Undergrad How Does Dirac Notation Simplify Quantum Mechanics for Harmonic Oscillators?

[jstrunk]

I am completely baffled by bit of notation in Quantum Mechanics Concepts and Applications by Zitteli. He is trying to get the differential equation for the ground state of a harmonic oscillator using the algebraic method as opposed to Schrodinger's equation. I suspect he is compressing a lot of steps into one and probably also abusing the notation. Can someone clarify this, or point me to a source that develops Dirac notation slowly and clearly?

The annihilation operator is \hat a = \frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)
The eigenstates are given by \hat a\left. {\left| n \right.} \right\rangle = \sqrt n \left. {\left| {n - 1} \right.} \right\rangle
The ground state is given by \hat a\left. {\left| 0 \right.} \right\rangle = 0
He deduces the differential equation like this
<br /> \begin{array}{l}<br /> \left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\<br /> x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0<br /> \end{array}<br />
The part I don't understand is
\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right)
I would expect this \left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x + x_0^2\frac{d}{{dx}}} \right)} 0dx = 0
The book went to a lot of trouble to show that n has to be a non-negative integer, but even if we somehow allow n=0 to morph into a continuous variable {\psi _0}\left( x \right), I don't get his answer, I get the following

\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x{\psi _0}\left( x \right) + x_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx = \int_{ - \infty }^\infty {\left( {{x^2}{\psi _0}\left( x \right) + xx_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx

On the other hand, if we start with \hat a\left. {\left| 0 \right.} \right\rangle = 0 instead of \left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = 0 and allow n=0 to morph to {\psi _0}\left( x \right) I can easily get the desired answer
<br /> \begin{array}{l}<br /> \hat a\left. {\left| 0 \right.} \right\rangle = 0\\<br /> \frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)\left. {\left| {{\psi _0}\left( x \right)} \right.} \right\rangle = 0\\<br /> \frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\<br /> x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0<br /> \end{array}<br />

 

[hilbert2]

The vector $\left|\right.0 \left.\right>$ is not a wavefunction that is identically zero. It's the state vector of the ground state and its inner product with the position eigenstate $\left|\right. x\left.\right>$ is the ground state wavefunction:

$\left<\right.x\left.\right|\left.0\right> = \psi_0 (x)$,

which is a normalized Gaussian function, not zero.

 

[PeroK]

I get the following

\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x{\psi _0}\left( x \right) + x_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx = \int_{ - \infty }^\infty {\left( {{x^2}{\psi _0}\left( x \right) + xx_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx

Note that the left-hand side is expressed as state vectors in an abstract linear space. Not, in the linear space of position space wave-functions. The inner product as a integral is only defined once you have expressed states as position space wave functions.

To expand on the idea explained in post #2 above:

First, consider eigenstates of the position operator (or, in fact, any states represented by a real number):

If $\langle x$ is a bra and $y\rangle$ is a ket, then $\langle x | y\rangle$ is a complex number. Now, fix $y \in \mathbb{R}$ for each $x \in \mathbb{R}$ we have that $\langle x | y\rangle$ is a complex number.

This allows us to associate with the state $y \rangle$ a complex-valued function $f_y$ of a real variable $x$, defined by:

$f_y(x) = \langle x | y\rangle$

Now, let $n \rangle$ be an eigenstate of the Harmonic Oscillator. Similarly, we can associate with each eigenstate $n \rangle$ a complex-valued function by:

$\psi_n(x) = \langle x | n\rangle$

And, in particular:

$\psi_0(x) = \langle x | 0\rangle$

(Where $\langle x|$ is the bra associated with the eigenstate of the position operator with eigenvalue $x$.)

That's how in general abstract states get expressed as position space wave functions.

 

[jstrunk]

Thank you. Can you recommend a source that teaches Dirac notation as you have described it, rather than the loose way it is taught in the book I am using.Preferably one that is free or inexpensive.

 

[PeterDonis]

The part I don't understand is
$$
\langle x \vert X + x^2_0 \frac{d}{dx} \vert 0 \rangle = \left( x \psi_0(x) + x^2_0 \frac{d}{dx} \psi_0(x) \right)
$$

It might help to first expand out the LHS without changing notation. Since the operator between the bra and the ket is linear, we can write:

$$
\langle x \vert X + x^2_0 \frac{d}{dx} \vert 0 \rangle = \langle x \vert X \vert 0 \rangle + \langle x \vert x^2_0 \frac{d}{dx} \vert 0 \rangle
$$

Then, it turns out that both operators can be moved outside the bra-kets (or, to put it another way, they both commute with the operation of taking the bra-ket inner product), so you get

$$
X \langle x \vert 0 \rangle + x^2_0 \frac{d}{dx} \langle x \vert 0 \rangle
$$

Then all you have to do is switch notation from $\langle x \vert 0 \rangle$ to $\psi_0(x)$, which @hilbert2 already explained.

