# Doubt about fifth postulate of QM degenerate case

I think I didn't not really understand a comment found on "Quantum Mechanics" by Claude Cohen-Tannoudji.

Talking about the fifth postulate of quantum mechanics in the case in which $a_n$ is a degenerate eigenvalue is said

"We stress the fact, however, that is not an arbitrary ket of the eigensubspace $E_n$, but the part of $| \psi \rangle$ which belongs to $E_n$".

I did not understand very well probably. In that case I would say that immediately after the measure of $a_n$ the state of the system is a linear combination of the bases of $E_n$, eigensubspace related to $a_n$.

Fifth postulate:

Immediately after the measurement of an observable $A$ has yielded a value $a_n$, the state of the system is the normalized eigenstate $| u_n \rangle$.

• There is no standard way of numbering the postulates of QM. Could you include the "5th postulate" in your question? May 22, 2018 at 12:34
• I'm sorry, I added it. May 22, 2018 at 12:40
• You need to define the notation. Nobody can tell what this is about based on these two very brief snippets.
– user4552
May 22, 2018 at 13:07
• For further reference @Stefano Barone, please use "von Neumann projection postulate" or the "collapse postulate" in Quantum Mechanical slang. May 22, 2018 at 14:16
• What is your question? May 26, 2018 at 17:53

Let $\mathscr{H}$ be the relevant Hilbert space and $\Psi\in \mathscr{H}$. Let furthermore, $P : \mathscr{H}\to \mathscr{H}$ be a projection operator onto some subspace. It is a fact that $P$ defines the complementary projection $Q : \mathscr{H}\to \mathscr{H}$ given by $Q = \mathbf{1} - P$ where $\mathbf{1}$ is the Hilbert space identity transformation.

Notice that by definition a projection satisfies $P^2 = P$. It turns out that $Q$ is indeed a projection operator. We have the proof:

$$Q^2=(1-P)^2=1-2P+P^2=1-2P+P=1-P=Q,$$

where we've used that $P$ is a projection. Another fact is that $PQ\Psi =0$. This is true because we have

$$P(Q\Psi)=P(1-P)\Psi=P\Psi-P^2\Psi=P\Psi-P\Psi=0.$$

Similarly $QP\Psi = 0$ since $[P,Q]=0$.

But if $P$ and $Q$ are projections, they are projections onto where? Well, we can define

$$\mathscr{B}_p=P(\mathscr{H})$$

$$\mathscr{B}_Q=P(\mathscr{H}).$$

These are the subspaces of $\mathscr{H}$ onto which the projectors project. They are not independent however, since these are complementary projections. We have the following result:

$$\mathscr{H}=\mathscr{B}_P\oplus \mathscr{B}_Q.$$

Indeed, by definition of $Q$ we have that $\mathbf{1} = P+Q$ and hence any $\Psi$ is

$$\Psi=P\Psi+Q\Psi.$$

The decomposition is unique, however. Indeed if $\Psi=\Psi_P+\Psi_Q$ with $\Psi_P\in \mathscr{B}_P$ and $\Psi_Q\in \mathscr{B}_Q$ then since $PQ\Psi=QP\Psi=0$ by applying $P$ and $Q$ to the decomposition it follows $\Psi_P = P\Psi$ and $\Psi_Q = Q\Psi$.

This indeed shows that the decomposition is a direct sum decomposition.

Why bother with all this? Well, if $A$ is an observable, corresponding to each eigenvalue there is an eigenspace. The eigenspace corresponding to $\lambda$ has a projector $P_\lambda$ associated to it. Then $Q_\lambda = 1-P_\lambda$ projects out of it.

Thus, all of this allow us to say that a state $\Psi$ may have a "part of it" contained in the eigenspace of $A$ with eigenvalue $\lambda$ and this part is $P_\lambda \Psi$.

The properties of a projection are what allow us to say that $P_\lambda \Psi$ is its part inside that eigenspace.

What the postulate says is that immediately after the measurement the state is $$\frac{P_\lambda \Psi}{\sqrt{(P_\lambda \Psi, P_\lambda \Psi)}}$$

since this projection won't usually be normalized. In summary what is says is: the system doesn't collapse to an arbitrary element of $P_\lambda \mathscr{H}$ but rather to specificaly the part of $\Psi$ lying there properly normalized, which is given by the above formula.