# Trying to first understand position and momentum bases in Quantum Mechanics

In my lectures, I am told:

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

Which can only be valid if the overlap integral is:

$$\langle x \mid \psi \rangle = \int_{-\infty}^{\infty} \delta (x-x') \ \psi(x') \ dx' = \psi(x)$$

in which the complex conjugate of the delta is itself. Which is okay, but this would imply that..

$$|x \rangle = \langle x | = \delta(x-x')$$

My lecturer describes it roughly as "Quantum amplitude for being in position eigenstate $$| x \rangle$$ (at position $$x$$), given that one is in state $$|\psi \rangle$$". Why does something being in position eigenstate $$| x \rangle$$ imply that it is at position $$x$$? Since $$x$$ is arbitrary, this does, I agree, make some sense that this would describe $$\psi(x)$$, but the wording of it is hard for me to understand.

For instance, $$\delta(x'-x)$$ is a delta function, yet we're using it to describe $$\psi(x)$$. This is confusing to me because $$\delta(x'-x)$$ has a definite position, but $$\psi(x)$$ has a continuum of possible positions for a particle, not just "particle is at position $$x$$." I suppose the previous statement is arbitrary, so $$\psi(x)$$ can still represent a continuum of possible values $$x$$ could be? (*)

I also learned that $$\langle p \mid \psi \rangle = \bar \psi (p)$$, where $$\psi(x)$$ and $$\bar \psi(p)$$ are Fourier transform pairs.

This has been told to me to translate to rougly, as well: "Quantum amplitude for being in momentum eigenstate $$| p \rangle$$ (having momentum $$p$$) given that one is in state $$| \psi \rangle$$".

If the momentum eigenstate $$| p \rangle = \delta (p-p')$$, like before, why does something being in momentum eigenstate $$| p \rangle$$ imply that it has momentum $$p$$?

Does it mean that $$\hat p |p\rangle = p |p\rangle$$? I'm not sure if that's true with my delta function for $$|p\rangle$$.

I know I have a mess of questions, so I'll try and list them to be concise:

1. The question about why being in a eigenstate of position or momentum implies having a position or momentum of $$x$$ or $$p$$ respectively.

2. The question at (*) about how this "quantum amplitude for being in position $$x$$" can be an apt description of $$\psi(x)$$ (more specifically, intuitively how $$|x\rangle = \delta (x-x')$$ has one definite position yet $$\langle x \mid \psi \rangle = \psi(x)$$ which doesn't have a definite position).

3. If $$|x\rangle = \delta(x-x')$$ and $$|p\rangle = \delta(p-p')$$ then why does $$\langle x \mid p \rangle = \int_{-\infty}^{\infty} \delta(x-x') \ \delta(p-p') \ dx = \frac{1}{\sqrt{2 \pi \hbar}} e^{i p/\hbar x}$$ Where the RHS is clearly a plane wave?

• Feb 1, 2019 at 12:22
• Feb 1, 2019 at 12:28

this would imply that.. $$|x\rangle=\langle x|=\delta(x-x')$$

There are a few different things that do not make sense here. It might be easiest to work with this statement in terms of a geometric analogy. The state $$|x\rangle$$ is a member of a Hilbert space; you can think of it as a vector. In keeping with this analogy, $$\langle x|$$ is the corresponding member of the dual space: you can think of it as a function that takes a vector and returns a number, or equivalently, as the transpose of a vector. By saying $$|x\rangle=\langle x|$$, you are trying to say that, for example,

$$\begin{bmatrix}1&2&5\end{bmatrix}=\begin{bmatrix}1\\2\\5\end{bmatrix}$$

which is obviously incorrect, as those are two fundamentally different objects. By saying $$|x\rangle=\delta(x-x')$$, you are saying that a vector is equivalent to a number. By saying $$\langle x|=\delta(x-x')$$, you are saying that a function on vectors is equivalent to a number. None of these make sense, as they are all different types of objects.

So what is the proper way to make sense of the wavefunction? Let's begin with the question: how do you express an arbitrary state in a particular basis?

