2021 at 23:37
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    $\begingroup$ You should define your theory - both the quantization of what you call $\phi^4$ theory and your regularization at least - before your question can be answered. $\endgroup$ May 20, 2021 at 4:23
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    $\begingroup$ Usually in QFT the vacuum energy is zero by definition. $\endgroup$ May 20, 2021 at 4:55

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Energies are defined only up to a constant, so if you want to compute any given energy you need to specify the origin. The usual convention is to define the origin precisely at the ground state so, with this prescription, $E_0\equiv 0$, exactly. This has the advantage that the ground state becomes time-independent, which simplifies some formulas (like LSZ, otherwise you need to carry some factors of $e^{iE_0t}$ around).

If you want to put the origin somewhere else you need to redefine $H\mapsto H+\Delta$, where $\Delta$ is an arbitrary constant. With this redefinition, $E_0\equiv \Delta$, again exactly. But $\Delta$ is arbitrary, it is not physical nor computable. In more technical terms, it is regularization-scheme-dependent, so it is not measurable nor meaningful. (It is related to the renormalization of the partition function, i.e., the zero-point function $\langle 0|0\rangle=Z$. As $Z\sim e^{iHt}$, a linear shift in $H$ corresponds to a multiplicative shift in $Z$, and this shift is unobservable).

Note that the redefinition $H\mapsto H+\Delta$ is an unavoidable ambiguity in QFT because $H$ is a composite operator so it suffers ordering ambiguities. It is not something you can get rid of if you were clever enough.

If you want to define the origin of energies according to some other unspecified prescription, and you wish to compute $E_0$ for that choice, the algorithm is well-known. A particularly clean way is to compute the thermal partition function, i.e., you compactify the time-direction on a circle with radius $\beta$ and treat this direction as euclidean. This geometry computes $$ Z=\sum_{E}e^{-\beta E} $$ so the ground state energy can be extracted from the leading behaviour as $\beta\to+\infty$, where all excitations get exponentially suppressed.

So the computation of $E_0$ becomes the computation of the thermal partition function at large radius. This is not something that can be done exactly, but you can always resort to the usual methods, namely compute it in perturbation theory if possible, or put it on the lattice. I don't know any reference where this has been done explicitly for $\phi^4$ theory, precisely for the reason I described above: this is not a meaningful question -- its answer has no physical meaning.

(But of course $\phi^4_4$ might not even exist as a healthy QFT, so it is not clear whether the question makes sense even in principle. And if the theory does exist, the lattice might be entirely unsuitable to study it. The status of this theory in particular is kind of an open question. The ideas sketched above should work for similar theories when they are well-defined; whether they work for this theory specifically depends on details that are not yet quite understood.)

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  • $\begingroup$ Thank you for your reply. It makes perfect sense. How about the other question? Can we approximate the ground state within some rigorous error? $\endgroup$
    – Heisenberg
    May 23, 2021 at 9:29
  • $\begingroup$ Either using perturbation theory or lattice theory $\endgroup$
    – Heisenberg
    May 23, 2021 at 9:37
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Since the question is nonperturbative in nature, this strongly limits its applicability. For the $\phi^4$ model that means one can only consider the 2d and 3d versions. Also, these are models of QFT which require renormalization, and in particular for the $0$-point function, involving "subtracting an infinite quantity". So since, by definition $$ E_0= \text{wrong infinite value} - \infty\ , $$ the question as is does not even make sense. As mentioned in previous comments/answers, one can simply decide by decree that $E_0=0$. If one really wants to make sense of this, one would have to look at a family of Hamiltonians indexed by some parameter $s$, with a uniform way of doing the infinite subtraction. Then one would presumably get a $E_0(s)$ which may have been normalized by a condition such as $E_0(0)=0$ but would be worth computing for $s\neq 0$.

There are simpler nonrelativistic QFT models than $\phi^4$ where one has a rigorous nonperturbative definition of the ground state energy and a convergent expansion for it. By truncating this expansion you can get approximations like $\lambda_0$.

For the spin boson model this was done in my article "The Ground State Energy of the Massless Spin-Boson Model" in AHP 2011.

For the Nelson model a similar result was proved in my article with David Hasler "Analyticity of the Ground State Energy for Massless Nelson Models" in CMP 2012.

You can also look up papers citing the above two references in say Google Scholar and get an idea of what kind of bounds one can prove about ground state energies of many-body or field theoretic models.

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  • $\begingroup$ Im more interested in ground state rather than ground state energy. what do we know about ground state of phi to the 4 theory? any perturbative result on that? $\endgroup$
    – Heisenberg
    May 25, 2021 at 17:42
  • $\begingroup$ The question was about the ground state energy, now I see the goalposts have moved to the ground state itself. Well...it is the constant function equal to 1 in the Hilbert space $L^2(\mathscr{S}'(\mathbb{R}^{d-1}),d\nu)$ of physics.stackexchange.com/questions/633835/… $\endgroup$ May 25, 2021 at 18:11

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