perfect bayesian nash equilibrium is simply nash equilibrium Is it true that for two player zero sum game, Perfect Bayesian Nash equilibrium is simply Nash Equilibrium?

I am learning game theory and our lecturer does not explicitly cover it.
The quick answer is Yes.

A Bayesian Nash equilibrium (BNE) is Nash equilibrium (NE) generalized to situations where there is uncertainty over players' payoffs; that is, games with incomplete information. For example, in the following zero-sum game, $t$ is a "state variable" that may take values $\epsilon&gt;0$ or $-\epsilon$, each with probability $1/2$, say. Then the equilibrium you solve for would be a BNE.

On the other hand, if $t$ is not a random variable (or a degenerate random variable, as the jargons would have it), e.g. $t=0$ with probability $1$, then the equilibrium would simply be a NE.

In other words, a NE (of a complete information game) is a special case of a BNE (of a game of incomplete information, where there is only one payoff-relevant state, so to speak).

$$
\begin{array}{|c|c|c|}\hline
  &amp; H &amp; T \\\hline
H &amp; 1+t,-1+t &amp; -1+t, 1+t \\\hline
T &amp; -1,1 &amp; 1,-1\\\hline
\end{array}
$$

Adding perfection to BNE puts more restrictions on the players' off equilibrium beliefs about their opponents' strategies. In the case of matching pennies (the game above with $t=0$), which I assume you have in mind when saying "zero-sum games", this consideration wouldn't matter.