 

[PeroK]

Thank you. Can you recommend a source that teaches Dirac notation as you have described it, rather than the loose way it is taught in the book I am using.Preferably one that is free or inexpensive.

I think Dirac notation grows on you the more you use it. It is useful to be able to dot a few mathematical i's and cross a few t's here and there but the essence of the Dirac notation is to be able to get on with QM in a notation that ultimately helps a lot rather than hinders.

The big step in my opinion in what Zitteli is doing is:

$\langle x|\frac{d}{dx}|0 \rangle = \frac{d}{dx} \langle x|0 \rangle = \frac{d}{dx}\psi_0(x)$

Does he show a proof of that somewhere? I checked Sakurai and he justifies that quite nicely.

But that is different from Dirac notation, per se. That's about justifiying (or proving) the various symbolic manipulations you get using the notation.

On the other hand, you could go with the flow and learn the QM for now; accepting the natural seeming identities such as the above. Then, when you are more familiar with everything, if you want, go back and look for more formal mathematical justification for these manipulations. There's something to be said for that approach, perhaps.

 

[jstrunk]

I work through almost every piece of math in the text and every exercise at the end of each chapter.
I have never seen anything like that.
I understand your suggestion but I am already carrying a lot of other confusion that I am hoping will clear up later.
Thanks for your help, but sometimes I think it would be easier to figure out QM from the raw experimental data like the pioneers did back in the 1930s than to understand the formalism they came up with.

 

[vanhees71]

I am completely baffled by bit of notation in Quantum Mechanics Concepts and Applications by Zitteli. He is trying to get the differential equation for the ground state of a harmonic oscillator using the algebraic method as opposed to Schrodinger's equation. I suspect he is compressing a lot of steps into one and probably also abusing the notation. Can someone clarify this, or point me to a source that develops Dirac notation slowly and clearly?

The annihilation operator is \hat a = \frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)
The eigenstates are given by \hat a\left. {\left| n \right.} \right\rangle = \sqrt n \left. {\left| {n - 1} \right.} \right\rangle
The ground state is given by \hat a\left. {\left| 0 \right.} \right\rangle = 0
He deduces the differential equation like this
<br /> \begin{array}{l}<br /> \left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\<br /> x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0<br /> \end{array}<br />

If this is really a correct citation of this text, look for a better one. He mixes different kinds of operators from the beginning of this passage on. Either you use the Dirac formalism (representation free "covariant" representation of QT) or a concrete representation (in your case the position representation aka. wave mechanics).

In the Dirac formalism there are abstract operators $\hat{x}$ and $\hat{p}$ acting on an abstract Hilbert space. Then there is wave mechanics using the Hilbert space of square integrable functions with position "eigenvalues" as independent variables. You come from the abstract to the position representation by using the generalized position eigenvectors $|x \rangle$, and the wave function is defined by
$$\psi(x)=\langle x|\psi \rangle.$$
Now I denote the operators of observables in wave mechanics with bold-faced symbols and the abstract ones with hats. Then you have
$$\mathbf{x} \psi(x)=x \psi(x), \quad \mathbf{p} \psi(x)=-\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x} \psi(x).$$
The relation between these operators in the position representation and the abstract ones is
$$\mathbf{x} \psi(x)=\langle x|\hat{x}|\psi \rangle=\langle \hat{x} x|\psi \rangle=x \langle x|\psi \rangle=x \psi(x)$$
and
$$\mathbf{p} \psi(x)=\langle x|\hat{p}|\psi \rangle=\langle \hat{p} x|\psi \rangle = - \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x} \langle x|\psi \rangle=- \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x} \psi(x).$$
Note that you can derive the here used non-trivial relation
$$\hat{p} |x \rangle=+\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x} |x \rangle$$
solely from the commutator relation
$$[\hat{x},\hat{p}] = \mathrm{i} \hbar \hat{1}.$$

Note that the abstract states do not depend on $x$ and thus taking a derivative wrt. $x$ would simply get 0. The book is thus misleading at best! I cannot stress more the importance of getting it straight in ones mind that the abstract state kets are not identical with wave functions in the position representation, but there is a one-to-one mapping between these entities. The wave function is a kind of generalized components with respect to a (generalized) basis with a continuous index $x \in \mathbb{R}$.

