Why does decltype(declval

I was trying to detect the presence of a member function baz() in a template parameter:

template<typename T, typename = void>
struct ImplementsBaz : public std::false_type { };

template<typename T>
struct ImplementsBaz<T, decltype(&T::baz)> : public std::true_type { };

But it always produces false:

struct Foo {};
struct Bar { void baz() {} };

std::cout << ImplementsBaz<Foo>::value << std::endl;  // 0
std::cout << ImplementsBaz<Bar>::value << std::endl;  // also 0

Using declval and calling the method does work, though:

template<typename T>
struct ImplementsBaz<T, decltype(std::declval<T>().baz())> : public std::true_type { };

Of course, now this can only detect a baz function with 0 arguments. Why is the specialization correctly selected when using declval<T>().baz(), but not decltype(&T::baz)?