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I sligthly confused about Dirac notation. As of now I always thought that

$$ |ψ,t⟩ = |ψ(x),t⟩ = |ψ(x,t)⟩ . $$

However, now I found out that

$$⟨x|ψ,t⟩=ψ(x,t).$$

What does this notation actually mean?

Qmechanic
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Mark
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  • When we write $|\psi \rangle$, then that state does not have a representation. Taking the inner product, $\langle x|\psi \rangle$ is defined as $\psi(x)$. – Jens Roderus Mar 29 '18 at 14:05
  • But in that case the notation |ψ(x,t)⟩ makes no sense, right? – Mark Mar 29 '18 at 14:07
  • I don't think so. – Jens Roderus Mar 29 '18 at 14:12
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    Think of it like this: $|\psi\rangle$ is a vector and $\psi(x)$ is a component of that vector in a particular basis. – DanielSank Mar 29 '18 at 14:54
  • @Mark I think DanielSank's comment gets at why some of your top set of equations are incorrect notationally. $|\psi>$ is an abstract vector quantity and should be expressible in any number of different bases. $|\psi(x)>$ doesn't make sense (assuming x is position) because $\psi$ is just a label for this vector, not a function or other mathematical object. – Tyberius Mar 29 '18 at 15:20
  • This is listed as a "hot question" - does this imply that people here love and scramble over for the chance to answer basic textbook questions? – ina Mar 29 '18 at 19:26
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    @ina. No, it means there's a lot of views of it, lots of people interested, and 4 people decided to award points to the questioner. To me it goes to show that there's always an interest in in having other people try to explain that notation, that maybe you'll understand something better. – Bob Bee Mar 30 '18 at 03:36

1 Answers1

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What does this notation actually mean?

This is a ket labelled by $\psi$:

$$|\psi\rangle$$

This is a ket valued function of the time parameter $t$ labelled by $\psi(t)$

$$|\psi(t)\rangle$$

that returns a ket given a value of $t$.

The contraction of a bra and ket is a complex number

$$\langle \psi_1|\psi_2\rangle = c_{12}$$

The contraction of a bra and a ket valued function of time is complex valued function of time:

$$\langle \alpha|\psi(t)\rangle = \psi_\alpha(t)$$

Consider the ket valued function of the coordinate $x$

$$|x\rangle$$

which for a given $x$ coordinate, returns the eigenket of the position observable $\hat X$ with eigenvalue $x$

$$\hat X|x\rangle = x|x\rangle\,\quad \langle x |\hat X = x\langle x |$$

Then the contraction of the ket valued function of $t$, $|\psi(t)\rangle$, and the bra valued function of $x$, $\langle x|$, is a complex valued function of $x$ and $t$

$$\langle x|\psi(t)\rangle = \psi(x,t)$$

which is known as the (coordinate space) wavefunction.

I'm not sure what to make of something like $|\psi(x,t)\rangle$ unless, in this case, $x$ is considered a parameter like $t$

Hal Hollis
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  • Very good answer. I would have added that the role of the eigen-bra of a particular operator in the bra-ket product is to offer a representation of the ket on a space of functions. – DanielC Mar 29 '18 at 15:16