For a better presentation of the formalism, including a less then usual handwaving foundation of the notion of "generalized eigenvectors" in terms of the modern rigged-Hilbert-space formulation, see the excellent textbook by Ballentine, Quantum Mechanics.

 

[jstrunk]

Ballentine seems kind of advanced. I have studied Fourier Analysis and Hamiltonian/Lagrangian Mechanics to at least the level that would be expected for an undergraduate, but I haven't studied Group Theory or Measure Theory. A little bit of Measure Theory came up in the book I am using. It was passed over with some handwaving and I was fairly comfortable with that but if Ballentine is going to really get into it then that would be a problem.

Also, are there a lot of exercises with detailed solutions? One of the good things about Zettili is the abundance of exercises with fairly detailed solutions.

 

[jstrunk]

I think I am getting a glimmer of what's going on here based on what people have said in this post and on some general knowledge I have of QM from outside of the text I am using.

I will try to describe the glimmer and let me know if its on the right track. I am sure the language is ridiculous but I just need to know if the essence of the idea there.

The book has so far covered Matrix, Position, and Momentum Representations and it has always been stated or been fairly obvious which Representation was intended. So when shown an expression like \left\langle x \right.\left. {\left| y \right.} \right\rangle I know y is really a position wavefunction if the understood context is Position Representation and its really a column matrix if in the Matrix Representation and the whole expression is really the appropriate matrix or integral inner product. But \left. {\left| n \right.} \right\rangle is a new thing. Its not in any of these Representations. Up to this point, Zettili hasn't discussed the measurement process much, but I guess \left. {\left| 0 \right.} \right\rangle represents the ground state of a completely abstract ideal of the whatever it is that underlies the various concrete representations. There isn't any way to write another expression showing what it “really is”. It doesn't make sense to put it in a matrix or integral inner product. For this abstract representation, \left\langle x \right.\left. {\left| 0 \right.} \right\rangle = {\psi _0}\left( x \right), if it can be considered an inner product at all, is an inner product in the space of possible representations. The bra x specifies that the measurement we are making will be a position measurement and causes the abstract ideal to manifest as a position space wavefunction {\psi _0}\left( x \right). If this is at all close to being correct, I will have to reevaluate everything I learned in the last 6 months.

 

[PeroK]

I think I am getting a glimmer of what's going on here based on what people have said in this post and on some general knowledge I have of QM from outside of the text I am using.

I will try to describe the glimmer and let me know if its on the right track. I am sure the language is ridiculous but I just need to know if the essence of the idea there.

The book has so far covered Matrix, Position, and Momentum Representations and it has always been stated or been fairly obvious which Representation was intended. So when shown an expression like \left\langle x \right.\left. {\left| y \right.} \right\rangle I know y is really a position wavefunction if the understood context is Position Representation and its really a column matrix if in the Matrix Representation and the whole expression is really the appropriate matrix or integral inner product. But \left. {\left| n \right.} \right\rangle is a new thing. Its not in any of these Representations. Up to this point, Zettili hasn't discussed the measurement process much, but I guess \left. {\left| 0 \right.} \right\rangle represents the ground state of a completely abstract ideal of the whatever it is that underlies the various concrete representations. There isn't any way to write another expression showing what it “really is”. It doesn't make sense to put it in a matrix or integral inner product. For this abstract representation, \left\langle x \right.\left. {\left| 0 \right.} \right\rangle = {\psi _0}\left( x \right), if it can be considered an inner product at all, is an inner product in the space of possible representations. The bra x specifies that the measurement we are making will be a position measurement and causes the abstract ideal to manifest as a position space wavefunction {\psi _0}\left( x \right). If this is at all close to being correct, I will have to reevaluate everything I learned in the last 6 months.

Interesting. Some fundamental points:

$|\alpha \rangle, \ |x \rangle, \ |p \rangle, \ |n \rangle, \ |0 \rangle, \ |anything \rangle$

Are all states of a system, represented by vectors in an abstract Linear (Hilbert) Space. Or, "kets" in Dirac notation, which is simply a notation for doing Linear Algebra that is particularly useful when doing QM.

What each symbol ($\alpha, \ x, \ 0$) means depends on the context. Generally, the convention is that $\alpha$ is a general state, $x$ is an eigenstate of the position operator, $p$ is an eigenstate of the the momentum operator, and $n$ is the eigenstate of a system with a countable basis of eigenstates, such as the Harmonic Oscillator. But, it's up to the author and they should be careful to define what each symbol stands for in each context.

