# Two-particle wave function.

#### majed_q8i

Ever two-particle wave function is a product of two one-particle wave functions.
Is this true?
if not, can you give me a example ?
Thank you :)

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#### Meir Achuz

Homework Helper
Gold Member
You can have $$\psi(x_1,x_2)$$ that does not equal
$$\phi_1(x_1)\phi_2(x_2)$$,
or you can have a wave function that is a sum of products.

#### majed_q8i

in griffiths, 2nd edition, page 203
says that there exist entangled states that cannot be decomposed to the product of two one-particle wave function.

can you describe entangled states of this type?
I mean write down the equation.

Thanks!

#### Gokul43201

Staff Emeritus
Gold Member
An example, as stated in that same page in Griffiths is the spin singlet state |ud - du>. See pgs 184-185.

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#### RedX

Ever two-particle wave function is a product of two one-particle wave functions.
Is this true?
if not, can you give me a example ?
Thank you :)
If that were true, then any function in 2 variables can be decomposed as a product of two functions in one variable.

But such a function would have the property that it has no zeroes, or infinitely many zeroes, since if h(x0,y0)= f(x0)g(y0)=0, then either f(x0)=0 or g(y0)=0, which implies h(x0,y) and h(x,y0) are zero for all y and x, respectively.

#### alxm

Ever two-particle wave function is a product of two one-particle wave functions. Is this true? if not, can you give me a example ?
It's not true any time those two particles are interacting. If your wave function is a product of two single-particle functions, then every observable quantity is a product of the probabilities for the respective particles. I.e. they're statistically uncorrelated and so, independent of each other.

#### RedX

It's not true any time those two particles are interacting. If your wave function is a product of two single-particle functions, then every observable quantity is a product of the probabilities for the respective particles. I.e. they're statistically uncorrelated and so, independent of each other.
Is the converse true? That is, if they're statistically uncorrelated, can they be written as a direct product?

#### alxm

Is the converse true? That is, if they're statistically uncorrelated, can they be written as a direct product?
Statistical non-correlation means that P(A|B) = P(A)P(B), so yes.

I might want to add to the earlier that any set of interacting (correlated) particles may still be written as a linear expansion of different single-particle functions though. (Handwaving: Throw out the interaction terms from the Hamiltonian and solve for the single particles which can be used as a basis for the interacting Hamiltonian. E.g. a Slater determinant)

#### RedX

Statistical non-correlation means that P(A|B) = P(A)P(B), so yes.

I might want to add to the earlier that any set of interacting (correlated) particles may still be written as a linear expansion of different single-particle functions though. (Handwaving: Throw out the interaction terms from the Hamiltonian and solve for the single particles which can be used as a basis for the interacting Hamiltonian. E.g. a Slater determinant)
oh yes, that is almost by definition. If you have P(x,y), and P(x|y=a)=P(x,a)=f(x)g(a), then it must be true that P(x,y)=f(x)g(y), or that the state can be written:

$$|\psi>=|\sqrt{f}>\otimes|\sqrt{g}>$$

#### RedX

oh, I think you and I made a slight error. It's not:

P(A|B) = P(A)P(B)

but

$$P(A \cap B)=P(A)P(B)$$

If you wanted to use conditional probability, then the equation would be:

P(A|B) = P(A)

Anyways, I don't think this changes the concept, just the notation:

$$P( (x=a) \cap (y=b))=P(a,b)=f(a)g(b)$$, then it must be true that P(x,y)=f(x)g(y), or that the state can be written:

$$|\psi>=|\sqrt{f}>\otimes|\sqrt{g}>$$

Where here the x,y, positions are uncorrelated, and:

$$(<x|\otimes <y|)|\psi>=\sqrt{f(x)} \sqrt{g(y)}$$

I guess technically one should show that if $$\psi(x,y)$$ can't be written as a direct product, i.e., $$\psi(x,y)\neq f(x)g(y)$$, then it's modulus squared can't be written as one either. But that's easy, since if it's modulus squared could be written as one, then just by taking the square root contradicts that $$\psi(x,y)$$ can't be written as a direct product.