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I came across this expression:

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

How can it be justified? I understand the LHS as an inner product, and the RHS just as a function of the parameter $x$.

David Z
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Frank
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    An inner product is also just a number, a number that is a function of the two inputs. Fix one input, in this case $\psi$, and you have left a function of one input. – ACuriousMind Mar 16 '15 at 15:27
  • Yeah - I can understand that :-) But an inner product is usually a sum of products (not always, I understand that too), whereas the RHS probably is not (a sum of products). – Frank Mar 16 '15 at 15:32
  • It's still a number. How the RHS or LHS are calculated in detail in their respective frameworks doesn't matter, they're still equal as numbers. – ACuriousMind Mar 16 '15 at 15:38
  • $x$ is an element of a vector space and $\psi$ is an element of the dual space, so $\psi(x)$ makes perfect sense. The equation provides a definition of $<x\mid\psi>$. – WillO Mar 16 '15 at 15:38
  • the left hand side is, by definition, the application of the $\delta(x-\cdot)$ distribution to the square-integrable function $\psi$ (typical notation in physics, that mingles ill-defined mathematical concepts altogether) – yuggib Mar 16 '15 at 15:41
  • Could somebody explain why he/she downvoted this question?? – Frank Mar 16 '15 at 15:43
  • To be precise, $\psi$ is also a vector, that can be expressed in an (infinite) basis of a (vector) function space, right (the Hilbert vector space of L2-integrable functions). Thought about it that way, the inner product is more natural than the function application actually... – Frank Mar 16 '15 at 15:45
  • @frank I did because it was an expression I had trouble with (years ago on self study ) and I thought other students on the same level would encounter it. I thought the question was a useful one. The op has a range of answers here to choose from here and advanced students will find them useful –  Mar 16 '15 at 16:04
  • @frank please excuse me It was orginally downvoted when i first saw it and I upvoted it for the reason in my last post. I misread your question as who downvoted it. hope this makes sense. up/down.....left/right...always had a problem with these concepts....sorry frank –  Mar 16 '15 at 16:09
  • @irish physics: no problem! :-) I also thought the question would be useful to others. – Frank Mar 16 '15 at 16:17
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    possible duplicate of Meaning of inner product $\langle \vec{r} | \psi(t)\rangle $ – joshphysics Mar 16 '15 at 16:57

3 Answers3

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The LHS is an inner product while the RHS is the evaluation of a function from an $L^2$ space at the point $x$. To somehow link the two you need to be able to write the RHS as an integral, so you need a "function" $\delta_x$ such that $$\langle x|\psi\rangle = \int\overline{\delta_x(s)}\psi(s)\text ds = \psi(x).$$ There is no such function, but the map $\psi\mapsto \psi(x)$ is a well-defined linear functional on test functions. However, it fails to be continuous and therefore it lives outside of the topological dual of the Hilbert space, which by Riesz representation theorem is isomorphic to itself. $\delta_x$ turns out to be a distribution, the Dirac $\delta$ "function". If you then "enlarge" your Hilbert space to contain distributions (usually one considers the Schwartz space in the $L^2$ space and assumes that all bras and kets come from its dual space, cf. rigged Hilbert space).

Phoenix87
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    the map you described is not a well-defined linear functional on $L^2$ (i.e. is neither on the algebraic dual), for a function in $L^2$ is really an equivalence class up to measure zero sets, so you cannot well define the behavior of such function in a point... – yuggib Mar 16 '15 at 16:25
  • Going to integrals and $L^2$ space seems kind of un-necessarily specific here. – DanielSank Mar 16 '15 at 16:51
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Paraphrasing WillO in the comments above, $x$ is an element of a vector space, and $\psi$ is an element of the dual space, so it follows immediately that: $\psi(x) = \langle x\ |\ \psi\rangle$. More precisely, "$x$" is a vector in the infinite vector space of positions, whose basis are Dirac deltas, while $\psi$ is a complex function of these vectors, itself a vector in a (vector) space of functions with suitable properties (it lives in a Hilbert space, not in just any function space, which is required to ensure that states can be normalized to 1 over all space, to be physically interpretable, among other things).

Frank
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    this is, mathematically, very very very unprecise. Let's say that if $\psi$ is a function of rapid decrease, than there exists an object of its (topological) dual such that the equation is verified; this object is the Dirac's delta distribution – yuggib Mar 16 '15 at 15:46
  • yuggib - I'm aware of the details - I tried to amend my answer to reflect more mathematical details. What should I add/remove/explain more? – Frank Mar 16 '15 at 16:20
  • you have to be careful with duality, there are two concepts of duality (topological dual and algebraic dual) and none of the two allows you to define dirac's delta as functionals on $L^2$ (see my comment on another answer). the expression you wrote has meaning only in certain spaces, the most usual being the functions of rapid decrease for $\psi$ and its topological dual the tempered distributions (where $\delta$ is well-defined) – yuggib Mar 16 '15 at 16:30
  • Yes, I'm aware of tempered distributions from a class on distributions. Makes sense. Thanks! – Frank Mar 16 '15 at 17:32
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The notation $\langle x\,\lvert\,\psi\rangle$ represents a number. But it's written that way (as opposed to being written, say, $y$) so that it suggests a procedure for transforming a different number, $x$, into the number $\langle x\,\lvert\,\psi\rangle$. Physically one can think of the procedure as something like

  1. measure the square root of the probability density for a particle to appear near position $x$
  2. measure the relative phase using some suitable interference experiment
  3. choose some convention for unmeasurable degrees of freedom (i.e. fix the gauge)

Never mind that all this is not really practical... regardless, it is a map $\mathbb{R}\to\mathbb{C}$ so it makes sense to denote it as a function, $\psi$.

We choose $\psi$ to denote the function because the nature of this function describes the abstract quantum state $\lvert\psi\rangle$. A different quantum state will, in general, be described by a different function, even using the same procedure (unless you consider the quantum state part of the procedure).

If you ask me, it's kind of limiting to learners that $\psi$ is used for both the wavefunction and the quantum state. So it may be clearer to consider an example where this is not the case, e.g. the 1D harmonic oscillator, where the states are labeled $\lvert\, n\rangle$ and the corresponding functions are labeled $\varphi_n$.

David Z
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