# In quantum mechanics, is $|\psi\rangle$ equal to $\psi(x)$?

So I'm going through my notes and I think I've confused myself. We often imply $$|\psi\rangle \to \psi(x)\\ \langle\psi| \to \psi(x)^*$$ for instance when we talk about eigenvalue equations we interpret $$\hat{H}|\psi\rangle =E|\psi\rangle$$ as simply $$\hat{H}\psi(x)=E\psi(x)$$ but I don't understand why we say $$|\psi\rangle \to \psi(x)$$ because if that was the case then $$\langle\psi|\psi\rangle=|\psi|^2$$ when clearly $$\langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1$$ I'm obviously missing something simple, could anyone point out where I'm going wrong?

• What does the arrow $\to$ mean? Do you think you could make the statement $|\psi\rangle\to\psi(x)$ mathematically precise? What is the exact relation between $|\psi\rangle$ and $\psi(x)$? – AccidentalFourierTransform Sep 30 '18 at 0:18
• Not sure to be honest, in my head I treat it as a equality but I don't think it actually is. I meant it as $|\psi\rangle$ 'can be replaced with' $\psi$ to answer questions – user7971589 Sep 30 '18 at 0:33
• @user7971589 Typically, when I have seen that arrow used in this context, it is explained in the following way: $x\to y$ means that $y$ is the representation of $x$ in a particular (previously-specified) basis. So, for example, if we're working in the position basis, we can say that $|\psi\rangle \to \psi(x)$, $\hat{p}\to -i\hbar\frac{\partial}{\partial x}$, and $\hat{x}\to x$. In the momentum basis, we can say that $|\psi\rangle \to \psi(p)$, $\hat{p}\to p$, and $\hat{x}\to i \hbar\frac{\partial}{\partial p}$. – probably_someone Sep 30 '18 at 0:45
• @user7971589 This is a general point, but whenever you're learning something and you think something fuzzy like "I think X isn't Y, but X can be replaced with Y?", that should be a mental stop sign to go back to the basics and figure out what's really going on! It's best to stop these misconceptions the moment they appear. – knzhou Sep 30 '18 at 14:19

To start, the kets are vectors, which means if we want an explicit realization of them, we would need to write them with respect to some basis. The first basis most people see is the position basis, where the basis kets are the states of definite position. Then, an arbitrary state $$|\psi\rangle$$ can be written as

$$|\psi\rangle = \int_{-\infty}^{\infty}dx \ \psi(x)|x\rangle,$$

where $$\psi(x)$$ is the familiar position space wavefunction. This should help explain why the last line has the integral in it, it comes from this superposition of states of definite position.

Now for rewriting $$\hat{H}|\psi\rangle = E|\psi\rangle$$, the exact same thing happens, but the reason there aren't any integrals is because the $$\hat{H}$$ in $$\hat{H}\psi(x) = E\psi(x)$$ is written as an differential operator acting on position space now, rather than an operator that acts in some way on an arbitrary ket. This might be a slight abuse of notation, but it is usually clear from context what basis the Hamiltonian is written in in a given equation.

• Just to add, the position basis is not always used. $|\psi\rangle\to\psi(k)$ is also very common, where $k$ is the wavevector and is tantamount to working in momentum space. This is why for general things, $|\psi\rangle$ is preferred, as it exists without any regard to a specific basis. – bRost03 Sep 30 '18 at 0:27
• Thankyou this helped a lot. To clarify then, in a different basis would the definition be $|\psi\rangle=\int_{-\infty}^{\infty}dk\psi(k)|k\rangle$ – user7971589 Sep 30 '18 at 0:49
• Yup! That's correct – bRost03 Sep 30 '18 at 1:49

The state of a quantum system is an element of a vector space (a Hilbert space, specifically). The notation $$|\psi\rangle$$ is used to label a particular vector in this space. Operators such as $$\hat{H}$$ map states in the Hilbert space to other states in the Hilbert space, so $$\hat{H} |\psi\rangle = |\phi\rangle$$, where $$|\phi\rangle$$ is another state in the Hilbert space. Some vectors might be eigenvectors of an operator, in which case we might write $$\hat{H}|\psi\rangle = E |\psi\rangle$$ (where $$E$$ is just a regular number, the eigenvalue).

If we are considering a wavefunction that describes the real-space motion of a particle, then we might be interested in the spread of the wavefunction over space. The Hilbert space in this case is infinite-dimensional. A natural choice for the basis vectors in which to expand $$|\psi\rangle$$ is the position eigenstates $$|x\rangle$$, where $$x$$ is any position in real space. $$|\psi\rangle$$ can always be expressed as a superposition of different basis states with different amplitudes in each basis state: $$|\psi\rangle = \int dx \ \psi(x)|x\rangle$$ where $$\psi(x)$$ is now a function that maps positions to some complex amplitude.

We can formally relate $$\psi(x)$$ to $$|\psi\rangle$$ as follows: take the inner product of the expansion of $$|\psi\rangle$$ in the position basis with any particular position eigenstate $$|x'\rangle$$: $$\langle x' | \psi \rangle = \int dx \ \psi(x) \langle x' | x \rangle = \psi(x')$$ where we rely on the fact that different position eigenstates are orthogonal (ie., $$\langle x' | x \rangle = \delta(x-x')$$).

