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When an observable/selfadjoint operator $\hat{A}$ has only discrete eigenvalues, the eigenvectors are orthogonal each other. Similarly, when an observable $\hat{A}$ has only continuous eigenvalues, the eigenvectors are orthogonal each other.

But what if $\hat{A}$ has both of discrete eigenvalues and continuous ones?


CONTEXT:

According to my teacher, an observable $\hat{A}$ can have discrete eigenvalues and continuous ones simultaneously.

$$\hat{A} |n\rangle = \alpha _n |n\rangle$$

$$\hat{A} |\xi\rangle = \xi |\xi\rangle$$

Completeness is this.

$$\sum _n |n\rangle\langle n| + \int d\xi |\xi\rangle\langle\xi| = \hat{1}$$

Now the following are true.

$$\langle n|m\rangle = \delta _{nm}$$

$$\langle \xi | \xi '\rangle = \delta (\xi - \xi ')$$


QUESTION:

Are $|n\rangle$ and $|\xi\rangle$ orthogonal each other? I mean, is the equation below true?

$$\langle n|\xi\rangle = 0$$

I want this to be true. I have to use this to prove the expectation value formula $$E[A] = \frac{\langle \psi|\hat{A}|\psi\rangle}{\langle \psi|psi\rangle}.$$

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  • $\begingroup$ The case of continuous eigenvalues already includes the case of both discrete and continuous eigenvalues. $\endgroup$ – mastrok Apr 24 '17 at 6:05
  • $\begingroup$ A sidenote to this discussion is that there is freedom in choosing the eigenvectors from a degenerate subspace. They can be chosen either way, though the practical advantage lies with choosing them orthogonal. $\endgroup$ – dmckee Apr 24 '17 at 6:05
  • $\begingroup$ @mastrok Thank you for your comment. It certainly seems to be true, come to think of it. $\endgroup$ – ynn Apr 24 '17 at 6:12
  • $\begingroup$ @dmckee Thank you for your comment. It seems a bit difficult for me, but it would help me for further understanding :) $\endgroup$ – ynn Apr 24 '17 at 6:12
  • $\begingroup$ Say you have exactly two eigenvectors $|a_i\rangle$ and $|a_j\rangle$ with the same eigenvalue $a$. Then $|a_k\rangle = (|a_i\rangle + |a_j\rangle)/\sqrt{2}$ is another vector with eigenvlaue $a$ (check me on that). Now, any two of $|a_i\rangle$, $|a_j\rangle$, and $|a_k\rangle$ (or indeed any other linear combination of $|a_i\rangle$ and $|a_j\rangle$) form a spanning set for the subspace containing the three vectors and can be selected as a basis for that sub-space. But there is no win in choosing a set that is not orthogonal. $\endgroup$ – dmckee Apr 24 '17 at 6:17
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You need to formalize the notion of discrete/continuous. If we assume that this is a well defined property of the system then there must exist an observable $D$ that has the same eigenstates as $A$ with eigenvalues $0$ for discrete eigenstates and $1$ for continuous eigenstates. You can then prove that a discrete eigenstate $\left|n\right>$ and a continuous eigenstate $\left|\xi\right>$ are orthogonal when $n = \xi$ (otherwise with different eigenvalues we would already know that they have to be orthogonal), using the fact the eigenvalues of $D$ of these states are different.

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  • $\begingroup$ Thank you for your answer. Is "are orthogonal when n = ξ" a mistype? I thought it would be "are orthogonal when n ≠ ξ". $\endgroup$ – ynn Apr 24 '17 at 6:00
  • $\begingroup$ @ynn If the two eigenvectors are different then it's trivial that they are orthogonal. This follows from computing $\left<\xi\right|A\left|n\right>$ by letting $A$ act on the ket and the bra which have to yield the same result, but if the eigenvalues are different then they can only be the same if the inner product between the two states is zero. So, what remains to be done is the case when the two eigenvalues are the same. But in that case you use the same argument but now with $A$ replaced by $D$ as the two states then have different eigenvalues for that operator. $\endgroup$ – Count Iblis Apr 24 '17 at 8:23
  • $\begingroup$ Thank you. Now I understand what you were saying. And I've got what I wanted. $\endgroup$ – ynn Apr 24 '17 at 16:12

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