Suppose we have a finite orthonormal basis of states $$|\phi_1\rangle,|\phi_2\rangle,...,|\phi_n\rangle$$. If we have an arbitrary state $$|\psi\rangle$$, then we can express this state as a linear combination of basis states:

$$|\psi\rangle=c_1|\phi_1\rangle+c_2|\phi_2\rangle+...+c_n|\phi_n\rangle$$

The reason we can do this is because of the definition of a basis, from linear algebra: a basis is a minimal set of vectors whose span (set of linear combinations) is the entire vector space in question. Now, from this, we can easily see that, for any $$k$$:

$$\langle\phi_k|\psi\rangle=c_k$$

The reason we can say this is because our basis is orthonormal: it's both orthogonal (meaning $$\langle \phi_j|\phi_k\rangle=0$$ for $$j\neq k$$) and normal (meaning $$\langle \phi_k|\phi_k\rangle=1$$ for all $$k$$). Therefore, we can write:

$$|\psi\rangle = \langle\phi_1|\psi\rangle|\phi_1\rangle+\langle\phi_2|\psi\rangle|\phi_2\rangle+...+\langle\phi_n|\psi\rangle|\phi_n\rangle=\sum_{k=1}^n\langle\phi_k|\psi\rangle|\phi_k\rangle$$

This gives us a general procedure for expanding a function in a finite basis. This procedure also extends to expanding a state in a countably infinite basis; the finite sum at the end simply becomes an infinite sum.

When working with uncountably infinite bases like the position basis, where there is one basis state $$|x\rangle$$ for any real number $$x$$, we need to be a bit more careful. In particular, the position basis is not orthonormal. The inner product of two basis states $$|x\rangle$$ and $$|x'\rangle$$ is given by:

$$\langle x|x'\rangle=\delta(x-x')$$

So we still have an orthogonal basis ($$\langle x|x'\rangle =0$$ for $$x\neq x'$$), but it's not a normal basis ($$\langle x|x\rangle$$ is infinite).

In any case, the procedure to expand a state in the position basis is similar, but instead of a sum, we have the uncountable analogue of a sum, which is an integral:

$$|\psi\rangle = \int_{-\infty}^\infty \langle x|\psi\rangle |x\rangle dx$$

Instead of having a finite or countable set of coefficients $$c_k$$, we now have a coefficient for every real number $$x$$. It is convenient to express these coefficients as a function which takes a real number $$x$$ and returns the appropriate coefficient $$\langle x|\psi\rangle$$. This is what we mean when we say $$\langle x|\psi\rangle = \psi(x)$$, which means we can say:

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

This is how the wavefunction $$\psi(x)$$ is related to the state vector $$|\psi\rangle$$. To recover your statement, we simply take the overlap of some particular position basis state $$|x\rangle$$ with our state $$|\psi\rangle$$, and expand appropriately:

\begin{align} \langle x|\psi\rangle&=\langle x|\int_{-\infty}^\infty\langle x'|\psi\rangle |x'\rangle dx'\\ &=\int_{-\infty}^\infty \langle x'|\psi\rangle\langle x|x'\rangle dx'\\ &=\int_{-\infty}^\infty \psi(x')\delta(x-x')dx'\\ &=\psi(x) \end{align}

which is exactly what should be happening, given the definition.

1. The question about why being in a eigenstate of position or momentum implies having a position or momentum of x or p respectively.

One of the fundamental postulates of quantum mechanics is the following: measurements of a particular observable give you eigenvalues of the operator associated with that observable. The position operator is associated with the position observable; measuring a particular position means that you have measured a particular eigenvalue of an operator. Therefore, the state post-measurement must be the eigenstate associated with that particular eigenvalue. Conversely, an eigenstate of the position operator is associated with a particular eigenvalue of the position operator, which is, in turn, associated with being measured at a particular position.

1. The question at (*) about how this "quantum amplitude for being in position x" can be an apt description of ψ(x) (more specifically, intuitively how |x⟩=δ(x−x′) has one definite position yet ⟨x∣ψ⟩=ψ(x) which doesn't have a definite position).

Another fundamental postulate of quantum mechanics is the following: the probability that a state $$|\psi\rangle$$ will be in a particular eigenstate $$|\phi\rangle$$ after measurement of the corresponding observable is $$|\langle \phi|\psi\rangle|^2$$. Hence, the probability that $$|\psi\rangle$$ will be in a particular position eigenstate $$|x\rangle$$ after measurement is $$|\langle x|\psi\rangle|^2=|\psi(x)|^2$$. Using the answer to the first question, this is the same as the probability that the object described by $$|\psi\rangle$$ will be measured in position $$x$$.

