In physics and mathematics, the Golden–Thompson inequality is a trace inequality between exponentials of symmetric and Hermitian matrices proved independently by Golden (1965) and Thompson (1965). It has been developed in the context of statistical mechanics, where it has come to have a particular significance.
Contents
Statement
The Golden–Thompson inequality states that for (real) symmetric or (complex) Hermitian matrices A and B, the following trace inequality holds:
tr
e
A
+
B
≤
tr
(
e
A
e
B
)
.
{\displaystyle \operatorname {tr} \,e^{A+B}\leq \operatorname {tr} \left(e^{A}e^{B}\right).}
This inequality is well defined, since the quantities on either side are real numbers. For the expression on the right hand side of the inequality, this can be seen by rewriting it as
tr
(
e
A
/
2
e
B
e
A
/
2
)
{\displaystyle \operatorname {tr} (e^{A/2}e^{B}e^{A/2})}
using the cyclic property of the trace.
Let
‖
⋅
‖
{\displaystyle \|\cdot \|}
denote the Frobenius norm, then the Golden–Thompson inequality is equivalently stated as
‖
e
A
+
B
‖
≤
‖
e
A
e
B
‖
.
{\displaystyle \|e^{A+B}\|\leq \|e^{A}e^{B}\|.}
Motivation
The Golden–Thompson inequality can be viewed as a generalization of a stronger statement for real numbers. If a and b are two real numbers, then the exponential of a+b is the product of the exponential of a with the exponential of b:
e
a
+
b
=
e
a
e
b
.
{\displaystyle e^{a+b}=e^{a}e^{b}.}
If we replace a and b with commuting matrices A and B, then the same inequality
e
A
+
B
=
e
A
e
B
{\displaystyle e^{A+B}=e^{A}e^{B}}
holds.
This relationship is not true if A and B do not commute. In fact, Petz (1994) proved that if A and B are two Hermitian matrices for which the Golden–Thompson inequality is verified as an equality, then the two matrices commute. The Golden–Thompson inequality shows that, even though
e
A
Generalizations
Other norms
In general, if A and B are Hermitian matrices and
‖
⋅
‖
{\displaystyle \|\cdot \|}
is a unitarily invariant norm, then (Bhatia 1997, Theorem IX.3.7)
‖
e
A
+
B
‖
≤
‖
e
A
e
B
‖
.
{\displaystyle \|e^{A+B}\|\leq \|e^{A}e^{B}\|.}
The standard Golden–Thompson inequality is a special case of the above inequality, where the norm is the Frobenius norm.
The general case is provable in the same way, since unitarily invariant norms also satisfy the Cauchy-Schwarz inequality. (Bhatia 1997, Exercise IV.2.7)
Indeed, for a slightly more general case, essentially the same proof applies. For each
p
≥
1
Multiple matrices
The inequality has been generalized to three matrices by Lieb (1973) and furthermore to any arbitrary number of Hermitian matrices by Sutter, Berta & Tomamichel (2016). A naive attempt at generalization does not work: the inequality
tr
(
e
A
+
B
+
C
)
≤
|
tr
(
e
A
e
B
e
C
)
|
{\displaystyle \operatorname {tr} (e^{A+B+C})\leq |\operatorname {tr} (e^{A}e^{B}e^{C})|}
is false. For three matrices, the correct generalization takes the following form:
tr
e
A
+
B