Quantum mechanics math question:

Suppose that there is eigenstate $|q \rangle$ where $q$ is position observable .

The question is,

1) What is eigenstate? How is this different from eigenvector? From this, how is $| q \rangle$ defined?

2) Why is $\langle q'|q \rangle=0$ whenever $q\neq q'$? I heard that this is due to orthogonal nature, but can anyone explain this more thoroughly?

3) Why is $|\psi \rangle =\int_{-\infty}^{\infty} \langle q|\psi \rangle |q \rangle dq$ true?

4) I am still not getting it. Why is $\langle q' |q \rangle $ referring to the probability of finding position $q'$ in particular state? Can anyone explain this mathematically? I am still not getting how multiplying eigenstate by the transpose of eigenstate produce probability.....

5) How is $\int |q \rangle\langle q|dq$ different from $\int \langle q |q \rangle dq$


1 Answer 1


1) Eigenstates and eigenvectors are the same thing. More specifically, an eigenstate is an eigenvector of some linear operator.

2) Technically, $\langle q'|q \rangle =\delta (q'-q)$ Here's a simple explanation: suppose a system is described by the state vector $|q \rangle$: "a particle is at position $q$". The probability (amplitude) for the particle to be at $q'$, or in the state $|q'\rangle$, is then zero: $\langle q'|q \rangle =0$.

3) This is due to the continuous closure relation: $1=\int |q \rangle\langle q|dq$. "Multiply" this by $|\psi \rangle$.

4) Interpreting $\langle b | a \rangle $ as a probability amplitude (a complex number) is one of the axioms of quantum mechanics. Mathematically, it is a "Fourier" coefficient: you're expressing a state as a linear combination of eigenstates.

5) First, $| \psi \rangle$ is in the vector space while $\langle \psi |$ is in the dual space. $| \psi \rangle \langle \psi |$ and $\langle \psi | \psi \rangle$ are NOT the same: $| \psi \rangle \langle \psi |$ is an operator while $\langle \psi | \psi \rangle$ is a complex number (probability amplitude).

$| \psi \rangle \langle \psi |$ is called a projection operator, analogous to projecting one vector onto another in $\mathbb{R}^3$. Here's an example:

Let $\{| \psi_n \rangle \}$ be a set of (orthonormal) eigenstates: $\langle \psi_i |\psi_j \rangle = \delta_{ij}$. Suppose that a system is "prepared" so that it is in a superposition of $| \psi_1 \rangle, | \psi_2 \rangle, | \psi_3 \rangle$. The system's state vector $|\psi \rangle$ is described as a linear combination of $| \psi_1 \rangle, | \psi_2 \rangle, | \psi_3 \rangle$: $|\psi \rangle = a_1 | \psi_1 \rangle +a_2 | \psi_2 \rangle +a_3 | \psi_3 \rangle$.

The projection operator $P_2=| \psi_2 \rangle \langle \psi_2 |$ "projects" $|\psi \rangle$ onto $| \psi_2 \rangle$:

$P_2 |\psi \rangle= | \psi_2 \rangle \langle \psi_2 |(a_1 | \psi_1 \rangle +a_2 | \psi_2 \rangle +a_3 | \psi_3 \rangle = a_1 \langle \psi_2 | \psi_1 \rangle |\psi_2 \rangle + a_2 \langle \psi_2 | \psi_2 \rangle |\psi_2 \rangle +a_3 \langle \psi_2 | \psi_3 \rangle |\psi_2 \rangle =0 +a_2 |\psi_2 \rangle +0 = a_2 |\psi_2 \rangle$

If it helps, think of a vector in $\mathbb{R}^3$: $\vec{v} = 3\hat{i} +4\hat{j} -5\hat{k}$. To "project" this onto the $\hat{j}$-component you can take a dot product of $\vec{v}$ and $\hat{j}$: $\hat{j} \cdot \vec{v} = 4$. The projection operator above does the same thing.

  • $\begingroup$ I edited my question. Can you have a look over it? $\endgroup$
    – W12
    Oct 15, 2012 at 3:29
  • $\begingroup$ I made a slight error: probability amplitude is not the same thing as probability. Fixed. $\endgroup$
    – BobaFret
    Oct 15, 2012 at 3:52
  • $\begingroup$ Thank you very much. What about 5)? I guess two are equal... $\endgroup$
    – W12
    Oct 15, 2012 at 4:02
  • $\begingroup$ BTW, I will register my account soon, and I'll upvote your answer as soon as possible. Thanks! $\endgroup$
    – W12
    Oct 15, 2012 at 4:03
  • $\begingroup$ About $|q\rangle \langle q|$: what does this operator mean? I do get that it results in an operator, but unsure of what physical meaning it has. The same with $|\psi \rangle \langle \psi |$ $\endgroup$
    – W12
    Oct 15, 2012 at 5:48

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