1. If $$|x\rangle = \delta(x-x')$$ and $$|p\rangle = \delta(p-p')$$ then why does $$\langle x \mid p \rangle = \int_{-\infty}^{\infty} \delta(x-x') \ \delta(p-p') \ dx = \frac{1}{\sqrt{2 \pi \hbar}} e^{i p/\hbar x}$$ Where the RHS is clearly a plane wave?

The action of the momentum operator $$\hat{p}$$ on the state $$|\psi\rangle$$ can be represented in the position basis as a function of the position wavefunction of $$\psi$$:

$$\hat{p}|\psi\rangle\to -i\hbar\frac{d\psi(x)}{dx}$$

where the symbol "$$\to$$" is used to emphasize that these two statements are not strictly equivalent, but rather that the right-hand side is one of many possible representations of the left-hand side. For example, in the momentum basis, the representation of the momentum operator is a function of the momentum wavefunction of $$|\psi\rangle$$:

$$\hat{p}|\psi\rangle\to p\psi(p)$$

Anyway, suppose we want to find the representation of the momentum eigenstate $$|p\rangle$$ in position space. This is equivalent to finding the position wavefunction of the momentum eigenstate, which for clarity we will call $$\psi_p(x)=\langle x|p\rangle$$. Based on the definition of an eigenstate, we know that, regardless of representation,

$$\hat{p}|p\rangle=p|p\rangle$$

where $$p$$ is the momentum eigenvalue associated with the momentum eigenstate $$|p\rangle$$. If we put this equation into the momentum representation, we have the following differential equation:

$$-i\hbar\frac{d\psi_p(x)}{dx}=p\psi_p(x)$$

Solving this differential equation gives you:

$$\psi_p(x)=\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}$$

where the factor in front is to ensure that the wavefunction is normalized. Therefore, based on the definition of $$\psi_p(x)$$, we have:

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}$$

Incidentally, this relation helps you transform a position wavefunction into a momentum wavefunction, and vice-versa. We simply insert a basis expansion into the definition of the position wavefunction as follows:

$$\psi(x)=\langle x|\psi\rangle=\int_{-\infty}^\infty \langle x|p\rangle \langle p|\psi\rangle dp=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}\psi(p)dp$$

This shows that the position and momentum wavefunctions are related by a Fourier transform.

• This answer has opened my eyes like crazy. Thank you so much for taking the time. A couple things I'd like ask: 1) So all bra vectors are actually just linear functionals? I was told that $\langle \psi |$ represented the complex conjugate of $| \psi \rangle = \psi(x)$ 2) Why is $\langle x \mid x' \rangle = \delta(x-x')$? 3) Why do we need to have a basis state for each real number? I get that it's supposed to represent a position (which relies on space, which is continuous), but that doesn't make it clear to me as why it needs all elements in $\mathbb R$ Feb 1, 2019 at 17:26
• @sangstar A bra is the Hermitian conjugate of a ket, not just the complex conjugate (in finite-dimensional vector spaces, the Hermitian conjugate is equivalent to taking the complex conjugate of the transpose of a vector). I also wouldn't call the bra a functional; it's a function that takes vectors to (complex) numbers. The wavefunction of a bra is the complex conjugate of the wavefunction of the corresponding ket, though, just like the coefficients of the conjugate transpose of a vector are the complex conjugates of the coefficients of the original vector. Feb 1, 2019 at 17:32
• 4) Is the statement $| \psi \rangle = \int_{-\infty}^{\infty} \psi(x) \| x \rangle$ saying that, $\langle x \mid$ is a linear functional (mapping to $\mathbb C$ though) that when applied to $\|\psi \rangle$ returns the position in meters of its basis? Like $\langle 4 m \mid \psi \rangle = 4 m$? Trying to wrap my head around it. And then the original integral at the top is the sum of all those field elements (along with 4 meters) from $-\infty$ to $\infty$ attached as scalars to the position basis? I know this is a very confused question, so please ask if clarification is needed and where. Feb 1, 2019 at 17:35