# What is the value of a quantum field?

As far as I'm aware (please correct me if I'm wrong) quantum fields are simply operators, constructed from a linear combination of creation and annihilation operators, which are defined at every point in space. These then act on the vacuum or existing states. We have one of these fields for every type of particle we know. E.g. we have one electron field for all electrons. So what does it mean to say that a quantum field is real or complex valued. What do we have that takes a real or complex value? Is it the operator itself or the eigenvalue given back after it acts on a state? Similarly when we have fermion fields that are Grassmann valued what is it that we get that takes the form of a Grassmann number?

The original reason I considered this is that I read boson fields take real or complex values whilst fermionic fields take Grassmann variables as their values. But I was confused by what these values actually tell us.

-

## 1 Answer

When physicists say that a quantum field $\phi(x)$ is real-valued, they are usually referring to Feynman's path integral formulation of quantum field theory, which is equivalent to Schwinger's operator formulation.

The values of a field $\phi(x)$ in the path integral formulations are numbers. E.g.:

• If the numbers are real, we say that the field $\phi(x)$ is real-valued. (Such a field $\phi(x)$ typically corresponds to a Hermitian field operator $\hat{\phi}(x)$ in the operator formalism.)

• If the numbers are complex, we say that the field $\phi(x)$ is complex-valued.

• If the numbers are Grassmann-odd, we say that the field $\phi(x)$ is Grassmann-odd. (The numbers in this case are so-called supernumbers. See also this Phys.SE post.)

-
Thank you very much :). – Siraj R Khan Dec 31 '12 at 21:28

1
Incidentally, was my initial interpretation of a quantum field (as a linar combination of creation/annihilation operators at every point) correct as well, or are there subtleties that I'm missing? – Siraj R Khan Jan 1 at 3:01
You are essentially correct, though there are difficulties due to interactions in perturbation theory. Typically the fields in the interacting case have a nonzero amplitude to produce multi-particle states from the vacuum. You could think of it like this: the bare electron field operator creates a bare electron, which is a superposition of a real "dressed" electron and a bunch of other states. Renormalisation rescales the field so that the single-particle amplitude is simple, and you do extra things (LSZ reduction) to remove the multi-particle contributions from physical amplitudes. – Michael Brown Jan 1 at 3:13
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9247723817825317, "perplexity_flag": "head"}