So $$\psi(x) = \langle x | \psi \rangle$$. That is, the amplitude of the wavefunction at a position $$x$$ is given by the projection of the state $$|\psi\rangle$$ onto the basis state defined at position $$x$$, $$|x\rangle$$.

This is something that also confused me for a while when I was first learning about quantum mechanics. First, I think it's important to understand the concept of different 'spaces'. Physicists deal with many different spaces: the usual old 3D real space we live in, and lots and lots of abstract 'spaces'.

The Hilbert space of a particle is the 'space' of all possible states the particle can be in. It is NOT the same 3D space that the particle actually 'moves around in'. The Hilbert space also obeys the usual laws of linear algebra, and it is infinite dimensional (every bit as infinite dimensional as real space is three dimensional).

In particular, the dimensionality is equal to the number of points in space, and the set of basis vectors is $$\{\,|x\rangle\,|\,x\in \mathbb{R}\}$$. The meaning of $$|x\rangle$$ as a state is the particle is located at x". $$|x\rangle$$ is also a basis vector of the Hilbert space; note that there is one for every x. So $$|2.77\rangle$$, $$|-\sqrt{3}\rangle$$, $$|\pi\rangle$$, and $$|42\rangle$$ are all basis vectors, as are an infinite other ones.

The important equation is actually $$\langle x|\psi\rangle = \psi(x)$$. This is the 'proper' version of your $$|\psi \rangle \rightarrow \psi(x)$$. The vector $$|\psi\rangle$$ lives in Hilbert space and is the abstract mathematical object representing the state of the particle. The equation $$\langle x|\psi\rangle = \psi(x)$$ essentially says that the scalar product of $$| x\rangle$$ and $$|\psi\rangle$$, or, 'the component of the vector $$|\psi\rangle$$ along the $$|x\rangle$$ direction' is equal to some number which we call $$\psi(x)$$. Because of the infinite dimension, you can take all of these scalar products (i.e. all the components of $$|\psi\rangle$$ along all the possible 'directions') and piece them together into a continuous function, the wave function.

Technically speaking if $$\vert \psi\rangle\to \psi(x)$$ then $$\langle\psi\vert\to \int \,dx\, \psi(x)^*$$. The integral symbol guarantees that glueing a bra and a ket together, $$\langle \phi \vert \psi\rangle$$, is a scalar product and a (generally complex) number, gives the same as the number computed as $$\int dx \phi^*(x)\psi(x)$$.

More specifically, $$\psi(x)$$ is really $$\langle x\vert\psi\rangle$$, as per this question. Using $$\langle x\vert\psi\rangle =\psi(x)$$ allows to interpret $$\psi(x)$$ as the component of $$\vert\psi\rangle$$ along the basis vector $$\vert x\rangle$$. If you buy this then you can also express $$\vert \psi\rangle$$ in the momentum basis, where $$\psi(p)=\langle p\vert\psi\rangle$$.

There are even states - like spin states $$\vert \pm\rangle$$, for which there is no "spatial" wavefunction, i.e. $$\langle x\vert +\rangle$$ doesn't make sense as the spin degree of freedom does have physical dimensions.

The Dirac notation is thus a way of emphasizing the vectorial nature of states, which can be combined and multiplied by scalars just like vectors can. It also emphasizes that expressing $$\psi(x)$$ or $$\psi(p)$$, i.e. the wavefunction in position or momentum space, is basically a choice of basis, much like choosing to express a vector in spherical or Cartesian coordinates. Indeed, to go from one basis to the other we need the transition formula $$\langle x\vert p\rangle \sim e^{-i p x/\hbar}$$.

It's actually not the case that $$\langle \psi | = \psi(x)^*$$. A more accurate statement would be as follows.

$$\langle \psi |\cdots := \int_{-\infty}^\infty dx\ \psi(x)^*\cdots$$

This bra $$\langle \psi |$$ can thus be thought of as a function that maps ket states $$| \phi \rangle$$ to a complex number.

$$|\phi\rangle \overset{\langle \psi |}{\mapsto} \langle \psi |\phi \rangle := \int_{-\infty}^\infty dx\ \psi(x)^* \phi(x) \in \mathbb C$$

When the state space is the projectivized Hilbert space of $$L^2$$ functions on $$ℝ^1$$, $$|\psi\rangle$$ is by definition the name of the equivalence class represented by the function $$\psi$$. $$\langle\phi|\psi\rangle$$ is by definition the same thing as $$\int|\phi(x)\psi(x)|$$ (when the state space is the projectivized Hilbert space of $$L^2$$ functions on $$ℝ^1$$".).

• I don't understand what you mean by "in this context" here, because it leads to a very strange notion of "by definition." – probably_someone Sep 30 '18 at 0:31
• @probably_someone: "in this context" means "when the state space is the (projectivized) Hilbert space of $L^2$ functions on (in this case) ${\mathbb R}^1$". – WillO Sep 30 '18 at 0:50