In particular, $|n \rangle$ is not a new concept, but simply another state - in this case an eigenstate of the Harmonic Oscillator.

With every ket $|\alpha \rangle$ there is associated a bra $\langle \alpha |$. This is the dual of the ket, using the concept from Linear Algebra that each vector can be seen as a linear functional on the linear space.

The inner product $\langle \alpha | \beta \rangle$ can be seen as a bra acting on a ket. This is standard linear algebra in Dirac notation.

Note: you must lose the idea that only an integral of two functions is an inner product. The inner product is defined on the abstract Hilbert Space and is denoted by $\langle \alpha | \beta \rangle$.

The inner product (of two states) does not represent a measurement, but is analogous to the usual idea of the inner product in vector spaces of the amount that one vector overlaps another, or defines the "angle" between two vectors. In particular, if $\langle \alpha | \beta \rangle = 0$, then $| \alpha \rangle$ and $|\beta \rangle$ are "orthogonal".

In fact, everything is defined on the abstract Hilbert space and then has various representations in position space, momentum space or as a matrix.

 

[PeroK]

To analyse what you wrote:

So when shown an expression like ⟨x|y⟩\left\langle x \right.\left. {\left| y \right.} \right\rangle I know y is really a position wavefunction

These are not wavefunctions. These are states. Probably eigenstates of the position operator.

whole expression is really the appropriate matrix or integral inner product.

It's better to think of $\langle x | y \rangle$ as the inner product, which has an equivalent form in each representation - such as an integral in the position space representation.

But |n⟩\left. {\left| n \right.} \right\rangle is a new thing. Its not in any of these Representations.

It's nothing new, conceptually. Just a new state. As such, it has a representation as a wave function in position space etc.

|0⟩\left. {\left| 0 \right.} \right\rangle represents the ground state of a completely abstract ideal

Yes, I'd say "... is the ground state in an abstract Hilbert space". The more you think of the state as detached from a specific representation the better. It's the same idea as thinking of a vector free from any basis or a linear operator free from any matrix representation.

For this abstract representation, ⟨x|0⟩=ψ0(x)\left\langle x \right.\left. {\left| 0 \right.} \right\rangle = {\psi _0}\left( x \right), if it can be considered an inner product at all, is an inner product in the space of possible representations.

It definitely is the inner product. It's not the space of all possible representations. It's the space of all possible states, free from any particular representation. At the end of the day, it's just a HIlbert space.

The bra x specifies that the measurement we are making will be a position measurement

You need to rethink this. Measurements are represented by operators. The inner product of two states (a bra and a ket in Dirac notation) is just like the inner product of two vectors.

Not least, the inner product gives a complex number. A measurement gives another state (vector/ket).

 

[jstrunk]

OK, then I am hopelessly lost and the only thing I can do is find a new book or move on with the book I have and hope it clears up by magic.
Thanks for your help.

 

[vanhees71]

Another good QM book using the Dirac notation from the very beginning (as one should, because it's the most clear exposition of the fundamentals of QT, while matrix and wave mechanics are easily derived from them when needed) is

J. J. Sakurai, Modern Quantum Mechanics (revised edition, coedited by Tuan; the newer 2nd edition coedited by Napolitano is also ok but contains a misleading chapter on relativistic quantum mechanics).

 

[PeroK]

OK, then I am hopelessly lost and the only thing I can do is find a new book or move on with the book I have and hope it clears up by magic.
Thanks for your help.

I wouldn't say hopeless. But you have some misconceptions to iron out. Accepting that you don't totally understand is a necessary first step.

I'm not sure how many people nail QM first time. Are you learning on your own?

I generally seem to take three attempts at things. So, I'm not too bothered if I have to go back to square one a couple of times.

 

[jstrunk]

I sort of see where

<br /> <br /> \begin{array}{l}<br /> <br /> \left. {\left| \psi \right.} \right\rangle = \int_{ - \infty }^\infty {\psi \left( x \right)\left. {\left| x \right.} \right\rangle } dx\\<br /> <br /> \psi \left( x \right) = \left\langle x \right.\left. {\left| \psi \right.} \right\rangle<br /> <br /> \end{array}<br /> <br />

comes from by analogy with the discrete case

<br /> <br /> \begin{array}{l}<br /> <br /> \left. {\left| \psi \right.} \right\rangle = \sum\limits_{i = 1}^n {{\psi _i}} \left. {\left| i \right.} \right\rangle \\<br /> <br /> {\psi _i} = \left\langle i \right.\left. {\left| \psi \right.} \right\rangle<br /> <br /> \end{array}<br /> <br />

.
But it would help a lot if I had an concrete example of the continuous case to illustrate the formalism. I need to see what \psi \left( x \right) and \left. {\left| \psi \right.} \right\rangle look like in a particular example and where the come from. Then I need to apply the formulas to convert one to the other to get a feel for this. Just handing me a \psi \left( x \right) or \left. {\left| \psi \right.} \right\rangle won't cut it. Does anyone have an example like that?

