终于完成了..。
代码如下：

public List<Collaborator> getCollaborators(Long answerId) {
  CriteriaBuilder cb = entityManager.getCriteriaBuilder();
  CriteriaQuery<Collaborator> criteriaQuery = cb.createQuery(Collaborator.class);
  
  Root<Answer> answerRoot = criteriaQuery.from(Answer.class);
  SetJoin<Answer, Collaborator> answers = answerRoot.join(Answer_.collaborators);
  criteriaQuery.where(cb.equal(answerRoot.get(Answer_.id), answerId));
  
  return entityManager
    .createQuery(criteriaQuery.select(answers))
    .getResultList();
}

使用HQL：
您可以使用以下命令：

Criteria criteria = session.createCriteria(Answer.class);
criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
criteria.createAlias("collaborators", "collaborators");
criteria.add(Restrictions.eq("collaborators.id",desiredCollaboratorId);

以获取与某个合作者相关的所有答案。
还有这个：

Criteria criteria = session.createCriteria(Answer.class);
criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
criteria.setFetchMode("collaborators", FetchMode.JOIN)
criteria.add(Restrictions.idEq(desiredAnswerId));
dsrTrackingCriteria.setProjection(Projections.property("collaborators"));

以使所有协作者与某个答案相关联。
使用JPA2 Criteria API，您可以执行以下操作：

CriteriaBuilder cb = em.getCriteriaBuilder(); //creted from EntityManager instance

CriteriaQuery<Long> cq = cb.createQuery(Collaborator.class);
Root<Answer> rootAnswer = cq.from(Answer.class);
Join<Collaborator,Answer> joinAnswerCollaborator = rootAnswer.join("collaborators"); //(or rootAnswer.join(Answer_.collaborators); if you've created the metamodel with JPA2

使用Criteria Builder：

Join<CLASS_A, CLASS_B> join = root.join(WHAT_UVE_DECLARED_IN_MAPPEDBY, JoinType.INNER);
searchCriteria.add(criteriaBuilder.like(join.get("FIELD_IN_SUBCLASS").as(String.class), "%blabla%"));

Join<Answer , Collaborator> join = root.join("collaborators",JoinType.INNER);

predicates.add(criteriaBuilder.equal(join.get("id"),id));