# Simple question about vector spaces and bases in QM

1. Jan 20, 2012

### Waxbear

When reading in Griffiths and on Wikipedia about the vector space formulation of wavefunctions, i am constantly faced with the statement that a vector can be expressed in different bases, but that it's still the same vector. However, I'm having a hard time imagining what it is about a vector that makes it the same vector, independent of the base you express it in. As i see it, the base and coefficients completely describe the vector, so how can you say it's the same vector when you change the base and coefficients. In other words, what property of the vector is the same in all bases? (I guess this is more of a Linear algebra question really.)

2. Jan 20, 2012

### Ken G

Yes, this is something that is easier to understand in more typical vector spaces, where a vector represents an arrow in some Euclidean space. A typical way to write an arrow is in coordinate form, but this requires choosing a basis. A basis is a set of "basis vectors", call them e1 and e2. Then we can write the full vector v with coordinates x and y as v = x e1 + y e2. This expression shows that the vector v involves a combination of the coordinates and the basis vectors-- it is not either one by itself. This combination allows for a sense of v being the "same thing" in some other basis, call the other basis e'1 and e'2, if we can say x e1 + y e2 = x' e'1 + y' e'2. That's what is the "same thing" about v-- we can put an equals sign between these two coordinate expressions. (You can check that if you do a rotation to both the coordinates and the basis vectors, this equality will hold.)

So that's a mathematical way to show it using coordinates, but there's also a graphical way to see it, which is to imagine that a vector is an arrow. An arrow has magnitude and direction, so it is clear that neither of those properties care how you choose your basis vectors, and the coordinates are just the projections of the arrow onto the basis vectors. Different types of vector spaces have different ways to define what a "projection" entails, but the net result is the same-- you are projecting the "same vector" onto different basis states to get the different ways of coordinatizing that vector. The point is that the vector is not the list of numbers we use to coordinatize it, for the latter depends on the basis but the former does not. Physics usually uses vector spaces that also have a metric, so a concept of a "dot product." Then, you can say that "dot products" between two vectors always gives the same scalar, regardless of the basis used to do the dot product (indeed sometimes there is a graphical way to do the dot product without even using a basis at all). In quantum mechanics, dot products involve overlap integrals, and to my knowledge you generally need to choose a basis to calculate these. Does anyone know of a "graphical" way to do dot products in quantum mechanics without choosing a basis? (Other than the trivial case where both vectors are eigenstates of the same operator, since eigenstates are orthonormal.)

Last edited: Jan 20, 2012
3. Jan 21, 2012

### Waxbear

Thank your for your answer. I didn't know the equation you wrote in the first paragraph (equality between bases), very enlightening. I had the picture of the vector having a direction and magnitude regardless of base, however i was unable to imagine how you could talk about direction without referring to a certain basis. I guess it's just a shortcoming of the graphical way to imagine a vector.

I don't think there is a graphical way of doing dot products in QM, since vectors can be infinite-dimensional. Also, my book (Griffiths) uses the analogy of dot products in 2d euclidean space to introduce dot products in an arbitrary space, but the book also mentions that there is no way to graphically represent dot products in an n-dimensional Hilbert space.

4. Jan 21, 2012

### Ken G

OK, I didn't think so.