 

[kith]

The easiest way to build intuition about how ket vectors work is to look at the spin-1/2 case. There, you don't have differential operators and infinite-dimensional operator representations but simple linear algebra in two dimensions. Armed with the intuition from the spin-1/2 case, you won't have trouble to understand the difference between a wavefunction and the corresponding ket vector.

The first ~30 pages of Sakurai introduce QM this way. Another source which I can recommend from first-hand experience is Cohen-Tannoudjii (although you may need to jump around a bit because he elaborates on a number of sidetracks).

 

[PeroK]

I sort of see where

<br /> <br /> \begin{array}{l}<br /> <br /> \left. {\left| \psi \right.} \right\rangle = \int_{ - \infty }^\infty {\psi \left( x \right)\left. {\left| x \right.} \right\rangle } dx\\<br /> <br /> \psi \left( x \right) = \left\langle x \right.\left. {\left| \psi \right.} \right\rangle<br /> <br /> \end{array}<br /> <br />

comes from by analogy with the discrete case

<br /> <br /> \begin{array}{l}<br /> <br /> \left. {\left| \psi \right.} \right\rangle = \sum\limits_{i = 1}^n {{\psi _i}} \left. {\left| i \right.} \right\rangle \\<br /> <br /> {\psi _i} = \left\langle i \right.\left. {\left| \psi \right.} \right\rangle<br /> <br /> \end{array}<br /> <br />

.
But it would help a lot if I had an concrete example of the continuous case to illustrate the formalism. I need to see what \psi \left( x \right) and \left. {\left| \psi \right.} \right\rangle look like in a particular example and where the come from. Then I need to apply the formulas to convert one to the other to get a feel for this. Just handing me a \psi \left( x \right) or \left. {\left| \psi \right.} \right\rangle won't cut it. Does anyone have an example like that?

First, note that is is purely conventional to write:

$\sum\limits_{i = 1}^n {{a_i}}$

Instead of:

$\sum\limits_{i = 1}^n {a(i)}$

In fact, mathematically, one way to look at a complex sequence $a_1, a_2, a_3 \dots$ is to view $a$ as a function of the natural numbers:

$a: \mathbb{N} \rightarrow \mathbb{C}$

By convention, we write this function and its values as $a_i$ rather than $a(i)$.

Equally, you could write the values of a function, $f$, of a continuous variable $x$ as $f_x$, although by convention, we normally write this as $f(x)$.

Note that if we have a function of two variables we can look at it two ways:

$f:\mathbb{R}^2 \rightarrow \mathbb{C}$ with function values denoted by $f(x, y)$

Or, we can look at this as a set of functions. For each $y \in \mathbb{R}$ we have a function:

$f_y:\mathbb{R} \rightarrow \mathbb{C}$ with function values denoted by $f_y(x) = f(x, y)$

I'm not sure if that mathematical digression is any help! Hopefully.

For the continuous case, the function could be anything, as long as it's normalised. In the same way that the coefficients can be anything, as long as the sum of the squares of their absolute values is $1$. For example:

$\psi(x) = \frac{1}{\sqrt{\pi}}exp(-x^2)$

Correction: it should, of course, be:

$\psi(x) = (\frac{1}{\pi})^{\frac14} exp(\frac{-x^2}{2})$

What this means in your example is that the weighting of each position eigenstate $|x \rangle$ is $\frac{1}{\sqrt{\pi}}exp(-x^2)$ and by convention we write this as:

$|\psi \rangle = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}}exp(-x^2) |x \rangle dx$

Now, if we take the bra of a position eigenstate $\langle y |$

$\langle y |\psi \rangle = \langle y|\int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}} exp(-x^2) |x \rangle dx = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}}exp(-x^2)\langle y |x \rangle dx = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}}exp(-x^2) \delta(x-y) dx = \frac{1}{\sqrt{\pi}}exp(-y^2)$

What this means is that if you define a state as a continuous distribution (function) of position eigenstates, then the position space wave-function is the same function.