Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I'm reading a QM book. It first says for wave function:

"The state of a physical system (or particle) is completely specified by an entity associated with it called a wave function, Ψ , that in general depends on the spatial coordinates of the system and time. The square modulus of this wave function is the probability density for finding the system with a specified set of values for the spatial and temporal coordinates"

But later it says:

"At any given instant in time, the wave function Ψ of a particle (or an isolated system) can be expressed as a linear superposition of a complete orthonormal set of wave functions Ψn"


"$a_n = |c_n|^2$ represents the probability that the system will be found in state Ψn"


I'm confused.

We, already, can get probabilities of system from wave function itself. What is that orthonormal wave functions thing?

share|cite|improve this question
Are you asking why we would want to express the wavefunction like that or did you never take a linear algebra class? – Jordan Jul 29 '14 at 7:06

2 Answers 2

The former and the latter are really the same: "$c_n=\psi(x)$".

If you want to measure positions, then possible outcome states are $|x\rangle$, therefore you write $$ |\psi\rangle = \sum_x|x\rangle\langle x|\psi\rangle:=\sum_x\psi(x)|x\rangle :=\sum_xc_x|x\rangle $$ This tells you, the probability to find the particle at position $x$, i.e. to measure it in the state $|x\rangle$ is $|c_x|^2=|\psi(x)|^2$

If you perform another type of measurement, then the set of possible outcomes is described by other states, call them $|\lambda_n\rangle$. Then you write $$ |\psi\rangle = \sum_n|\lambda_n\rangle\langle\lambda_n|\psi\rangle :=\sum_n\tilde c_n|\lambda_n\rangle $$ Therefore the probability to find your system in state $|\lambda_n\rangle$ is $|\tilde c_n|^2$.

share|cite|improve this answer
Thanks. But first it says: a wave function of a particle gives the probabilities of its position. But later it says: for every possible position there is a wave function. Isn't that what is says? Can you please clarify this? – user50322 Jul 29 '14 at 8:14
I think in your book they are being imprecise about what they call wave function. In the first part they call the coefficients $\psi(x,t)$ wave function, in the second they call $|\psi(t)\rangle$ wave function. The relation between them is, as written above, that the former are coefficients of the latter for a specific choice of basis. I'd stick to calling $|\psi\rangle$ wave function. – Stan Jul 29 '14 at 8:22

Your book is almost correct. Remember $ \psi $ is the wave function that represents the state of the system and is the solution of the Schrodinger's equation.

Now, the Schrodinger's Equation is a linear partial differential equation (PDE) and the solution has several interesting properties: It has infinitely many particular solutions and we consider only those which are physically interesting (this is a fancy way of saying that we take those conditions that satisfy the boundary condition). The solutions are also 'orthogonal' and 'complete'.

This enables us to write the wave function $ \psi $ as a sum (more precisely, as a linear combination) of these orthogonal functions $\psi_{n} $(more precisely, stationary states).

The $ |c_{n}|^2\ 's$ are the coefficients $\psi_{n} $ in of the linear combination mentioned earlier and are interpreted as

probability that a measurement of the energy would yield the value $ E_{n} $.

Most books interpet it as the "probability of finding the particle in the nth stationary state $ \psi_{n}$" but Griffiths insists that it is wrong and gives the above explanation.

I suggest you read Chapter 2 from Griffiths' book. Section 2.1 and 2.2 will answer your question more lucidly.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.