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Which of the following corresponds to a $ \psi(x)$, a wavefunction written in the position basis: $ x| \psi\rangle $ or $ \langle x| \psi\rangle $? If it is the second expression (which my textbook asserts), what is the meaning of the first expression?

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physics.stackexchange.com/q/65794 –  Wildcat Jul 21 '13 at 16:30
    
I should have also asked how to interpret $ \int\langle \psi*|x|\psi\rangle dx $ –  Noah Jul 21 '13 at 20:42
    
@Noah Can you define $\langle\psi *|$? –  Will Jul 21 '13 at 21:00
    
I think you might mean $\langle \psi |\hat{x}| \psi \rangle = \int dx \langle \psi | x \rangle x \langle x|\psi \rangle = \int dx~ \psi^*(x) ~x~\psi(x)$. This is just the expected value of $x$ for state $\psi$. –  Will Jul 21 '13 at 21:20
    
Yes. My mistake. –  Noah Jul 22 '13 at 1:12

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It is the second, $\psi(x) = \langle x|\psi\rangle$ which is correct. The first, if $x$ is the position operator, is just the position operator acting on the state $|\psi\rangle$. The abstract state $|\psi\rangle$ can be expanded in any basis, using a completion relation: $$|\psi\rangle = \underbrace{\sum_i |i\rangle \langle i|}_{1~=~identity}\psi\rangle$$ (where the sum can mean sum or integral, depending on this situation) using some complete basis $\{|i\rangle\}$. An example of this is the position basis, $\{|x\rangle\}$, from which we have $$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$ This shows us that the wavefunction $\psi(x) = \langle x|\psi\rangle$, is the coefficient from the expansion of the state $|\psi\rangle$ in the position basis $|x\rangle$, at position $x$.


Edit to respond to a question asked by Noah in response to Matt's answer: "But doesn't an operator acting on the state project the state onto the eigenstates of the operator and is that the same as representing the state in a new basis?"

As I have shown above, the "projection" you are talking about is given by applying the identity in terms of the position basis $\mathbf{1=\int dx~|x\rangle\langle x|}$.

Let's see what happens when we apply $\hat{x}$ instead: Just like state vectors can be expanded in a complete basis, so too can operators. In general, if we take our complete basis $\{|i\rangle\}$ we can write the operator $\hat{A}$ in terms of matrix elements $$\hat{A} = \sum_{i,j} |i\rangle\langle i |\hat{A}|j\rangle\langle j |$$ where the matrix elements are $A_{ij} = \langle i |\hat{A}|j\rangle$. For the case of the position operator $\hat{x}$ and using the position basis $\{|x\rangle\}$ $$\hat{x} = \int dx~dy~ |x\rangle\langle x |\hat{x}|y\rangle\langle y |\\ = \int dx~ x~|x\rangle\langle x |$$ that is, it is diagonal in the position basis (obviously). Using this we see $$\hat{x}|\psi\rangle = \int dx~ x~|x\rangle\langle x |\psi\rangle$$ So we see that application of $\hat{x}$ can be thought of as a kind of projection, weighted by $x$. So we see that it is not the same as representing the state in the $x$ basis, which is actually $$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$ as above.

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$\psi(x)\equiv\left\langle{x}\,\middle|\,\psi\right\rangle$ is the correct notation. $x\left|\psi\right\rangle$ means that the position operator is acting on the state. People sometimes put a hat on operators to remove ambiguity: $\hat{x}\left|\psi\right\rangle$.

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The expression $x|\psi\rangle$ could also mean a real number $x$ multiplying the state $|\psi\rangle$. –  joshphysics Jul 21 '13 at 16:28
    
But doesn't an operator acting on the state project the state onto the eigenstates of the operator and is that the same as representing the state in a new basis? –  Noah Jul 21 '13 at 17:26
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An operator just performs an operation on a specific state. What basis the state is irrelevant and is generally selected for convenience. The projection onto a state occurs during the application of the "bra". –  Matt Jul 21 '13 at 17:43
    
@Noah - see my edit in my answer. I try to explain this in more detail. –  Will Jul 21 '13 at 18:12

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